Class 12: Algebra of Matrices – Exercise 5.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Compute the indicated products:}$

$\displaystyle \text{(i) } \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & b- \\ b & a \end{bmatrix}$

$\displaystyle \text{(ii) } \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \end{bmatrix}$

$\displaystyle \text{(iii) } \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & b- \\ b & a \end{bmatrix}$

$\displaystyle = \begin{bmatrix} a \times a + b \times b & a \times (-b)+ b \times a\\ -b \times a + a \times b & -b \times (-b) + a \times a \end{bmatrix}$

$\displaystyle = \begin{bmatrix} a^2 + b^2 & -ab + ab \\ -ab + ab & b^2 + a^2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & b^2 + a^2 \end{bmatrix}$

$\displaystyle \text{(ii) }$

$\displaystyle \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 \times 1 + (-2) \times ( -3) & 1 \times 2 + (-2) \times 2 & 1 \times 3 + (-2) \times (-1) \\ 2 \times 1 + 3 \times (-3) & 2 \times 2 + 3 \times 2 & 2 \times 3 + 3 \times (-1)\end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1+ 6 & 2-4 & 3+2 \\ 2-9 & 4+6 & 6-3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 7 & -2 & 5 \\ -7 & 10 & 3 \end{bmatrix}$

$\displaystyle \text{(iii) }$

$\displaystyle \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 \times 1 + 3 \times 0 +4 \times 3 & 2 \times (-3) + 3 \times 2 + 4 \times 0 & 2 \times 5 + 3 \times 4 + 4 \times 5\\ 3 \times 1 + 4 \times 0 5 \times 3 & 3 \times (-3) + 4 \times 2 + 5 \times 0 & 3 \times 5 + 4 \times 4 + 5 \times 5\\ 4 \times 1 + 5 \times 0 + 6 \times 3 & 4 \times ( -3) + 5 \times 2 + 6 \times 0 & 4 \times 5 + 5 \times 4 + 6 \times 5 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2+ 0 + 12 & -6+6+0 & 10 + 12 + 20 \\ 3+ 0 + 15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 2: Show that } AB \neq BA \text{ in each of the following cases}$

$\displaystyle \text{(i) } A = \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$

$\displaystyle \text{(ii) } A = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$

$\displaystyle \text{(iii) } A = \begin{bmatrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle AB = \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$

$\displaystyle \text{Therefore } AB \neq BA. \text{ Hence proved. }$

$\displaystyle \text{(ii) }$

$\displaystyle AB = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2 +0+4 & 4+3+4 & 6+0+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$

$\displaystyle \text{Therefore } AB \neq BA. \text{ Hence proved. }$

$\displaystyle \text{(iii) }$

$\displaystyle AB = \begin{bmatrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0+3+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 1+0+0 & 0+0+0 \\ 0+1+0 & 4+0+0 & 0+0+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 3 & 1 & 0 \\ 1 & 1 & 0 \\ 1 & 4 & 0 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 5 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 & 0 \\ 1 & 1 & 0 \\ 4 & 1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0+1+0 & 0+1+1 & 0+0+0 \\ 1+0+0 & 3+0+0 & 0+0+0 \\ 0+5+4 & 0+5+1 & 0+0+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 3 & 0 \\ 9 & 6 & 0 \end{bmatrix}$

$\displaystyle \text{Therefore } AB \neq BA. \text{ Hence proved. }$

$\\$

$\displaystyle \text{Question 3: Compute the product } AB \text{ and } BA \text{ whichever exists in each of the following cases:}$

$\displaystyle \text{(i) } A = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$

$\displaystyle \text{(ii) } A = \begin{bmatrix} 3 & 2 \\ -1 & 0 \\ -1 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix}$

$\displaystyle \text{(iii) } A = \begin{bmatrix} 1 & -1 & 2 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix}$

$\displaystyle \text{(iv) } \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} c \\ d \end{bmatrix} + \begin{bmatrix} a & b & c & d \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle AB = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+ 6 & 4+9 & 6+ 3 \end{bmatrix}= \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

$\displaystyle BA \text{ does not exist because the number of columns in } B \text{ is greater than rows number of } A.$

$\displaystyle \text{(ii) }$

$\displaystyle AB = \begin{bmatrix} 3 & 2 \\ -1 & 0 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 12+0 & 15+2 & 18 +4 \\ -4+0 & -5+0 & -6+0 \\ -4+0 & -5+1 & -6+2 \end{bmatrix} = \begin{bmatrix} 12 & 17 & 22\\ -4 & -5 & -6\\ -4 & -4 & -4 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ -1 & 0 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 12-5-6 & 8+0+6 \\ 0-1-2 & 0+0+2 \end{bmatrix} = \begin{bmatrix} 1 & 14 \\ -3 & 2 \end{bmatrix}$

$\displaystyle \text{(iii) }$

$\displaystyle AB = \begin{bmatrix} 1 & -1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix} = \begin{bmatrix} 0-1+6+6 \end{bmatrix} = \begin{bmatrix} 11 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 0 \\ 1 \\ 3 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & -1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & -1 & 2 & 3 \\ 3 & -3 & 6 & 9 \\ 2 & -2 & 4 & 6 \end{bmatrix}$

$\displaystyle \text{(iv) }$

$\displaystyle \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} c \\ d \end{bmatrix} + \begin{bmatrix} a & b & c & d \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix} = \begin{bmatrix} ac+bd \end{bmatrix} + \begin{bmatrix} a^2+b^2+c^2+d^2 \end{bmatrix} = \begin{bmatrix} ac+bd + a^2+b^2+c^2+d^2 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 4: Show that } AB \neq BA \text{ of the following cases:}$

$\displaystyle \text{(i) } A = \begin{bmatrix} 1 & 3 & - 1 \\ 2 & -1 & - 1 \\ 3 & 0 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \end{bmatrix}$

$\displaystyle \text{(ii) } A = \begin{bmatrix} 10 & -4 & 1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} \text{ and } B = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle AB = \begin{bmatrix} 1 & 3 & - 1 \\ 2 & -1 & - 1 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2-3+6 & 3+6-9 & -1-3+4 \\ -4+1+6 & 6-2-9 & - 2+1+4 \\ -6+0+6 & 9+0-9 & -3+0+4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 0 & 0 \\ 3 & -5 & 3 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} -2 & 3 & -1 \\ -1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 3 & - 1 \\ 2 & -1 & - 1 \\ 3 & 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2+6-3 & -6-3+0 & 2-3+1 \\ -1+4-3 & -3-2+0 & 1-2+1 \\ -6+18-12 & -18-9+0 & 6-9+4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & -9 & 0 \\ 0 & -5 & 0 \\ 0 & -27 & 1 \end{bmatrix}$

$\displaystyle \text{Therefore } AB \neq BA.$

$\displaystyle \text{(ii) }$

$\displaystyle AB = \begin{bmatrix} 10 & -4 & 1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 10-12-1 & 20-16-3 & 10-8-2 \\ -11+15+0 & -22+20+0 & -11+10+0 \\ 9-15+1 & 18-20+3 & 9-10+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}$

$\displaystyle BA = \begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix} \begin{bmatrix} 10 & -4 & 1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 10-22+9 & -4+10-5 & -1+0+1 \\ 30-44+10 & -12+20-10 & -3+0+2 \\ 10-33+18 & -4+15-10 & -1+0+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -3 & 1 & 0 \\ 4 & -2 & -1 \\ -5 & 1 & 1 \end{bmatrix}$

$\displaystyle \text{Therefore } AB = BA.$

$\\$

$\displaystyle \text{Question 5: Evaluate the following: }$

$\displaystyle \text{(i) } \Bigg( \begin{bmatrix} 1 & 3 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix} \Bigg) \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix}$

$\displaystyle \text{(ii) } \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

$\displaystyle \text{(iii) } \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{bmatrix} \Bigg( \begin{bmatrix} 1 & 0 & 2 \\ 2 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 2 \end{bmatrix} \Bigg)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \Bigg( \begin{bmatrix} 1 & 3 \\ -1 & 4 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ -1 & 1 \end{bmatrix} \Bigg) \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & 1 \\ -2 & -3 \end{bmatrix} \begin{bmatrix} 4 & 5 & 6 \\ 0 & 1 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+2 & 12+4 & 20+6 \\ -2-6 & -6-12 & -10-18 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 6 & 16 & 26 \\ -8 & -18 & -28 \end{bmatrix}$

$\displaystyle \text{(ii) }$

$\displaystyle \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1+4+0 & 0+0+3 & 2+2+6 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 5 & 3 & 10 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 10 & 12 & 60 \end{bmatrix}$

$\displaystyle \text{(iii) }$

$\displaystyle \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{bmatrix} \Bigg( \begin{bmatrix} 1 & 0 & 2 \\ 2 & 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 1 & 0 & 2 \end{bmatrix} \Bigg)$

$\displaystyle = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & -1 & 0 \\ 1 & 0 & -1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1-1 & -1+0 & 0+1 \\ 0+2 & 0+0 & 0-2 \\ 2+3 & -2+0 & 0-3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & -1 & 1 \\ 2 & 0 & -2 \\ 5 & -2 & -3 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 6: If } A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \text{ and } C=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \text{ then show that } \\ A^2 = B^2 = C^2 = I_2.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1+0 & 0+1 \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle B^2 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1+0 & 0-0 \\ 0-0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle C^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+1 & 0+0 \\ 0+0 & 1+0 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle \text{Hence, } A^2 = B^2 = C^2 = I_2$

$\\$

$\displaystyle \text{Question 7: If } A = \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \text{ and } B =\begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} , \text{ find } 3A^2 - 2B + I. \hspace{1.0cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle 3A^2 - 2B + I$

$\displaystyle = 3\begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 3 & 2 \end{bmatrix} - 2 \begin{bmatrix} 0 & 4 \\ -1 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = 3\begin{bmatrix} 1 & -4 \\ 12 & 1 \end{bmatrix} - \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 3 & -12 \\ 36 & 3 \end{bmatrix} - \begin{bmatrix} 0 & 8 \\ -2 & 14 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & -20 \\ 38 & -10 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 8: If } A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} , \text{ prove that} (A-2I)(A-3I) = O$

$\displaystyle \text{Answer:}$

$\displaystyle (A-2I)(A-3I) = \Bigg( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\Bigg) \Bigg(\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 3\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Bigg)$

$\displaystyle = \Bigg( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\Bigg) \Bigg(\begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \Bigg)$

$\displaystyle = \Bigg( \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \Bigg) \Bigg(\begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \Bigg)$

$\displaystyle = \begin{bmatrix} 2-2 & 2 \\ -1 & -1 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 9: If } A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \text{ show that } A^2 = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \text{ and } A^3 = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = A \cdot A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1+0 & 1+1 \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$

$\displaystyle A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1+0 & 1+2 \\ 0+0 & 0+1 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 10: If } A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}, \text{ show that } A^2 = O.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = A \cdot A = \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix} \begin{bmatrix} ab & b^2 \\ -a^2 & -ab \end{bmatrix}$

$\displaystyle = \begin{bmatrix} a^2b^2 - a^2b^2 & ab^3-ab^3 \\ -a^3b+a^3b & -a^2b^2+a^2b^2 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 11: If } A = \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ - \sin 2 \theta & \cos 2 \theta \end{bmatrix}, \text{ find } A^2. \hspace{3.0cm} \text{[CBSE 2000C]}$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = A \cdot A = \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ - \sin 2 \theta & \cos 2 \theta \end{bmatrix} \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ - \sin 2 \theta & \cos 2 \theta \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos^2 2 \theta - \sin^2 2 \theta & \cos 2 \theta \sin 2 \theta + \cos 2 \theta \sin 2 \theta \\ - \cos 2 \theta \sin 2 \theta - \sin 2 \theta \cos 2 \theta & -\sin^2 2 \theta + \cos^2 2 \theta \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos ( 2 \times 2 \theta) & 2 \sin 2 \theta \cos 2 \theta \\ - 2 \sin 2 \theta \cos 2 \theta & \cos ( 2 \times 2 \theta) \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos 4 \theta & 2 \sin 4 \theta \\ - \sin 4 \theta & \cos 4 \theta \end{bmatrix}$

$\\$

$\displaystyle \text{Question 12: If } A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5 \end{bmatrix}, \text{ show that } \\ AB = BA = O_{3 \times 3}.$

$\displaystyle \text{Answer:}$

$\displaystyle AB = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2-3+5 & 6+9-15 & 10+15-25 \\ 1+4-5 & -3-12+15 & -5-20+25 \\ -1-3+4 & 3+9-12 & 5+15-20 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle = O_{3 \times 3}$

$\displaystyle BA = \begin{bmatrix} -1 & 3 & 5 \\ 1 & -3 & -5 \\ -1 & 3 & 5 \end{bmatrix} \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2-3+5 & 3+12-15 & 5+15-20 \\ 2+3-5 & -3-12+15 & -5-15+20 \\ -2-3+5 & 3+12-15 & 5+15-20 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle = O_{3 \times 3}$

$\displaystyle \text{Therefore } AB = BA = O_{3 \times 3}$

$\\$

$\displaystyle \text{Question 13: If } A = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix} \text{ and } B = \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix}, \text{ show that } \\ AB = BA = O_{3 \times 3}.$

$\displaystyle \text{Answer:}$

$\displaystyle AB = \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix} \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0+abc-abc & 0+b^2c-b^2c & 0+bc^2-bc^2 \\ -a^2c+0+a^2c & -abc+0+abc & -ac^2+0+ac^2 \\ a^2b-a^2b+0 & ab^2-ab^2+0 & abc-abc+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle = O_{3 \times 3}$

$\displaystyle BA = \begin{bmatrix} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{bmatrix} \begin{bmatrix} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0-abc+abc & a^2c+0-a^2c & -a^2b+a^2b+0 \\ 0-b^2c+b^2c & abc+0-abc & -ab^2+ab^2+0 \\ 0-bc^2+bc^2 & ac^2+0-ac^2 & -abc+abc+0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

$\displaystyle = O_{3 \times 3}$

$\displaystyle \text{Therefore } AB = BA = O_{3 \times 3}$

$\\$

$\displaystyle \text{Question 14: If } A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \text{ and } B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}, \text{ show that } \\ AB = A \text{ and } BA = B.$

$\displaystyle \text{Answer:}$

$\displaystyle AB = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+3-5 & -4-9+10 & -8-12+15 \\ -2-4+5 & 2+12-10 & 4+16-15 \\ 2+3-4 & -2-9+8 & -4-12+12 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} = A$

$\displaystyle BA = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+2-4 & -6-8+12 & -10-10+16 \\ -2-3+4 & 3+12-12 & 5+15-16 \\ 2+2-3 & -3-8+9 & -5-10+12 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B$

$\\$

$\displaystyle \text{Question 15: If } A = \begin{bmatrix} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5 \end{bmatrix} \text{ and } B = \begin{bmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{bmatrix}, \text{ compute } \\ A^2-B^2.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 - B^2 = \begin{bmatrix} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5 \end{bmatrix} \begin{bmatrix} -1 & 1 & -1 \\ 3 & -3 & 3 \\ 5 & 5 & 5 \end{bmatrix} - \begin{bmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{bmatrix} \begin{bmatrix} 0 & 4 & 3 \\ 1 & -3 & -3 \\ -1 & 4 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1+3-5 & -1-3-5 & 1+3-5 \\ -3-9+15 & 3+9+15 & -3-9+15 \\ -5+15+25 & 5-15+25 & -5+15+25 \end{bmatrix} - \begin{bmatrix} 0+4-3 & 0-12+12 & 0-12+12 \\ 0-3+3 & 4+9-12 & 3+9-12 \\ 0+4-4 & -4-12+12 & -3-12+16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1 & -9 & -1 \\ 3 & 27 & 3 \\ 35 & 15 & 35 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2 & -9 & -1 \\ 3 & 26 & 3 \\ 35 & 15 & 34 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 16: For the following matrices verify the associativity of matrix} \\ \text{multiplication i.e. } (AB)C = A(BC).$

$\displaystyle \text{(i) } A = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} , B = \begin{bmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

$\displaystyle \text{(ii) } A = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} , B = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle (AB)C = \Bigg(\begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

$\displaystyle = \Bigg(\begin{bmatrix} 1-2+0 & 0+4+0 \\ -1-0+0 & 0+0+3 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 & 4 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} -1 -4 \\ -1 -3 \end{bmatrix} = \begin{bmatrix} -5 \\ -4 \end{bmatrix}$

$\displaystyle A(BC) = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \Bigg( \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 2 \\ 2 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \Bigg)$

$\displaystyle = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 -0 \\-1 - 2 \\ 0 - 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 0 \\ -1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ - 3 \\ - 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1-6-0 & -1-0-3 \end{bmatrix} = \begin{bmatrix} -5 \\ -4 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\displaystyle \text{(ii) }$

$\displaystyle (AB)C = \Bigg(\begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{bmatrix} \Bigg) \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 10 & -5 & 11 \\ 5 & -2 & 5 \\ 5 & -4 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{bmatrix}$

$\displaystyle A(BC) = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \Bigg(\begin{bmatrix} 1 & 0 \\ -1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 1 \\ 0 & 0 & 1 \end{bmatrix} \Bigg)$

$\displaystyle = \begin{bmatrix} 4 & 2 & 3 \\ 1 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} -2 & 2 & -1 \\ 3 & 0 & 3 \\ -1 & 4 & -2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -5 & 20 & -4 \\ -1 & 10 & -2 \\ -7 & 10 & -5 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 17: For the following matrices verify the distributivity of matrix multiplication} \\ \text{over matrix addition i.e. } A(B + C) = AB + AC.$

$\displaystyle \text{(i) } A = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} , B = \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix} \text{ and } C = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$

$\displaystyle \text{(ii) } A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} , B = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \text{ and } C = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle A(B+C) = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \Bigg( \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \Bigg) = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 6 & 0 \end{bmatrix}$

$\displaystyle AB + AC = \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -3 & -1 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} -1 & 2 \\ 2 & -2 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 6 & 0 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\displaystyle \text{(ii) }$

$\displaystyle A(B+C) = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \Bigg( \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix} \Bigg) = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{bmatrix}$

$\displaystyle AB + AC = \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ 1 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -1 & 1 \\ 1 & 2 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 2 & 2 \\ 1 & 4 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 18: If } A = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} , B = \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \\ \text{verify that } A(B-C) = AB - AC.$

$\displaystyle \text{Answer:}$

$\displaystyle A(B-C) = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \Bigg( \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix} \Bigg)$

$\displaystyle = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & -6 \\ -1 & 0 & 3 \\ -1 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{bmatrix}$

$\displaystyle AB - AC = \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 5 & -4 \\ -2 & 1 & 3 \\ -1 & 0 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 0 & -2 \\ 3 & -1 & 0 \\ -2 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 5 & 2 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & 5 & -8 \\ 2 & 14 & -15 \\ -3 & -9 & 13 \end{bmatrix} - \begin{bmatrix} 1 & 7 & 0 \\ 4 & 14 & 6 \\ -3 & -10 & -3 \end{bmatrix} = \begin{bmatrix} 1 & -2 & -8 \\ -2 & 0 & -21 \\ 0 & 1 & 16 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 19: Compute the elements } a_{43} \text{ and } a_{22} \text{ of the matrix}$
$\displaystyle A = \begin{bmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 & 2 & - 2 \\ 3 & -3 & 4 & -4 & 0 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle A = \begin{bmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 & 2 & - 2 \\ 3 & -3 & 4 & -4 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 1 & 0 \\ 2 & 0 & 2 \\ 0 & 3 & 2 \\ 4 & 0 & 4 \end{bmatrix} \begin{bmatrix} -3 & 5 & -6 & 8 & -4 \\ 6 & -9 & 11 & -14 & 6 \\ 9 & -5 & 8 & -4 & -8 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 6 & -9 & 11 & -14 & 6 \\ 12 & 0 & 4 & 8 & -24 \\ 36 & -37 & 49 & -50 & 2 \\ 24 & 0 & 8 & 16 & -48 \end{bmatrix}$

$\displaystyle \therefore a_{43} = 8 , a_{22}= 0$

$\\$

$\displaystyle \text{Question 20: If } A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \text{ and } I \text{ is the identity matrix of order } 3, \text{ show that } \\ A^3 = pI+qA+rA^2.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ rp & p+rq & q+r^2 \end{bmatrix}$

$\displaystyle A^3 = A^2 \cdot A = \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ rp & p+rq & q+r^2 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix}$

$\displaystyle = \begin{bmatrix} p & q & r \\ rp & p+rq & q+r^2 \\ pq+r^2p& rp+q^2+r^2q & p+2rq+r^3 \end{bmatrix}$

$\displaystyle pI+qA+rA^2 = p \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + q \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ p & q & r \end{bmatrix} + r \begin{bmatrix} 0 & 0 & 1 \\ p & q & r \\ rp & p+rq & q+r^2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} p & 0 & 0 \\ 0 & p & 0 \\ 0 & 0 & p \end{bmatrix} + \begin{bmatrix} 0 & q & 0 \\ 0 & 0 & q \\ pq & q^2 & qr \end{bmatrix} + \begin{bmatrix} 0 & 0 & r \\ rp & rq & r^2 \\ r^2p & rp+r^2q & qr+r^3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} p & q & r \\ rp & p+rq & q+r^2 \\ pq+r^2p & q^2+rp+r^2q & p+2qr+r^3 \end{bmatrix}$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 21: If } \omega \text{ is a complex cube root of unity, show that}$

$\displaystyle \Bigg( \begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{bmatrix} + \begin{bmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS} = \Bigg( \begin{bmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{bmatrix} + \begin{bmatrix} \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \\ \omega & \omega^2 & 1 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\displaystyle = \Bigg( \begin{bmatrix} 1+ \omega & \omega + \omega^2 & \omega^2+1 \\ \omega + \omega^2 & \omega^2+1 & 1+\omega \\ \omega^2+\omega & 1+\omega^2 & \omega+1 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\displaystyle = \Bigg( \begin{bmatrix} -\omega^2 & -1 & -\omega \\ -1 & -\omega & -\omega^2 \\ -1 & -\omega & -\omega^2 \end{bmatrix} \Bigg) \begin{bmatrix} 1 \\ \omega \\ \omega^2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -\omega^2-\omega-\omega^3 \\ -1-\omega^2-\omega^4 \\ -1-\omega^2-\omega^4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -\omega (\omega+1 +\omega^2) \\ -1-\omega^2-\omega^3 \omega \\ -1-\omega^2-\omega^3 \omega \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -\omega (0) \\ -1-\omega^2- \omega \\ -1-\omega^2- \omega \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

$\displaystyle = \text{RHS}$

$\\$

$\displaystyle \text{Question 22: If } A = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}, \text{ show that } A^2 = I_3$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix} \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+3-5 & -6-12+15 & -10-15+20 \\ -2-4+5 & 3+16-15 & 5+20-20 \\ 2+3-4 & -3-12+12 & -5-15+16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & -3 & -5 \\ -1 & 4 & 5 \\ 1 & -3 & -4 \end{bmatrix}$

$\displaystyle = A$

$\displaystyle \text{Hence Proved.}$

$\\$

$\displaystyle \text{Question 23: If } A = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}, \text{ show that } A^2 = I_3$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 = \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix} \begin{bmatrix} 4 & -1 & -4 \\ 3 & 0 & -4 \\ 3 & -1 & -3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 16-3-12 & -4+0+4 & -16+4+12 \\ 12+0-12 & -3+0+4 & -12+0+12 \\ 12-3-9 & -3+0+3 & -12+4+9 \end{bmatrix}$

$\displaystyle =\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle = I_3$

$\\$

$\displaystyle \text{Question 24: }$

$\displaystyle \text{(i) If } \begin{bmatrix} 1 & 1 & x \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0, \text{ find } x.$

$\displaystyle \text{(ii) If } \begin{bmatrix} 2 & 3 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}= \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} = 0, \text{ find } x. \hspace{3.0cm} \text{[CBSE 2012]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \begin{bmatrix} 1 & 1 & x \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 1+2x & 2+x & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 1+2x + 2+x + 3 \end{bmatrix} = 0$

$\displaystyle \Rightarrow 3x+6 = 0$

$\displaystyle \Rightarrow x= - 2$

$\displaystyle \text{(ii) }$

$\displaystyle \begin{bmatrix} 2 & 3 \\ 7 & 5 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}= \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 2-6 & -6+12 \\ 5-14 & -15+28 \end{bmatrix}= \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} -4 & 6 \\ -9 & 13 \end{bmatrix}= \begin{bmatrix} -4 & 6 \\ -9 & x \end{bmatrix} = 0$

$\displaystyle \Rightarrow x=13$

$\\$

$\displaystyle \text{Question 25: If }\begin{bmatrix} x & 4 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0, \text{ find } x.$

$\displaystyle \text{Answer:}$

$\displaystyle \begin{bmatrix} x & 4 & 1 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x+4+0 & x+0+2 & 2x+8-4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x+4 & x+2 & 2x+4 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ -1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x^2 + 4x + 4x+8 - 2x - 4 \end{bmatrix} = 0$

$\displaystyle \Rightarrow 2x^2 + 6x +4 = 0$

$\displaystyle \Rightarrow x^2 + 3x + 2 = 0$

$\displaystyle \Rightarrow ( x+1)(x+2)=0$

$\displaystyle \Rightarrow x = -1 \text{ or } x = - 2$

$\\$

$\displaystyle \text{Question 26: } \begin{bmatrix} 1 & -1 & x \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = 0, \text{ find } x.$

$\displaystyle \text{Answer:}$

$\displaystyle \begin{bmatrix} 1 & -1 & x \end{bmatrix} \begin{bmatrix} 0 & 1 & -1 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} -2+x & x & -4+x \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} x-4+x \end{bmatrix} = 0$

$\displaystyle \Rightarrow 2x = 4$

$\displaystyle \Rightarrow x = 2$

$\\$

$\displaystyle \text{Question 27: If } A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \text{ then prove that } \\ A^2 - A + 2I = O$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 - A + 2I = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2+2 & 0+0 \\ 0+0 & -2+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 28: If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \text{ then find } \lambda \text{ so that } \\ A^2 =5 A + \lambda I.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 =5 A + \lambda I$

$\displaystyle \Rightarrow \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} = \begin{bmatrix} 15+ \lambda & 5 \\ -5 & 10+ \lambda \end{bmatrix}$

$\displaystyle \Rightarrow 8 = 15+ \lambda$

$\displaystyle \Rightarrow \lambda = - 7$

$\displaystyle \Rightarrow 3 = 10+ \lambda$

$\displaystyle \Rightarrow \lambda = - 7$

$\\$

$\displaystyle \text{Question 29: If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \text{ show that } A^2 -5 A + 7 I_2=O. \hspace{2.0cm} \text{[CBSE 2003, 2007]}$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 -5 A + 7 I_2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} - 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 8-15+7 & 5-5+ 0 \\ -5+5+0 & 3-10+7 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 30: If } A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, \text{ show that } A^2 -2 A + 3 I_2=O.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 -2 A + 3 I_2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} - 2 \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} + 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix} - \begin{bmatrix} 4 & 6 \\ 2 & 4 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1-4+3 & 6-6+0 \\ -2+2+0 & -3+0+3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 31: Show that the matrix } A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \text{ satisfies the equation } \\ A^3-4A^2+A=O \hspace{2.0cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle A^3-4A^2+A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} - 4 \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} - 4 \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}+ \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$

$\displaystyle = + \begin{bmatrix} 26 & 45 \\ 15 & 26 \end{bmatrix} - \begin{bmatrix} 28 & 48 \\ 16 & 28 \end{bmatrix}+ \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 26-28+2 & 45-48+3 \\ 15-16+1 & 26-28+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 32: Show that the matrix } A = \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} \text{ is a root of the equation } \\ A^2-12A-I=O.$

$\displaystyle \text{Answer:}$

$\displaystyle A^2-12A-I = \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} - 12 \begin{bmatrix} 5 & 3 \\ 12 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 61 & 36 \\ 144 & 85 \end{bmatrix} - \begin{bmatrix} 60 & 36 \\ 144 & 84 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 61-60-1 &36-36+ 0 \\ 144-144+0 & 85-84-1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 33: If } A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}, \text{ find } A^2 - 5A - 14I. \hspace{3.0cm} \text{[CBSE 2004]}$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 - 5A - 14I = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} - 5 \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} - 14 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} - \begin{bmatrix} 15 & -25 \\ -20 & 10 \end{bmatrix} - \begin{bmatrix} 14 & 0 \\ 0 & 14 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 29-15-14 & -25+25+0 \\ -20+20+0 & 24-10-14 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\\$

$\displaystyle \text{Question 34: If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, \text{ show that } A^2 - 5A + 7I =O. \text{ Use this to find } A^4.$

$\displaystyle \text{Answer:}$

$\displaystyle A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$

$\displaystyle A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

$\displaystyle A^2 - 5A + 7I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - 5\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

$\displaystyle \text{Thus, } A \text{ satisfies } A^2 - 5A + 7I = O$

$\displaystyle \text{Finding } A^4:$

$\displaystyle A^4 = (A^2)^2 = (5A - 7I)^2 = 25A^2 - 70A + 49I$

$\displaystyle A^2 = 5A - 7I,$

$\displaystyle A^4 = 25(5A - 7I) - 70A + 49I = (125 - 70)A + (-175 + 49)I$

$\displaystyle A^4 = 55A - 126I$

$\displaystyle A^4 = 55\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} - 126\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 39 & 55 \\ -55 & -16 \end{bmatrix}$

$\displaystyle A^2 - 5A + 7I = O \quad \text{and} \quad A^4 = \begin{bmatrix} 39 & 55 \\ -55 & -16 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 35: If } A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \text{ find } k \text{ such that } A^2 = kA -2I_2. \hspace{3.0cm} \text{[CBSE 2003]}$

$\displaystyle \text{Answer:}$

$\displaystyle A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}, \quad \text{we need to find } k \text{ such that } A^2 = kA - 2I_2.$

$\displaystyle A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} = \begin{bmatrix} 9 - 8 & -6 + 4 \\ 12 - 8 & -8 + 4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}$

$\displaystyle kA - 2I_2 = k \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$

$\displaystyle \text{Equating } A^2 \text{ and } kA - 2I_2:$

$\displaystyle \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k - 2 & -2k \\ 4k & -2k - 2 \end{bmatrix}$

$\displaystyle \text{Comparing entries: }$

$\displaystyle 1 = 3k - 2 \;\; \Rightarrow \;\; k = 1,$

$\displaystyle -2 = -2k \;\; \Rightarrow \;\; k = 1,$

$\displaystyle 4 = 4k \;\; \Rightarrow \;\; k = 1,$

$\displaystyle -4 = -2k - 2 \;\; \Rightarrow \;\; k = 1.$

$\displaystyle \boxed{k = 1}$

$\\$

$\displaystyle \text{Question 36: If } A = \begin{bmatrix} 1 & 0 \\ -1 & 7 \end{bmatrix}, \text{ find } k \text{ such that } A^2 -8A+kI=O. \hspace{3.0cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle A=\begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix},\qquad \text{find }k\text{ such that }A^2-8A+kI=O.$

$\displaystyle A^2= \begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix} \begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix} = \begin{bmatrix}1 & 0 \\ -8 & 49\end{bmatrix}$

$\displaystyle A^2-8A+kI = \begin{bmatrix}1 & 0 \\ -8 & 49\end{bmatrix} -8\begin{bmatrix}1 & 0 \\ -1 & 7\end{bmatrix} +k\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$

$\displaystyle = \begin{bmatrix}1 & 0 \\ -8 & 49\end{bmatrix} +\begin{bmatrix}-8 & 0 \\ 8 & -56\end{bmatrix} +\begin{bmatrix}k & 0 \\ 0 & k\end{bmatrix} = \begin{bmatrix}k-7 & 0 \\ 0 & k-7\end{bmatrix}.$

$\displaystyle \text{For this to be the zero matrix we need } k-7=0. \text{ Hence } \boxed{k=7.}$

$\\$

$\displaystyle \text{Question 37: If } A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \text{and} f(x) = x^2 - 2x - 3 \text{ show that } f(A) = O. \hspace{0.5cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle A=\begin{bmatrix}1 & 2\\[4pt] 2 & 1\end{bmatrix},\qquad f(x)=x^2-2x-3.$

$\displaystyle \text{Compute } A^2:$

$\displaystyle A^2=\begin{bmatrix}1 & 2\\[4pt] 2 & 1\end{bmatrix} \begin{bmatrix}1 & 2\\[4pt] 2 & 1\end{bmatrix} =\begin{bmatrix}1+4 & 2+2\\[4pt] 2+2 & 4+1\end{bmatrix} =\begin{bmatrix}5 & 4\\[4pt] 4 & 5\end{bmatrix}.$

$\displaystyle \text{Now evaluate } f(A)=A^2-2A-3I:$

$\displaystyle A^2-2A-3I =\begin{bmatrix}5 & 4\\[4pt] 4 & 5\end{bmatrix} -2\begin{bmatrix}1 & 2\\[4pt] 2 & 1\end{bmatrix} -3\begin{bmatrix}1 & 0\\[4pt] 0 & 1\end{bmatrix}$

$\displaystyle =\begin{bmatrix}5 & 4\\[4pt] 4 & 5\end{bmatrix} -\begin{bmatrix}2 & 4\\[4pt] 4 & 2\end{bmatrix} -\begin{bmatrix} 3 & 0 \\[4pt] 0 & 3\end{bmatrix} =\begin{bmatrix} 0 & 0 \\[4pt] 0 & 0 \end{bmatrix}.$

$\displaystyle \boxed{\,f(A)=O\,}$

$\\$

$\displaystyle \text{Question 38: If } A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \text{ and } I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \text{ then find } \lambda, \mu \text{ so that} \\A^2= \lambda A + \mu I$

$\displaystyle \text{Answer:}$

$\displaystyle A=\begin{bmatrix}2 & 3\\[4pt] 1 & 2\end{bmatrix},\qquad \text{find }\lambda,\mu \text{ such that }A^2=\lambda A+\mu I.$

$\displaystyle A^2= \begin{bmatrix}2 & 3\\[4pt] 1 & 2\end{bmatrix} \begin{bmatrix}2 & 3\\[4pt] 1 & 2\end{bmatrix} = \begin{bmatrix}2\cdot2+3\cdot1 & 2\cdot3+3\cdot2\\[4pt] 1\cdot2+2\cdot1 & 1\cdot3+2\cdot2\end{bmatrix} = \begin{bmatrix}7 & 12\\[4pt] 4 & 7\end{bmatrix}$

$\displaystyle \lambda A+\mu I =\lambda\begin{bmatrix}2 & 3\\[4pt] 1 & 2\end{bmatrix} +\mu\begin{bmatrix}1 & 0\\[4pt] 0 & 1\end{bmatrix} =\begin{bmatrix}2\lambda+\mu & 3\lambda\\[4pt] \lambda & 2\lambda+\mu\end{bmatrix}$

$\displaystyle \text{Equating entries of } A^2 \text{ and } \lambda A+\mu I\:$

$\displaystyle 12=3\lambda\quad\Rightarrow\quad \lambda=4,$

$\displaystyle 4=\lambda\quad\Rightarrow\quad \lambda=4\ (\text{consistent}),$

$\displaystyle 7=2\lambda+\mu \quad\Rightarrow\quad 7=8+\mu \quad\Rightarrow\quad \mu=-1.$

$\displaystyle \lambda=4,\qquad \mu=-1$

$\\$

$\displaystyle \text{Question 39: Find the value of } x\text{ for which the matrix product }$

$\displaystyle \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} \text{ equal to an identity matrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle \begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix} \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} -2x+ 0 + 7x & 28x+0-28x & 14x+0-14x \\ 0+0+0 & 0+1-0 & 0+0-0 \\ -x-0+x & 14x-2-4x & 7x-0-2x \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 5x & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10x-2 & 5x \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle \therefore 10x-2 = 0 \Rightarrow x = \frac{1}{5}$

$\\$

$\displaystyle \text{Question 40: Solve the matrix equations:}$

$\displaystyle \text{(i) } \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ 5 \end{bmatrix} =0$

$\displaystyle \text{(ii) } \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0$

$\displaystyle \text{(iii) } \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =0$

$\displaystyle \text{(iv) } \begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ 8 \end{bmatrix} =0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} x \\ 5 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} x-2 & -3 \end{bmatrix} \begin{bmatrix} x \\ 5 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} x^2-2x-15 \end{bmatrix}=0$

$\displaystyle \Rightarrow x^2-2x-15 = 0$

$\displaystyle \Rightarrow (x-5)(x+3)=0$

$\displaystyle \Rightarrow x = 5 \text{ or } x = - 3$

$\displaystyle \text{(ii) }$

$\displaystyle \begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} 6 & 2 & 4 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} 4+4x \end{bmatrix} =0$

$\displaystyle \Rightarrow 4x = - 4$

$\displaystyle \Rightarrow x = - 1$

$\displaystyle \text{(iii) }$

$\displaystyle \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} x-2 & -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} x^2-2x-40+2x-8 \end{bmatrix} =0$

$\displaystyle \Rightarrow x^2 - 48=0$

$\displaystyle \Rightarrow x = \pm 4\sqrt{3}$

$\displaystyle \text{(iv) }$

$\displaystyle \begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ 8 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x-9 & 4x \end{bmatrix} \begin{bmatrix} x \\ 8 \end{bmatrix} =0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x^2-9x+32x \end{bmatrix} = 0$

$\displaystyle \Rightarrow 2x^2+23x= 0$

$\displaystyle \Rightarrow x ( 2x+23) = 0$

$\displaystyle \Rightarrow x = 0 \text{ or } x = \frac{-23}{2}$

$\\$

$\displaystyle \text{Question 41: If } A = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3 \end{bmatrix}, \text{ compute } A^2 - 4A+3I_3$

$\displaystyle \text{Answer:}$

$\displaystyle A^2 - 4A+3I_3$

$\displaystyle = \begin{bmatrix} 1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3 \end{bmatrix} - 4 \begin{bmatrix} 1 & 2 & 0 \\ 3 & -4 & 5 \\ 0 & -1 & 3 \end{bmatrix} + 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1+6+0 & 2-8-0 & 0+10+0 \\ 3-12+0 & 6+16-5 & 0-20+15 \\ 0-3+0 & 0+4-3 & 0-5+9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12 \end{bmatrix} + \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 7 & -6 & 10 \\ -9 & 17 & -5 \\ -3 & 1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 0 \\ 12 & -16 & 20 \\ 0 & -4 & 12 \end{bmatrix} + \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 7-4+3 & -6-8+0 & 10-0+0 \\ -9-12+0 & 17+16+3 & -5-20+0 \\ -3-0+0 & 1+4+0 & 4-12+3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 6 & -14 & 10 \\ -21 & 36 & -25 \\ -3 & 5 & -5 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 42: If } f(x) = x^2-2x, \text{ find } f(A), \text{ where } A = \begin{bmatrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{bmatrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle f(A) = A^2 - 2A$

$\displaystyle = \begin{bmatrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{bmatrix} - 2 \begin{bmatrix} 0 & 1 & 2 \\ 4 & 5 & 0 \\ 0 & 2 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0+4+0 & 0+5+4 & 0+0+6 \\ 0+20+0 & 4+25+0 & 8+0+0 \\ 0+8+0 & 0+10+6 & 0+0+9 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & 9 & 6 \\ 20 & 29 & 8 \\ 8 & 16 & 9 \end{bmatrix} - \begin{bmatrix} 0 & 2 & 4 \\ 8 & 10 & 0 \\ 0 & 4 & 6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4-0 & 9-2 & 6-4 \\ 20-8 & 29-10 & 8-0 \\ 8-0 & 16-4 & 9-6 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & 7 & 2 \\ 12 & 19 & 8 \\ 8 & 12 & 3 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 43: If } f(x) = x^3+4x^2-x, \text{ find } f(A), \text{ where } A = \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle f(A) = A^3+4A^2-A$

$\displaystyle = \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} + 4 \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9-0 & 4-0+0 \\ 0-2+0 & 1+3-0 & 2-0+0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} + 4 \begin{bmatrix} 0+2+2 & 0-3-2 & 0+0+0 \\ 0-6+0 & 2+9-0 & 4-0+0 \\ 0-2+0 & 1+3-0 & 2-0+0 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix} + 4 \begin{bmatrix} 4 & -5 & 0 \\ -6 & 11 & 4 \\ -2 & 4 & 2 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0-10+0 & 4+15-0 & 8-0+0 \\ 0+22+4 & -6-33-4 & -12+0+0 \\ 0+8+2 & -2-12-2 & -4+0+0 \end{bmatrix} + \begin{bmatrix} 16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -10 & 19 & 8 \\ 26 & -43 & -12 \\ 10 & -16 & -4 \end{bmatrix} + \begin{bmatrix} 16 & -20 & 0 \\ -24 & 44 & 16 \\ -8 & 16 & 8 \end{bmatrix} - \begin{bmatrix} 0 & 1 & 2 \\ 2 & -3 & 0 \\ 1 & -1 & 0 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -10+16-0 & 19-20-1 & 8+0-2 \\ 26-24-2 & -43+44+3 & -12+16-0 \\ 10-8-1 & -16+16+1 & -4+8+0 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 6 & -2 & 6 \\ 0 & 4 & 4 \\ 1 & 1 & 4 \end{bmatrix}$
$\\$
$\displaystyle \text{Question 44: } \text{If } A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}, \text{ then show that } A \text{ is a root of the polynomial } f(x) = x^3 - 6x^2 + 7x + 2.$ $\displaystyle \text{Answer:}$
$\displaystyle f(x) = x^3 - 6x^2 + 7x + 2$
$\displaystyle A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$
$\displaystyle A^2 =\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$
$\displaystyle A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{bmatrix}$
$\displaystyle A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$
$\displaystyle f(A) = A^3 - 6A^2 + 7A + 2I_3$
$\displaystyle A^3 - 6A^2 + 7A + 2I_3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} -6\begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} +7\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} +2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix} + \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix} + \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 21-30+7+2 & 0-0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0-0+0+0 & 55-78+21+2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0$
$\displaystyle \text{Since } f(A) = 0,\ \text{A is a root of } f(x) = x^3 - 6x^2 + 7x + 2.$

$\\$

$\displaystyle \text{Question 45: } A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \text{ then prove that } A^2 - 4A - 5I = 0 \hspace{2.0cm} \text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \\ 2+2+4 & 4+1+4 & 4+2+2 \\ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$
$\displaystyle A^2 - 4A - 5I = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} -4\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} -5\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 9-4-5 & 8-8-0 & 8-8-0 \\ 8-8-0 & 9-4-5 & 8-8-0 \\ 8-8-0 & 8-8-0 & 9-4-5 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0$
$\displaystyle \text{Hence proved.}$
$\\$
$\displaystyle \text{Question 46: } A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \text{ show that } A^2 - 7A + 10I_3 = 0$

$\displaystyle \text{Answer:}$

$\displaystyle A = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 9+2+0 & 6+8+0 & 0+0+0 \\ 3+4+0 & 2+16+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+25 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix}$
$\displaystyle A^2 - 7A + 10I_3 = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} -7\begin{bmatrix} 3 & 2 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} +10\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 11 & 14 & 0 \\ 7 & 18 & 0 \\ 0 & 0 & 25 \end{bmatrix} - \begin{bmatrix} 21 & 14 & 0 \\ 7 & 28 & 0 \\ 0 & 0 & 35 \end{bmatrix} + \begin{bmatrix} 10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 11-21+10 & 14-14+0 & 0-0+0 \\ 7-7+0 & 18-28+10 & 0-0+0 \\ 0-0+0 & 0-0+0 & 25-35+10 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = 0$

$\\$

$\displaystyle \text{Question 47: } \text{Without using the concept of inverse of a matrix, find the matrix } \\ \begin{bmatrix} x & y \\ z & u \end{bmatrix} \text{ such that } \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} x & y \\ z & u \end{bmatrix} = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } \begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix} \begin{bmatrix} x & y \\ z & u \end{bmatrix} = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 5x-7z & 5y-7u \\ -2x+3z & -2y+3u \end{bmatrix} = \begin{bmatrix} -16 & -6 \\ 7 & 2 \end{bmatrix}$

$\displaystyle \text{The corresponding elements of two equal matrices are equal}$

$\displaystyle 5x - 7z = -16 \quad ...(1)$

$\displaystyle 5y - 7u = -6 \quad ...(2)$

$\displaystyle -2y + 3u = 2$

$\displaystyle \Rightarrow 3u = 2 + 2y$

$\displaystyle \Rightarrow u = \frac{2+2y}{3} \quad ...(3)$

$\displaystyle -2x + 3z = 7$

$\displaystyle \Rightarrow 3z = 7 + 2x$

$\displaystyle \Rightarrow z = \frac{7+2x}{3} \quad ...(4)$

$\displaystyle \text{Putting the value of } z \text{ in eq.(1), we get}$

$\displaystyle 5x - 7\left(\frac{7+2x}{3}\right) = -16$

$\displaystyle \Rightarrow 5x - \frac{49+14x}{3} = -16$

$\displaystyle \Rightarrow \frac{15x - 49 - 14x}{3} = -16$

$\displaystyle \Rightarrow x - 49 = -48$

$\displaystyle \Rightarrow x = -48 + 49$

$\displaystyle \Rightarrow x = 1$
$\displaystyle \text{Putting the value of } x \text{ in eq.(4), we get}$

$\displaystyle z = \frac{7 + 2(1)}{3}$

$\displaystyle \Rightarrow z = \frac{9}{3} = 3$

$\displaystyle \text{Putting the value of } u \text{ in eq.(2), we get}$

$\displaystyle 5y - 7\left(\frac{2 + 2y}{3}\right) = -6$

$\displaystyle \Rightarrow 5y - \frac{14 + 14y}{3} = -6$

$\displaystyle \Rightarrow \frac{15y - 14 - 14y}{3} = -6$

$\displaystyle \Rightarrow y - 14 = -18$

$\displaystyle \Rightarrow y = -18 + 14$

$\displaystyle \therefore y = -4$

$\displaystyle \text{Putting the value of } y \text{ in eq.(3), we get}$

$\displaystyle u = \frac{2 + 2(-4)}{3}$

$\displaystyle \Rightarrow u = -2$

$\displaystyle \therefore \begin{bmatrix} x & y \\ z & u \end{bmatrix} = \begin{bmatrix} 1 & -4 \\ 3 & -2 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 48: Find the matrix } A \text{ such that}$

$\displaystyle \text{(i) } \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} A = \begin{bmatrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix}$

$\displaystyle \text{(ii) } A \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

$\displaystyle \text{(iii) } \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} A = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$

$\displaystyle \text{(iv) } \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A$

$\displaystyle \text{(v) } \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & -4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix} \hspace{2.0cm} \text{[CBSE 2017]}$

$\displaystyle \text{(vi) } A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{bmatrix} \hspace{2.0cm} \text{[CBSE 2017]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Let } A = \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix}$

$\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix} = \begin{bmatrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} x+a & y+b & z+c \\ 0+a & 0+b & 0+c \end{bmatrix} = \begin{bmatrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} x+a & y+b & z+c \\ a & b & c \end{bmatrix} = \begin{bmatrix} 3 & 3 & 5 \\ 1 & 0 & 1 \end{bmatrix}$

$\displaystyle \text{The corresponding elements of two equal matrices are equal}$

$\displaystyle x+a = 3 \quad ...(1)$

$\displaystyle y+b = 3 \quad ...(2)$

$\displaystyle z+c = 5 \quad ...(3)$

$\displaystyle \Rightarrow a = 1,\; b = 0,\; c = 1$

$\displaystyle \text{Putting the value of } a \text{ in eq.(1), we get}$

$\displaystyle x + 1 = 3$

$\displaystyle \Rightarrow x = 3 - 1$

$\displaystyle \therefore x = 2$
$\displaystyle \text{Putting the value of } b \text{ in eq.(2), we get}$

$\displaystyle y + b = 3$

$\displaystyle y + 0 = 3$

$\displaystyle \therefore\ y = 3$

$\displaystyle \text{Putting the value of } c \text{ in eq.(3), we get}$

$\displaystyle z + 1 = 5$

$\displaystyle z = 5 - 1$

$\displaystyle \therefore\ z = 4$

$\displaystyle \therefore\ A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 0 & 1 \end{bmatrix}$

$\displaystyle \text{(ii) }$

$\displaystyle \text{(Let } A=\begin{bmatrix} w & x \\ y & z \end{bmatrix}$

$\displaystyle A\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

$\displaystyle \begin{bmatrix} w & x \\ y & z \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

$\displaystyle \begin{bmatrix} w+4x & 2w+5x & 3w+6x \\ y+4z & 2y+5z & 3y+6z \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$

$\displaystyle \text{The corresponding elements of two equal matrices are equal}$

$\displaystyle 3w+6x=-9 \quad \text{...(1)}$

$\displaystyle y+4z=2$

$\displaystyle \Rightarrow y=2-4z \quad \text{...(2)}$

$\displaystyle w+4x=-7$

$\displaystyle \Rightarrow w=-7-4x \quad \text{...(3)}$

$\displaystyle 2y+5z=4 \quad \text{...(4)}$

$\displaystyle \text{Putting the value of } w \text{ in eq.(1), we get}$

$\displaystyle 3(-7-4x)+6x=-9$

$\displaystyle -21-12x+6x=-9$

$\displaystyle -6x=12$

$\displaystyle \Rightarrow x=-2$
$\displaystyle \text{Putting the value of } x \text{ in eq.(3), we get}$

$\displaystyle w = -7 - 4(-2)$

$\displaystyle \Rightarrow w = -7 + 8$

$\displaystyle \Rightarrow w = 1$

$\displaystyle \text{Putting the value of } y \text{ in eq.(4), we get}$

$\displaystyle 2(2 - 4z) + 5z = 4$

$\displaystyle \Rightarrow 4 - 8z + 5z = 4$

$\displaystyle \Rightarrow 4 - 3z = 4$

$\displaystyle \Rightarrow -3z = 0$

$\displaystyle \Rightarrow z = 0$

$\displaystyle \text{Putting the value of } z \text{ in eq.(2), we get}$

$\displaystyle y = 2 - 4(0)$

$\displaystyle \Rightarrow y = 2$

$\displaystyle \therefore A = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$

$\displaystyle \text{(iii) }$

$\displaystyle \text{Let } A = [x\ y\ z]$

$\displaystyle \begin{bmatrix} 4 \\ 1 \\ 3 \end{bmatrix} [x\ y\ z] = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 4x & 4y & 4z \\ x & y & z \\ 3x & 3y & 3z \end{bmatrix} = \begin{bmatrix} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{bmatrix}$

$\displaystyle \text{The corresponding elements of two equal matrices are equal.}$

$\displaystyle 4x = -4 \quad ...(1)$

$\displaystyle 4y = 8 \quad ...(2)$

$\displaystyle 4z = 4 \quad ...(3)$

$\displaystyle x = -1$

$\displaystyle y = 2$

$\displaystyle z = 1$

$\displaystyle \therefore\ A = [-1\ 2\ 1]$

$\displaystyle \text{(iv) }$

$\displaystyle \ Let\ A = [x]$

$\displaystyle \Rightarrow [\,2\ \ 1\ \ 3\,] \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = A$

$\displaystyle \Rightarrow [\,2\ \ 1\ \ 3\,] \begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = [x]$

$\displaystyle \Rightarrow [\, -2 - 1 + 0 \ \ \ 0 + 1 + 3 \ \ \ -2 + 0 + 3 \,] \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = [x]$

$\displaystyle \Rightarrow [\, -3\ \ 4\ \ 1 \,] \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} = [x]$

$\displaystyle \Rightarrow [\, -3 + 0 - 1 \,] = [x]$

$\displaystyle \Rightarrow [-4] = [x]$

$\displaystyle \text{The corresponding elements of two equal matrices are equal.}$

$\displaystyle \therefore\ x = -4$

$\displaystyle \therefore\ A = [-4]$

$\displaystyle \text{(v) }$

$\displaystyle \ \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} A = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\displaystyle \text{Let } A = \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{bmatrix} \begin{bmatrix} x & y & z \\ a & b & c \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2x-a & 2y-b & 2z-c \\ x & y & z \\ -3x+4a & -3y+4b & -3z+4c \end{bmatrix} = \begin{bmatrix} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{bmatrix}$

$\displaystyle \text{By comparing the elements of second row, we get}$

$\displaystyle x = 1,\ y = -2,\ z = -5$

$\displaystyle \text{By comparing the elements of first row, we get}$

$\displaystyle 2x - a = -1$

$\displaystyle 2 - a = -1$

$\displaystyle a = 3$
$\displaystyle \text{Also,}$

$\displaystyle 2y - b = -8$

$\displaystyle \Rightarrow -4 - b = -8$

$\displaystyle \Rightarrow b = 4$

$\displaystyle \text{And,}$

$\displaystyle 2z - c = -10$

$\displaystyle \Rightarrow -10 - c = -10$

$\displaystyle \Rightarrow c = 0$

$\displaystyle \therefore A = \begin{bmatrix} 1 & -2 & -5 \\ 3 & 4 & 0 \end{bmatrix}$

$\displaystyle \text{(vi) }$

$\displaystyle A \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{bmatrix}$

$\displaystyle \text{Let } A = \begin{bmatrix} x & a \\ y & b \\ z & c \end{bmatrix}$

$\displaystyle \begin{bmatrix} x & a \\ y & b \\ z & c \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{bmatrix}$

$\displaystyle \begin{bmatrix} x + 4a & 2x + 5a & 3x + 6a \\ y + 4b & 2y + 5b & 3y + 6b \\ z + 4c & 2z + 5c & 3z + 6c \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{bmatrix}$

$\displaystyle x + 4a = -7 \quad \text{and} \quad 2x + 5a = -8$

$\displaystyle \Rightarrow a = -2 \quad \text{and} \quad x = 1$

$\displaystyle y + 4b = 2 \quad \text{and} \quad 2y + 5b = 4$

$\displaystyle \Rightarrow b = 0 \quad \text{and} \quad y = 2$

$\displaystyle z + 4c = 11 \quad \text{and} \quad 2z + 5c = 10$

$\displaystyle \Rightarrow c = 4 \quad \text{and} \quad z = -5$

$\displaystyle \therefore A = \begin{bmatrix} 1 & -2 \\ 2 & 0 \\ -5 & 4 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 49: } \text{Find a } 2 \times 2 \text{ matrix } A \text{ such that } A = \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = 6I_2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } A = \begin{bmatrix} w & x \\ y & z \end{bmatrix}$

$\displaystyle \begin{bmatrix} w & x \\ y & z \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 1 & 4 \end{bmatrix} = 6I_2$

$\displaystyle \Rightarrow \begin{bmatrix} w+x & -2w+4x \\ y+z & -2y+4z \end{bmatrix} = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} w+x & -2w+4x \\ y+z & -2y+4z \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

$\displaystyle \text{The corresponding elements of two equal matrices are equal.}$

$\displaystyle \therefore w + x = 6$

$\displaystyle \Rightarrow w = 6 - x \quad ...(1)$

$\displaystyle -2w + 4x = 0 \quad ...(2)$

$\displaystyle \text{Putting the value of } w \text{ in eq.(2), we get}$

$\displaystyle -2(6 - x) + 4x = 0$

$\displaystyle \Rightarrow -12 + 2x + 4x = 0$

$\displaystyle \Rightarrow -12 + 6x = 0$

$\displaystyle \Rightarrow 6x = 12$

$\displaystyle \Rightarrow x = 2$
$\displaystyle \text{Putting the value of } x \text{ in eq.(1), we get}$

$\displaystyle w = 6 - 2$

$\displaystyle \Rightarrow w = 4$

$\displaystyle \text{Now,}$

$\displaystyle y + z = 0$

$\displaystyle \Rightarrow y = -z \quad ...(3)$

$\displaystyle -2y + 4z = 6 \quad ...(4)$

$\displaystyle \text{Putting the value of } y \text{ in eq.(4), we get}$

$\displaystyle -2(-z) + 4z = 6$

$\displaystyle \Rightarrow 2z + 4z = 6$

$\displaystyle \Rightarrow 6z = 6$

$\displaystyle \Rightarrow z = 1$

$\displaystyle \text{Putting the value of } z \text{ in eq.(3), we get}$

$\displaystyle y = -1$

$\displaystyle \therefore A = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 50: } \text{If } A = \begin{bmatrix} 0 & 0 \\ 4 & 0 \end{bmatrix}, \text{ find } A^{16}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } A = \begin{bmatrix} 0 & 0 \\ 4 & 0 \end{bmatrix}$

$\displaystyle \text{Here,}$

$\displaystyle A^2 = AA$

$\displaystyle \Rightarrow A^2 = \begin{bmatrix} 0 & 0 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 4 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow A^2 = \begin{bmatrix} 0+0 & 0+0 \\ 0+0 & 0+0 \end{bmatrix}$

$\displaystyle \Rightarrow A^2 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle A^4 = A^2 A^2$

$\displaystyle \Rightarrow A^4 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow A^4 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle A^8 = A^4 A^4$

$\displaystyle \Rightarrow A^8 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow A^8 = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle A^{16} = A^8 A^8$

$\displaystyle \Rightarrow A^{16} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle \therefore A^{16} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle \text{Thus, } A^{16} \text{ is a null matrix.}$

$\\$

$\displaystyle \text{Question 51: } \text{If } A=\begin{bmatrix}0 & -x \\ x & 0\end{bmatrix},\ B=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \text{ and } x^2=-1,\ \text{then show that } \\ (A+B)^2 = A^2 + B^2.$

$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: } A=\begin{bmatrix}0&-x\\x&0\end{bmatrix},\ B=\begin{bmatrix}0&1\\1&0\end{bmatrix} \text{ and } x^2=-1$

$\displaystyle \text{To show: } (A+B)^2 = A^2 + B^2$

$\displaystyle \text{LHS:}$

$\displaystyle A+B =\begin{bmatrix}0&-x\\x&0\end{bmatrix} +\begin{bmatrix}0&1\\1&0\end{bmatrix} =\begin{bmatrix}0+0&-x+1\\x+1&0+0\end{bmatrix} =\begin{bmatrix}0&1-x\\x+1&0\end{bmatrix}$

$\displaystyle (A+B)^2 =\begin{bmatrix}0&1-x\\x+1&0\end{bmatrix} \begin{bmatrix}0&1-x\\x+1&0\end{bmatrix}$

$\displaystyle =\begin{bmatrix} 0+(1-x)(1+x) & 0+0 \\ 0+0 & (x+1)(1-x)+0 \end{bmatrix} =\begin{bmatrix}1-x^2&0\\0&1-x^2\end{bmatrix} \quad ... (1)$

$\displaystyle \text{RHS:}$

$\displaystyle A=\begin{bmatrix}0&-x\\x&0\end{bmatrix}$

$\displaystyle A^2=\begin{bmatrix}0&-x\\x&0\end{bmatrix} \begin{bmatrix}0&-x\\x&0\end{bmatrix} =\begin{bmatrix}-x^2&0\\0&-x^2\end{bmatrix}$

$\displaystyle B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$

$\displaystyle B^2=\begin{bmatrix}0&1\\1&0\end{bmatrix} \begin{bmatrix}0&1\\1&0\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$\displaystyle A^2+B^2 =\begin{bmatrix}-x^2+1&0\\0&-x^2+1\end{bmatrix} =\begin{bmatrix}1-x^2&0\\0&1-x^2\end{bmatrix} \quad ... (2)$

$\displaystyle B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

$\displaystyle B^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0+1 & 0+0 \\ 0+0 & 1+0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \quad ...(3)$

$\displaystyle \text{Adding (2) and (3), we get}$

$\displaystyle A^2 + B^2 = \begin{bmatrix} - x^2 & 0 \\ 0 & - x^2 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 - x^2 & 0 \\ 0 & 1 - x^2 \end{bmatrix} \quad ...(4)$

$\displaystyle \text{Comparing (1) and (4), we get}$

$\displaystyle (A + B)^2 = A^2 + B^2$

$\\$

$\displaystyle \text{Question 52: } A = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \text{ then verify that } A^2 + A = A(A + I), \\ \text{ where } I \text{ is the identity matrix.}$
$\displaystyle \text{Answer:}$
$\displaystyle A = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1+0+0 & 0+0-3 & -3+0-3 \\ 2+2+0 & 0+1+3 & -6+3+3 \\ 0+2+0 & 0+1+1 & 0+3+1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{bmatrix}$
$\displaystyle \text{L.H.S}$
$\displaystyle A^2 + A = \begin{bmatrix} 1 & -3 & -6 \\ 4 & 4 & 0 \\ 2 & 2 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1+1 & -3+0 & -6-3 \\ 4+2 & 4+1 & 0+3 \\ 2+0 & 2+1 & 4+1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{bmatrix}$
$\displaystyle \text{R.H.S}$
$\displaystyle A + I = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1+1 & 0+0 & -3+0 \\ 2+0 & 1+1 & 3+0 \\ 0+0 & 1+0 & 1+1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$
$\displaystyle A(A+I) = \begin{bmatrix} 1 & 0 & -3 \\ 2 & 1 & 3 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 2 & 0 & -3 \\ 2 & 2 & 3 \\ 0 & 1 & 2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 2+0+0 & 0+0-3 & -3+0-6 \\ 4+2+0 & 0+2+3 & -6+3+6 \\ 0+2+0 & 0+2+1 & 0+3+2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 2 & -3 & -9 \\ 6 & 5 & 3 \\ 2 & 3 & 5 \end{bmatrix}$
$\displaystyle \text{Therefore, LHS = RHS}$
$\displaystyle \text{Hence, } A^2 + A = A(A + I) \text{ is verified}$

$\\$

$\displaystyle \text{Question 53: } A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \text{ then find } A^2 - 5A - 14I. \text{ Hence, obtain } A^3.$
$\displaystyle \text{Answer:}$
$\displaystyle A = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} \begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 9+20 & -15-10 \\ -12-8 & 20+4 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix}$
$\displaystyle A^2 - 5A - 14I = \begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} -5\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix} -14\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
$\displaystyle \therefore A^2 - 5A - 14I = 0 \qquad (1)$
$\displaystyle \text{Premultiplying (1) by } A, \text{ we get}$
$\displaystyle A(A^2 - 5A - 14I) = A \cdot 0$
$\displaystyle \Rightarrow A^3 - 5A^2 - 14A = 0$
$\displaystyle \Rightarrow A^3 = 5A^2 + 14A$
$\displaystyle A^3 = 5\begin{bmatrix} 29 & -25 \\ -20 & 24 \end{bmatrix} +14\begin{bmatrix} 3 & -5 \\ -4 & 2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 145+42 & -125-70 \\ -100-56 & 120+28 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 187 & -195 \\ -156 & 148 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 54: }$
$\displaystyle \text{(i) If } P(x)=\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}, \text{ then show that } P(x)P(y)=P(x+y)=P(y)P(x).$
$\displaystyle \text{(ii) If } P = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \text{ and } Q = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}, \text{ prove that } PQ = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix} = QP.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given: } P(x)=\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$
$\displaystyle \text{Then,}$
$\displaystyle P(y)=\begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix}$
$\displaystyle \text{Now,}$
$\displaystyle P(x)P(y) =\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix} \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix}$
$\displaystyle = \begin{bmatrix} \cos x\cos y-\sin x\sin y & \cos x\sin y+\sin x\cos y \\ -\sin x\cos y-\cos x\sin y & -\sin x\sin y+\cos x\cos y \end{bmatrix}$
$\displaystyle = \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} \ldots (1)$
$\displaystyle \text{Also,}$
$\displaystyle P(x+y)= \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} \ldots (2)$
$\displaystyle \text{Also,}$
$\displaystyle P(x+y)= \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} \quad ...(2)$
$\displaystyle \text{Now,}$
$\displaystyle P(y)P(x)= \begin{bmatrix} \cos y & \sin y \\ -\sin y & \cos y \end{bmatrix} \begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$
$\displaystyle = \begin{bmatrix} \cos y \cos x - \sin y \sin x & \cos y \sin x + \sin y \cos x \\ -\sin y \cos x - \cos y \sin x & -\sin y \sin x + \cos y \cos x \end{bmatrix}$
$\displaystyle = \begin{bmatrix} \cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y) \end{bmatrix} \quad ...(3)$
$\displaystyle \text{From (1), (2) and (3), we get}$
$\displaystyle P(x)P(y)=P(x+y)=P(y)P(x)$

$\displaystyle \text{(ii) }$
$\displaystyle PQ = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}$
$\displaystyle = \begin{bmatrix} xa+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+yb+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+zc \end{bmatrix}$
$\displaystyle = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix} \quad ...(4)$
$\displaystyle QP = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$
$\displaystyle = \begin{bmatrix} ax+0+0 & 0+0+0 & 0+0+0 \\ 0+0+0 & 0+by+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+cz \end{bmatrix}$
$\displaystyle = \begin{bmatrix} xa & 0 & 0 \\ 0 & yb & 0 \\ 0 & 0 & zc \end{bmatrix} \quad ...(5)$
$\displaystyle \therefore \; PQ = QP$
$\\$
$\displaystyle \text{Question 55: } A=\begin{bmatrix} 2 & 0 & 1\\ 2 & 1 & 3\\ 1 & -1 & 0 \end{bmatrix}, \text{ find } A^2 - 5A + 4I \text{ and hence find a matrix } X \text{ such that } \\ A^2 - 5A + 4I + X = 0. \hspace{2.0cm} \text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given:}$
$\displaystyle A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$
$\displaystyle A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$
$\displaystyle \text{Now,}$
$\displaystyle A^2 - 5A + 4I =$
$\displaystyle \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} -5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} +4 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 5-10+4 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+4 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+4 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}$
$\displaystyle \text{Now, } A^2 - 5A + 4I + X = 0$
$\displaystyle \Rightarrow X = -(A^2 - 5A + 4I)$
$\displaystyle \therefore X = \begin{bmatrix} -1 & -1 & -3 \\ -1 & -3 & -10 \\ -5 & 4 & 2 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 3 & 10 \\ 5 & -4 & -2 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 56: } \text{If } A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}, \text{ prove that } A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \text{ for all positive integers } n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$
$\displaystyle \text{Step 1: If } n = 1, \text{ by definition of integral powers of matrix, we have}$
$\displaystyle A^1 = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = A$
$\displaystyle \text{So, the result is true for } n = 1.$
$\displaystyle \text{Step 2: Let the result be true for } n = m. \text{ Then,}$
$\displaystyle A^m = \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} \qquad (1)$
$\displaystyle \text{Now, we shall show that the result is true for } n = m + 1.$
$\displaystyle \text{Here,}$
$\displaystyle A^{m+1} = \begin{bmatrix} 1 & m+1 \\ 0 & 1 \end{bmatrix}$
$\displaystyle \text{By definition of integral power of matrix, we have}$
$\displaystyle A^{m+1} = A^m A$
$\displaystyle = \begin{bmatrix} 1 & m \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$
$\displaystyle \text{[From eq. (1)]}$
$\displaystyle = \begin{bmatrix} 1 + 0 & 1 + m \\ 0 + 0 & 0 + 1 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 1 & 1 + m \\ 0 & 1 \end{bmatrix}$
$\displaystyle \text{This shows that when the result is true for } n = m, \text{ it is also true for } n = m + 1.$
$\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for any positive integer } n.$

$\\$

$\displaystyle \text{Question 57: } \text{If } A=\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}, \text{ prove that } A^n=\begin{bmatrix} a^n & \dfrac{b(a^n-1)}{a-1} \\ 0 & 1 \end{bmatrix} \text{for every positive integer } n.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$
$\displaystyle \text{Step 1: If } n=1,\ \text{by definition of integral power of a matrix, we have}$
$\displaystyle A^1=\begin{bmatrix} a^1 & \dfrac{b(a^1-1)}{a-1} \\ 0 & 1 \end{bmatrix} =\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} = A$
$\displaystyle \text{So, the result is true for } n=1.$
$\displaystyle \text{Step 2: Let the result be true for } n=m.\ \text{Then,}$
$\displaystyle A^m=\begin{bmatrix} a^m & \dfrac{b(a^m-1)}{a-1} \\ 0 & 1 \end{bmatrix} \quad ...(1)$
$\displaystyle \text{Now, we shall show that the result is true for } n=m+1.$
$\displaystyle \text{Here,}$
$\displaystyle A^{m+1}=\begin{bmatrix} a^{m+1} & \dfrac{b(a^{m+1}-1)}{a-1} \\ 0 & 1 \end{bmatrix}$
$\displaystyle \text{By definition of integral power of matrix, we have}$
$\displaystyle A^{m+1} = A^{m}A$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^{m} & \dfrac{b(a^{m}-1)}{a-1} \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \quad [\text{From eq. (1)}]$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^{m}a+0 & \dfrac{a^{m}b + b(a^{m}-1)}{a-1} \\ 0+0 & 0+1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^{m+1} & \dfrac{a^{m+1}b - a^{m}b + a^{m}b - b}{a-1} \\ 0 & 1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^{m+1} & \dfrac{b(a^{m+1}-1)}{a-1} \\ 0 & 1 \end{bmatrix}$
$\displaystyle \text{This shows that when the result is true for } n=m, \text{ it is also true for } n=m+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,} \text{ the result is valid for any positive integer } n.$

$\\$

$\displaystyle \text{Question 58: } \text{If } A = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix}, \text{ then prove by the principle of mathematical induction that }$
$\displaystyle A^n = \begin{bmatrix} \cos n\theta & i \sin n\theta \\ i \sin n\theta & \cos n\theta \end{bmatrix} \text{ for all } n \in N. \hspace{2.0cm} \text{[CBSE 2005]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$
$\displaystyle \text{Step 1: If } n = 1, \text{ by definition of integral power of a matrix, we have}$
$\displaystyle A^1 = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos \theta & i \sin \theta \\ i \sin \theta & \cos \theta \end{bmatrix} = A$
$\displaystyle \text{Thus, the result is true for } n = 1.$
$\displaystyle \text{Step 2: Let the result be true for } n = m. \text{ Then,}$
$\displaystyle A^m = \begin{bmatrix} \cos m\theta & i \sin m\theta \\ i \sin m\theta & \cos m\theta \end{bmatrix}$
$\displaystyle \text{Now we shall show that the result is true for } n = m + 1.$
$\displaystyle \text{Here,}$
$\displaystyle A^{m+1} = \begin{bmatrix} \cos (m+1)\theta & i \sin (m+1)\theta \\ i \sin (m+1)\theta & \cos (m+1)\theta \end{bmatrix} \quad (1)$
$\displaystyle \text{By definition of integral power of matrix, we have}$
$\displaystyle A^{m+1} = A^m A$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos m\theta & i\sin m\theta \\ i\sin m\theta & \cos m\theta \end{bmatrix} \begin{bmatrix} \cos\theta & i\sin\theta \\ i\sin\theta & \cos\theta \end{bmatrix} \quad \text{[From eq. (1)]}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos m\theta \cos\theta + i\sin m\theta \, i\sin\theta & \cos m\theta \, i\sin\theta + i\sin m\theta \cos\theta \\ i\sin m\theta \cos\theta + \cos m\theta \, i\sin\theta & i\sin m\theta \, i\sin\theta + \cos m\theta \cos\theta \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos m\theta \cos\theta - \sin m\theta \sin\theta & i(\cos m\theta \sin\theta + \sin m\theta \cos\theta) \\ i(\sin m\theta \cos\theta + \cos m\theta \sin\theta) & \cos m\theta \cos\theta - \sin m\theta \sin\theta \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos(m\theta + \theta) & i\sin(m\theta + \theta) \\ i\sin(m\theta + \theta) & \cos(m\theta + \theta) \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos((m+1)\theta) & i\sin((m+1)\theta) \\ i\sin((m+1)\theta) & \cos((m+1)\theta) \end{bmatrix}$
$\displaystyle \text{This shows that when the result is true for } n = m \text{, it is true for } n = m + 1$
$\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for all } n \in N$

$\\$

$\displaystyle \text{Question 59: } A = \begin{bmatrix} \cos\alpha + \sin\alpha & \sqrt{2}\sin\alpha \\ -\sqrt{2}\sin\alpha & \cos\alpha - \sin\alpha \end{bmatrix}\text{Prove that}$
$\displaystyle A^n = \begin{bmatrix} \cos n\alpha + \sin n\alpha & \sqrt{2}\sin n\alpha \\ -\sqrt{2}\sin n\alpha & \cos n\alpha - \sin n\alpha \end{bmatrix}, \quad \text{for all } n \in \mathbb{N}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$

$\displaystyle \text{Step 1: } \text{If } n = 1, \text{ by definition of integral power of a matrix, we have}$

$\displaystyle A^1 = \begin{bmatrix} \cos 1\alpha + \sin 1\alpha & \sqrt{2}\sin 1\alpha \\ -\sqrt{2}\sin 1\alpha & \cos 1\alpha - \sin 1\alpha \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ -\sqrt{2}\sin \alpha & \cos \alpha - \sin \alpha \end{bmatrix} = A$

$\displaystyle \text{So, the result is true for } n = 1.$

$\displaystyle \text{Step 2: } \text{Let the result be true for } n = m. \text{ Then,}$

$\displaystyle A^m = \begin{bmatrix} \cos m\alpha + \sin m\alpha & \sqrt{2}\sin m\alpha \\ -\sqrt{2}\sin m\alpha & \cos m\alpha - \sin m\alpha \end{bmatrix}$

$\displaystyle \text{Now we shall show that the result is true for } n = m + 1.$

$\displaystyle \text{Here,}$

$\displaystyle A^{m+1} = \begin{bmatrix} \cos (m+1)\alpha + \sin (m+1)\alpha & \sqrt{2}\sin (m+1)\alpha \\ -\sqrt{2}\sin (m+1)\alpha & \cos (m+1)\alpha - \sin (m+1)\alpha \end{bmatrix}$

$\displaystyle \text{By definition of integral power of matrix, we have}$

$\displaystyle A^{m+1} = A^m A$

$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} \cos m\alpha + \sin m\alpha & \sqrt{2}\sin m\alpha \\ -\sqrt{2}\sin m\alpha & \cos m\alpha - \sin m\alpha \end{bmatrix} \begin{bmatrix} \cos \alpha + \sin \alpha & \sqrt{2}\sin \alpha \\ -\sqrt{2}\sin \alpha & \cos \alpha - \sin \alpha \end{bmatrix}$

$\displaystyle = \begin{bmatrix} (\cos m\alpha + \sin m\alpha)(\cos \alpha + \sin \alpha) - 2\sin m\alpha \sin \alpha & \sqrt{2}\sin \alpha (\cos m\alpha + \sin m\alpha) + \sqrt{2}\sin m\alpha (\cos \alpha - \sin \alpha) \\ -\sqrt{2}\sin m\alpha (\cos \alpha + \sin \alpha) - \sqrt{2}\sin \alpha (\cos m\alpha - \sin m\alpha) & -2\sin m\alpha \sin \alpha + (\cos m\alpha - \sin m\alpha)(\cos \alpha - \sin \alpha) \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos(m\alpha+\alpha) + \sin(m\alpha+\alpha) & \sqrt{2}\sin(m\alpha+\alpha) \\ -\sqrt{2}\sin(m\alpha+\alpha) & \cos(m\alpha+\alpha) - \sin(m\alpha+\alpha) \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \cos (m+1)\alpha + \sin (m+1)\alpha & \sqrt{2}\sin (m+1)\alpha \\ -\sqrt{2}\sin (m+1)\alpha & \cos (m+1)\alpha - \sin (m+1)\alpha \end{bmatrix}$

$\displaystyle \text{This shows that when the result is true for } n=m, \text{ it is also true for } n=m+1.$

$\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for all } n \in N.$

$\\$

$\displaystyle \text{Question 60: }$
$\displaystyle \text{Let } A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}. \text{ Use the principle of mathematical induction to show that }$
$\displaystyle A^n = \begin{bmatrix} 1 & n & \frac{n(n+1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix} \text{ for every positive integer } n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$
$\displaystyle \text{Step 1: If } n=1,\ \text{by definition of integral power of a matrix, we have}$
$\displaystyle A^1 = \begin{bmatrix} 1 & 1 & \dfrac{1(1+1)}{2} \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = A$
$\displaystyle \text{Thus, the result is true for } n=1.$
$\displaystyle \text{Step 2: Let the result be true for } n=m.\ \text{Then,}$
$\displaystyle A^m = \begin{bmatrix} 1 & m & \dfrac{m(m+1)}{2} \\ 0 & 1 & m \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \text{Now, we shall show that the result is true for } n=m+1.$
$\displaystyle \text{Here,}$
$\displaystyle A^{m+1} = \begin{bmatrix} 1 & m+1 & \dfrac{m+1(m+1+1)}{2} \\ 0 & 1 & m+1 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1 & m+1 & \dfrac{(m+1)(m+2)}{2} \\ 0 & 1 & m+1 \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \text{By definition of integral power of matrix, we have}$
$\displaystyle A^{m+1} = A^{m} A$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1 & m & \dfrac{m(m+1)}{2} \\ 0 & 1 & m \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \quad \text{[From eq. (1)]}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1+0+0 & 1+m+0 & 1+m+\dfrac{m(m+1)}{2} \\ 0+0+0 & 0+1+0 & 0+1+m \\ 0+0+0 & 0+0+0 & 0+0+1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1 & 1+m & \dfrac{2+2m+m^2+m}{2} \\ 0 & 1 & 1+m \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1 & 1+m & \dfrac{m^2+3m+2}{2} \\ 0 & 1 & 1+m \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} 1 & 1+m & \dfrac{(m+1)(m+2)}{2} \\ 0 & 1 & 1+m \\ 0 & 0 & 1 \end{bmatrix}$
$\displaystyle \text{This shows that when the result is true for } n = m, \text{ it is also true for } n = m + 1.$
$\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for any positive integer } n.$

$\\$

$\displaystyle \text{Question 61: } \text{If } B \text{ and } C \text{ are } n\text{-rowed square matrices and if } A = B + C,\ BC = CB,\ C^2 = O,\ \text{then show that for every } n \in \mathbb{N},\ A^{n+1} = B^n \big( B + (n+1)C \big).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } P(n) \text{ be the statement given by}$
$\displaystyle P(n):\ A^{n+1} = B^{n}\,(B + (n+1)C)$
$\displaystyle \text{For } n = 1, \text{ we have}$
$\displaystyle P(1):\ A^{2} = B(B + 2C)$
$\displaystyle \text{Here,}$
$\displaystyle \text{LHS} = A^{2}$
$\displaystyle = (B + C)(B + C)$
$\displaystyle = B(B + C) + C(B + C)$
$\displaystyle = B^{2} + BC + CB + C^{2}$
$\displaystyle = B^{2} + 2BC \quad [\because\ BC = CB \text{ and } C^{2} = 0]$
$\displaystyle = B(B + 2C) = \text{RHS}$
$\displaystyle \text{Hence, the statement is true for } n = 1.$
$\displaystyle \text{If the statement is true for } n = k, \text{ then}$
$\displaystyle P(k):\; A^{k+1} = B^k \bigl( B + (k+1)C \bigr) \qquad \text{...(1)}$
$\displaystyle \text{For } P(k+1) \text{ to be true, we must have}$
$\displaystyle P(k+1):\; A^{k+2} = B^{k+1} \bigl( B + (k+2)C \bigr)$
$\displaystyle \text{Now,}$
$\displaystyle A^{k+2} = A^{k+1} A$
$\displaystyle = \bigl[ B^k ( B + (k+1)C ) \bigr] (B + C) \qquad \text{[From eq. (1)]}$
$\displaystyle = \bigl[ B^{k+1} + (k+1) B^k C \bigr] (B + C)$
$\displaystyle = B^{k+1} (B + C) + (k+1) B^k C (B + C)$
$\displaystyle = B^{k+2} + B^{k+1} C + (k+1) B^k C B + (k+1) B^k C^2$
$\displaystyle = B^{k+2} + B^{k+1} C + (k+1) B^k B C \quad [\because BC = CB \text{ and } C^2 = 0]$
$\displaystyle = B^{k+2} + B^{k+1} C + (k+1) B^{k+1} C$
$\displaystyle = B^{k+2} + (k+2) B^{k+1} C$
$\displaystyle = B^{k+1} \bigl[ B + (k+2) C \bigr]$
$\displaystyle \text{So the statement is true for } n = k+1.$
$\displaystyle \text{Hence, by the principle of mathematical induction,}$
$\displaystyle P(n)\ \text{is true for all } n \in N.$

$\\$

$\displaystyle \text{Question 62: } \text{If } A = \text{diag} (a, b, c), \text{ show that } A^n = \text{diag} (a^n, b^n, c^n) \text{ for all positive integers } n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We shall prove the result by the principle of mathematical induction on } n.$
$\displaystyle \text{Step 1: If } n=1,\ \text{by definition of integral power of a matrix, we have}$
$\displaystyle A^1 = \begin{bmatrix} a^1 & 0 & 0 \\ 0 & b^1 & 0 \\ 0 & 0 & c^1 \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} = A$
$\displaystyle \text{So, the result is true for } n=1.$
$\displaystyle \text{Step 2: Let the result be true for } n=m.\ \text{Then,}$
$\displaystyle A^m = \begin{bmatrix} a^m & 0 & 0 \\ 0 & b^m & 0 \\ 0 & 0 & c^m \end{bmatrix} \quad \ldots (1)$
$\displaystyle \text{Now, we shall check if the result is true for } n=m+1.$
$\displaystyle \text{Here,}$
$\displaystyle A^{m+1} = \begin{bmatrix} a^{m+1} & 0 & 0 \\ 0 & b^{m+1} & 0 \\ 0 & 0 & c^{m+1} \end{bmatrix}$
$\displaystyle \text{By definition of integral power of matrix, we have}$
$\displaystyle A^{m+1} = A^m A$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^m & 0 & 0 \\ 0 & b^m & 0 \\ 0 & 0 & c^m \end{bmatrix} \begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix} \quad \text{[From eq. (1)]}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a a^m + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & b b^m + 0 & 0 + 0 + 0 \\ 0 + 0 + 0 & 0 + 0 + 0 & c c^m \end{bmatrix}$
$\displaystyle \Rightarrow A^{m+1} = \begin{bmatrix} a^{m+1} & 0 & 0 \\ 0 & b^{m+1} & 0 \\ 0 & 0 & c^{m+1} \end{bmatrix}$
$\displaystyle \text{This shows that when the result is true for } n = m,\ \text{it is also true for } n = m + 1.$
$\displaystyle \text{Hence, by the principle of mathematical induction, the result is valid for any positive integer } n.$

$\\$

$\displaystyle \text{Question 63: } \text{If } A \text{ is a square matrix, using mathematical induction prove that } (A^T)^n = (A^n)^T \text{ for all } n \in N.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the given statement } P(n), \text{ be given as}$
$\displaystyle P(n):\ (A^T)^n = (A^n)^T \text{ for all } n \in \mathbb{N}.$
$\displaystyle \text{We observe that}$
$\displaystyle P(1):\ (A^T)^1 = A^T = (A^1)^T$
$\displaystyle \text{Thus, } P(n) \text{ is true for } n = 1.$
$\displaystyle \text{Assume that } P(n) \text{ is true for } n = k \in \mathbb{N}.$
$\displaystyle \text{i.e., } P(k):\ (A^T)^k = (A^k)^T$
$\displaystyle \text{To prove that } P(k+1) \text{ is true, we have}$
$\displaystyle (A^T)^{k+1} = (A^T)^k (A^T)^1$
$\displaystyle = (A^k)^T (A^1)^T$
$\displaystyle = (A^{k+1})^T$
$\displaystyle \text{Thus, } P(k+1) \text{ is true, whenever } P(k) \text{ is true.}$
$\displaystyle \text{Hence, by the principle of mathematical induction, } P(n) \text{ is true for all } n \in N.$

$\\$

$\displaystyle \text{Question 64: } \text{A matrix } X \text{ has } (a+b) \text{ rows and } (a+2) \text{ columns, while the matrix } Y \text{ has } \\ (b+1) \text{ rows and } (a+3) \text{ columns. Both matrices } XY \text{ and } YX \text{ exist. \ \\ Find } a \text{ and } b. \\ \text{Can you say } XY \text{ and } YX \text{ are of the same type? Are they equal?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here,}$
$\displaystyle [X]_{(a+b)\times(a+2)}$
$\displaystyle [Y]_{(b+1)\times(a+3)}$
$\displaystyle \text{Since } XY \text{ exists, the number of columns in } X \text{ is equal to the number of rows in } Y.$
$\displaystyle \Rightarrow a+2=b+1 \quad \ldots(1)$
$\displaystyle \text{Similarly, since } YX \text{ exists, the number of columns in } Y \text{ is equal to the number of rows in } X.$
$\displaystyle \Rightarrow a+b=a+3$
$\displaystyle \Rightarrow b=3$
$\displaystyle \text{Putting the value of } b \text{ in (1), we get}$
$\displaystyle a+2=3+1$
$\displaystyle \Rightarrow a=2$
$\displaystyle \text{Since the order of the matrices } XY \text{ and } YX \text{ is not same, } XY \text{ and } \\ YX \text{ are not of the same type and they are unequal.}$

$\\$

$\displaystyle \text{Question 65: }$
$\displaystyle \text{Give examples of matrices:}$
$\displaystyle \text{(i) } A \text{ and } B \text{ such that } AB \ne BA.$
$\displaystyle \text{(ii) } A \text{ and } B \text{ such that } AB = O \text{ but } A \ne 0 \text{ and } B \ne 0.$
$\displaystyle \text{(iii) } A \text{ and } B \text{ such that } AB = O \text{ but } BA \ne O.$
$\displaystyle \text{(iv) } A, B \text{ and } C \text{ such that } AB = AC \text{ but } B \ne C \text{ and } A \ne 0.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$
$\displaystyle \ Let\ A=\begin{bmatrix}1 & -2\\ 3 & 2\end{bmatrix}\ \text{and}\ B=\begin{bmatrix}2 & 3\\ -1 & 2\end{bmatrix}$
$\displaystyle AB=\begin{bmatrix}1 & -2\\ 3 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\ -1 & 2\end{bmatrix}$
$\displaystyle \Rightarrow AB=\begin{bmatrix}2+2 & 3-4\\ 6-2 & 9+4\end{bmatrix}$
$\displaystyle \Rightarrow AB=\begin{bmatrix}4 & -1\\ 4 & 13\end{bmatrix}$
$\displaystyle \text{Now,}$
$\displaystyle BA=\begin{bmatrix}2 & 3\\ -1 & 2\end{bmatrix}\begin{bmatrix}1 & -2\\ 3 & 2\end{bmatrix}$
$\displaystyle \Rightarrow BA=\begin{bmatrix}2+9 & -4+6\\ -1+6 & 2+4\end{bmatrix}$
$\displaystyle \Rightarrow BA=\begin{bmatrix}11 & 2\\ 5 & 6\end{bmatrix}$
$\displaystyle \text{Thus, } AB\neq BA.$

$\displaystyle \text{(ii) }$
$\displaystyle \ \text{Let } A=\begin{bmatrix}0&2\\0&0\end{bmatrix}\ \text{and}\ B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\displaystyle AB=\begin{bmatrix}0&2\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0+0&0+0\\0+0&0+0\end{bmatrix}$
$\displaystyle =\begin{bmatrix}0&0\\0&0\end{bmatrix}=O$
$\displaystyle \text{Thus, } AB=O\ \text{while } A\neq 0\ \text{and } B\neq 0.$

$\displaystyle \text{(iii) }$
$\displaystyle \ \text{Let } A=\begin{bmatrix}0&1\\0&0\end{bmatrix}\ \text{and } B=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
$\displaystyle \therefore\ AB=O$
$\displaystyle \text{and } BA=\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$\displaystyle \Rightarrow\ BA=\begin{bmatrix}0+0&1+0\\0+0&0+0\end{bmatrix}$
$\displaystyle \Rightarrow\ BA=\begin{bmatrix}0&1\\0&0\end{bmatrix}$
$\displaystyle \text{Thus, } AB=O\ \text{but } BA\neq O.$

$\displaystyle \text{(iv) }$
$\displaystyle \ Let\ A=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ B=\begin{bmatrix}0&0\\0&1\end{bmatrix}\ \text{and}\ C=\begin{bmatrix}0&0\\0&2\end{bmatrix}$
$\displaystyle \therefore\ AB= \begin{bmatrix}1&0\\0&0\end{bmatrix} \begin{bmatrix}0&0\\0&1\end{bmatrix}$
$\displaystyle \Rightarrow AB= \begin{bmatrix} 0+0 & 0+0\\ 0+0 & 0+0 \end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\displaystyle \text{and } AC= \begin{bmatrix}1&0\\0&0\end{bmatrix} \begin{bmatrix}0&0\\0&2\end{bmatrix}$
$\displaystyle \Rightarrow AC= \begin{bmatrix} 0+0 & 0+0\\ 0+0 & 0+0 \end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$
$\displaystyle \text{Thus,}$
$\displaystyle AB = AC$
$\displaystyle \text{But } B \neq C \text{ and } A \neq 0.$
$\\$
$\displaystyle \text{Question 66: } \text{Let } A \text{ and } B \text{ be square matrices of the same order. Does } (A+B)^2 = A^2 + 2AB + B^2 \text{ hold? If not, why?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS} = (A + B)^2$
$\displaystyle = (A + B)(A + B)$
$\displaystyle = A(A + B) + B(A + B)$
$\displaystyle = A^2 + AB + BA + B^2$
$\displaystyle \text{We know that a matrix does not have commutative property. So,}$
$\displaystyle AB \neq BA$
$\displaystyle \text{Thus,}$
$\displaystyle (A + B)^2 \neq A^2 + 2AB + B^2$

$\\$

$\displaystyle \text{Question 67: }$
$\displaystyle \text{If } A \text{ and } B \text{ are square matrices of the same order, explain why in general}$
$\displaystyle \text{(i) } (A + B)^2 \neq A^2 + 2AB + B^2$
$\displaystyle \text{(ii) } (A - B)^2 \neq A^2 - 2AB + B^2$
$\displaystyle \text{(iii) } (A + B)(A - B) \neq A^2 - B^2$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$
$\displaystyle \text{ LHS} = (A + B)^2$
$\displaystyle = (A + B)(A + B)$
$\displaystyle = A(A + B) + B(A + B)$
$\displaystyle = A^2 + AB + BA + B^2$
$\displaystyle \text{We know that a matrix does not have commutative property. So,}$
$\displaystyle AB \ne BA$
$\displaystyle \text{Thus, Proven}$

$\displaystyle \text{(ii) }$
$\displaystyle \text{ LHS} = (A - B)^2$
$\displaystyle = (A - B)(A - B)$
$\displaystyle = A(A - B) - B(A - B)$
$\displaystyle = A^2 - AB - BA + B^2$
$\displaystyle \text{We know that a matrix does not have commutative property. So,}$
$\displaystyle AB \neq BA$
$\displaystyle \text{Thus, Proven}$

$\displaystyle \text{(iii) }$
$\displaystyle \text{LHS} = (A + B)(A - B)$
$\displaystyle = A(A - B) + B(A - B)$
$\displaystyle = A^2 - AB + BA - B^2$
$\displaystyle \text{We know that a matrix does not have commutative property. So,}$
$\displaystyle AB \neq BA$
$\displaystyle \text{Thus,}$
$\displaystyle (A + B)(A - B) \neq A^2 - B^2$
$\\$
$\displaystyle \text{Question 68: } \text{Let } A \text{ and } B \text{ be square matrices of order } 3 \times 3. \text{ Is } (AB)^2 = A^2 B^2 \text{? Give reasons.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Yes, } (AB)^2 = A^2 B^2 \text{ if } AB = BA.$
$\displaystyle \text{If } AB = BA, \text{ then}$
$\displaystyle (AB)^2 = (AB)(AB)$
$\displaystyle = A(BA)B \quad \text{(associative law)}$
$\displaystyle = A(AB)B$
$\displaystyle = A^2 B^2$

$\\$

$\displaystyle \text{Question 69: } \text{If } A \text{ and } B \text{ are square matrices of the same order such that } AB = BA,\ \text{then show that } (A + B)^2 = A^2 + 2AB + B^2.$

$\displaystyle \text{Answer:}$

$\displaystyle (A + B)^2 = (A + B)(A + B)$
$\displaystyle = A^2 + AB + BA + B^2$
$\displaystyle = A^2 + 2AB + B^2 \quad (\because AB = BA)$
$\displaystyle \text{Hence, } (A + B)^2 = A^2 + 2AB + B^2.$

$\\$

$\displaystyle \text{Question 70: } \text{Let } A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix}, \; B = \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ -2 & 4 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 2 \\ -3 & 5 \\ 5 & 0 \end{bmatrix}.$
$\displaystyle \text{Verify that } AB = AC \text{ though } B \neq C,\; A \neq O.$

$\displaystyle \text{Answer:}$

$\displaystyle Here,$
$\displaystyle A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix}, \; B = \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ -2 & 4 \end{bmatrix} \text{ and } C = \begin{bmatrix} 4 & 2 \\ -3 & 5 \\ 5 & 0 \end{bmatrix}$
$\displaystyle Now,$
$\displaystyle AB = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ 5 & 2 \\ -2 & 4 \end{bmatrix}$
$\displaystyle \Rightarrow AB = \begin{bmatrix} 3 + 5 - 2 & 1 + 2 + 4 \\ 9 + 15 - 6 & 3 + 6 + 12 \end{bmatrix}$
$\displaystyle \Rightarrow AB = \begin{bmatrix} 6 & 7 \\ 18 & 21 \end{bmatrix}$
$\displaystyle AC = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ -3 & 5 \\ 5 & 0 \end{bmatrix}$
$\displaystyle \Rightarrow AC = \begin{bmatrix} 4 - 3 + 5 & 2 + 5 + 0 \\ 12 - 9 + 15 & 6 + 15 + 0 \end{bmatrix}$
$\displaystyle \Rightarrow AC = \begin{bmatrix} 6 & 7 \\ 18 & 21 \end{bmatrix}$
$\displaystyle \text{So, } AB = AC \text{ though } B \neq C,\ A \neq O.$

$\\$

Question 71: Three shopkeepers A, B, and C go to a store to buy stationery. Shopkeeper A purchases 12 dozen notebooks, 5 dozen pens, and 6 dozen pencils. Shopkeeper B purchases 10 dozen notebooks, 6 dozen pens, and 7 dozen pencils. Shopkeeper C purchases 11 dozen notebooks, 13 dozen pens, and 8 dozen pencils. The cost of one notebook is 40 paise, one pen costs Rs. 1.25, and one pencil costs 35 paise. Using matrix multiplication, calculate the total bill amount for each shopkeeper individually.

$\displaystyle \text{Answer:}$

$\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Shopkeepers} & \text{Notebooks (in dozen)} & \text{Pens (in dozen)} & \text{Pencils (in dozen)} \\ \hline A & 12 & 5 & 6 \\ \hline B & 10 & 6 & 7 \\ \hline C & 11 & 3 & 8 \\ \hline \end{array}$
$\displaystyle \text{Here,}$
$\displaystyle \text{Cost of notebooks per dozen } = (12 \times 40)\text{ paise } = \text{Rs. }4.80$
$\displaystyle \text{Cost of pens per dozen } = \text{Rs. }(12 \times 1.25) = \text{Rs. }15$
$\displaystyle \text{Cost of pencils per dozen } = (12 \times 35)\text{ paise } = \text{Rs. }4.20$
$\displaystyle \begin{bmatrix} 12 & 5 & 6 \\ 10 & 6 & 7 \\ 11 & 13 & 8 \end{bmatrix} \begin{bmatrix} 4.80 \\ 15 \\ 4.20 \end{bmatrix} = \begin{bmatrix} 12 \times 4.80 + 5 \times 15 + 6 \times 4.20 \\ 10 \times 4.80 + 6 \times 15 + 7 \times 4.20 \\ 11 \times 4.80 + 13 \times 15 + 8 \times 4.20 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 57.60 + 75 + 25.20 \\ 48 + 90 + 29.40 \\ 52.80 + 195 + 33.60 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 157.80 \\ 167.40 \\ 281.40 \end{bmatrix}$
$\displaystyle \text{Thus, the bills of A, B and C are Rs. }157.80,\ \text{Rs. }167.40\ \text{and Rs. }281.40\text{ respectively.}$

$\\$

Question 72: The cooperative store of a particular school has 10 dozen physics books, 8 dozen chemistry books, and 5 dozen mathematics books. The selling price of each physics book is Rs. 8.30, each chemistry book is Rs. 3.45, and each mathematics book is Rs. 4.50. Find the total amount of money the store will receive by selling all the books.

$\displaystyle \text{Answer:}$

$\displaystyle \text{Stock of various types of books in the store is given by}$
$\displaystyle \text{Physics \ \ \ \ \ \ \ Chemistry \ \ \ \ \ \ \ Mathematics}$
$\displaystyle X=\begin{bmatrix}120 & 96 & 60\end{bmatrix}$
$\displaystyle \text{Selling price of various types of books in the store is given by}$
$\displaystyle Y=\begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix} \begin{array}{l} \text{Physics}\\ \text{Chemistry}\\ \text{Mathematics} \end{array}$
$\displaystyle \text{Total amount received by the store from selling all the items is given by}$
$\displaystyle XY=\begin{bmatrix}120 & 96 & 60\end{bmatrix} \begin{bmatrix}8.30\\3.45\\4.50\end{bmatrix}$
$\displaystyle =\left[(120)(8.30)+(96)(3.45)+(60)(4.50)\right]$
$\displaystyle =\left[996+331.20+270\right]$
$\displaystyle =\left[1597.20\right]$
$\displaystyle \text{Required amount }= \text{Rs }1597.20$

$\\$

$\displaystyle \text{Question 73: }$

$\displaystyle \text{In a legislative assembly election, a political group hired a public relations firm to promote} \\ \text{its candidates in three ways: telephone, house calls and letters.}$
$\displaystyle \text{The cost per contact (in paise) is given by the matrix } A \text{ as}$
$\displaystyle A = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{array}{l} \text{Telephone} \\ \text{House call} \\ \text{Letter} \end{array}$
$\displaystyle \text{The number of contacts of each type made in two cities } X \text{ and } Y \text{ is given by the matrix } B \text{ as}$
$\displaystyle \begin{array}{ccc} \text{Telephone} & \text{House call} & \text{Letter} \end{array}$
$\displaystyle B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \rightarrow X \\ \rightarrow Y \end{matrix}$
$\displaystyle \text{Find the total amount spent by the group in the two cities } X \text{ and } Y.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{The cost per contact (in paise) is given by}$
$\displaystyle A = \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix} \begin{array}{l} \text{Telephone} \\ \text{Housecall} \\ \text{Letter} \end{array}$
$\displaystyle \text{The number of contacts of each type made in the two cities X and Y is given by}$
$\displaystyle B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{array}{l} X \\ Y \end{array}$
$\displaystyle \text{Total amount spent by the group in the two cities X and Y is given by}$
$\displaystyle BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 40000 + 50000 + 250000 \\ 120000 + 100000 + 500000 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 340000 \\ 720000 \end{bmatrix}$
$\displaystyle \text{Thus,}$
$\displaystyle \text{Amount spent on } X = \text{Rs } 3400$
$\displaystyle \text{Amount spent on } Y = \text{Rs } 7200$

$\\$

Question 74: A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and the second bond pays 7% interest per year. Using the matrix method, determine how the amount of Rs 30,000 should be divided between the two bonds if the trust fund must earn a total annual interest of (i) Rs 1800 (ii) Rs 2000.

$\displaystyle \text{Answer:}$

$\displaystyle \text{If Rs } x \text{ are invested in the first type of bond and Rs } (30000 - x)$
$\displaystyle \text{are invested in the second type of bond, then the matrix}$
$\displaystyle A = [\, x \quad 30000 - x \,] \text{ represents investment and the matrix}$
$\displaystyle B = \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} \text{ represents rate of interest.}$
$\displaystyle (i)$
$\displaystyle [\, x \quad 30000 - x \,] \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = [\,1800\,]$
$\displaystyle \Rightarrow \left[ \frac{5x}{100} + \frac{7(30000 - x)}{100} \right] = 1800$
$\displaystyle \Rightarrow \frac{5x + 210000 - 7x}{100} = 1800$
$\displaystyle \Rightarrow 210000 - 2x = 180000$
$\displaystyle \Rightarrow 2x = 30000$
$\displaystyle \Rightarrow x = 15000$
$\displaystyle (ii)$
$\displaystyle [x\;\;30000-x] \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = \begin{bmatrix} 2000 \end{bmatrix}$
$\displaystyle \Rightarrow \left[ \frac{5x}{100} + \frac{7(30000-x)}{100} \right] = 2000$
$\displaystyle \Rightarrow \frac{5x + 210000 - 7x}{100} = 2000$
$\displaystyle \Rightarrow 210000 - 2x = 200000$
$\displaystyle \Rightarrow 2x = 10000$
$\displaystyle \Rightarrow x = 5000$
$\displaystyle \text{Thus,}$
$\displaystyle \text{Amount invested in the first bond } = \text{Rs }5000$
$\displaystyle (30000 - 5000) = \text{Rs }25000$

$\\$

$\displaystyle \text{Question 75: }$
$\displaystyle \text{To promote making of toilets for women, an organisation tried to generate awareness through} \\ \text{(i) house calls, (ii) letters, and (iii) announcements.}$
$\displaystyle \text{The cost for each mode per attempt is given as:}$
$\displaystyle \text{(i) } \text{Rs }50 \quad \text{(ii) } \text{Rs }20 \quad \text{(iii) } \text{Rs }40$
$\displaystyle \text{The number of attempts made in three villages } X, Y \text{ and } Z \text{ are given below:}$
$\displaystyle \begin{array}{c|ccc} & \text{(i)} & \text{(ii)} & \text{(iii)} \\ X & 400 & 300 & 100 \\ Y & 300 & 250 & 75 \\ Z & 500 & 400 & 150 \end{array}$
$\displaystyle \text{Find the total cost incurred by the organisation for the three villages separately, using matrices.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{According to the question,}$
$\displaystyle \text{Let } A \text{ be the matrix showing number of attempts made in three villages } X, Y \text{ and } Z.$
$\displaystyle A = \begin{bmatrix} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{bmatrix}$
$\displaystyle \text{And, } B \text{ be a matrix showing the cost for each mode per attempt.}$
$\displaystyle B = \begin{bmatrix} 50 \\ 20 \\ 40 \end{bmatrix}$
$\displaystyle AB = \begin{bmatrix} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{bmatrix} \begin{bmatrix} 50 \\ 20 \\ 40 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 400(50) + 300(20) + 100(40) \\ 300(50) + 250(20) + 75(40) \\ 500(50) + 400(20) + 150(40) \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 20000 + 6000 + 4000 \\ 15000 + 5000 + 3000 \\ 25000 + 8000 + 6000 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 30000 \\ 23000 \\ 39000 \end{bmatrix}$
$\displaystyle \text{Hence, the total cost incurred by the organisation for three villages separately is}$
$\displaystyle X : \text{Rs }30000$
$\displaystyle Y : \text{Rs } 23000$
$\displaystyle Z : \text{Rs }39000$

$\\$

Question 76:There are two families, A and B. Family A has four men, six women, and two children, while family B has two men, two women, and four children. The recommended daily intake is 2400 calories for men, 1900 calories for women, and 1800 calories for children, along with 45 grams of protein for men, 55 grams for women, and 33 grams for children. Represent this information using matrices and, by applying matrix multiplication, calculate the total daily requirements of calories and proteins for each family. From this problem, one can create awareness about the importance of a balanced and planned diet for maintaining good health. [CBSE 2015]

$\displaystyle \text{Answer:}$

$\displaystyle \text{According to the question,}$
$\displaystyle \text{Let } X \text{ be the matrix showing number of family members in family A and B.}$
$\displaystyle X = \begin{bmatrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix} \begin{array}{l} \text{Family A} \\ \text{Family B} \end{array}$
$\displaystyle \text{And, } Y \text{ be a matrix showing the recommended daily amount of calories.}$
$\displaystyle Y = \begin{bmatrix} 2400 \\ 1900 \\ 1800 \end{bmatrix}$
$\displaystyle \text{And, } Z \text{ be a matrix showing the recommended daily amount of proteins.}$
$\displaystyle Z = \begin{bmatrix} 45 \\ 55 \\ 33 \end{bmatrix}$
$\displaystyle \text{Now, the total requirement of calories of the two families will be shown by } XY.$
$\displaystyle XY = \begin{bmatrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix} \begin{bmatrix} 2400 \\ 1900 \\ 1800 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 9600 + 11400 + 3600 \\ 4800 + 3800 + 7200 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 24600 \\ 15800 \end{bmatrix}$
$\displaystyle \text{Also, the total requirement of proteins of the two families will be shown by } XZ.$
$\displaystyle XZ = \begin{bmatrix} 4 & 6 & 2 \\ 2 & 2 & 4 \end{bmatrix} \begin{bmatrix} 45 \\ 55 \\ 33 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 180 + 330 + 66 \\ 90 + 110 + 132 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 576 \\ 332 \end{bmatrix}$
$\displaystyle \text{Hence, the total requirement of calories and proteins for each of the two families is shown as:}$
$\displaystyle \begin{array}{c|c|c} & \text{Calories} & \text{Proteins} \\ \hline \text{Family A} & 24600 & 576 \\ \text{Family B} & 15800 & 332 \end{array}$

$\\$

$\displaystyle \text{Question 77: }$
$\displaystyle \text{In a parliament election, a political party hired a public relations firm to promote its } \\ \text{candidates in three ways - telephone, house calls and letters.}$
$\displaystyle \text{The cost per contact (in paisa) is given in matrix } A \text{ as}$
$\displaystyle A=\begin{bmatrix}140\\200\\150\end{bmatrix}\ \begin{array}{l}\text{Telephone}\\\text{House calls}\\\text{Letters}\end{array}$
$\displaystyle \text{The number of contacts of each type made in two cities } X \text{ and } Y \text{ is given in matrix } B \text{ as}$
$\displaystyle \begin{array}{ccc}\text{Telephone} & \text{House calls} & \text{Letters}\end{array}$
$\displaystyle B=\begin{bmatrix}1000 & 500 & 5000\\3000 & 1000 & 10000\end{bmatrix}\ \begin{array}{l}\text{City }X\\\text{City }Y\end{array}$
$\displaystyle \text{Find the total amount spent by the party in the two cities.}$
$\displaystyle \text{What should one consider before casting his/her vote - party's promotional activity} \\ \text{or their social activities?} \ \ \ \ \ \text{[CBSE 2015]}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{According to the question,}$
$\displaystyle \text{Let } A \text{ be the matrix showing the cost per contact (in paisa).}$
$\displaystyle A = \begin{bmatrix} 140 \\ 200 \\ 150 \end{bmatrix} \begin{array}{l} \text{Telephone} \\ \text{House calls} \\ \text{Letters} \end{array}$
$\displaystyle \text{And, } B \text{ be a matrix showing the number of contacts of each type made in two cities } X \text{ and } Y.$
$\displaystyle \begin{array}{ccc} \text{Telephone} & \text{House calls} & \text{Letters} \end{array}$
$\displaystyle B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{array}{l} \text{City X} \\ \text{City Y} \end{array}$
$\displaystyle \text{Now, the total amount spent by the party in the two cities will be shown by } BA.$
$\displaystyle BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 140 \\ 200 \\ 150 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 140000 + 100000 + 750000 \\ 420000 + 200000 + 1500000 \end{bmatrix}$
$\displaystyle = \begin{bmatrix} 990000 \\ 2120000 \end{bmatrix}$
$\displaystyle \text{Hence, the total amount spent by the party in the two cities is}$
$\displaystyle X : \text{Rs }9900$
$\displaystyle Y : \text{Rs }21200$
$\displaystyle \text{One should consider social activities of a party before casting his/her vote.}$

$\\$

Question 78: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each of them saves Rs. 15,000 per month, find their monthly incomes using the matrix method. This problem reflects the value of saving money for the future and financial planning. [CBSE 2016]

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the monthly incomes of Aryan and Babban be } 3x \text{ and } 4x \text{ respectively.}$
$\displaystyle \text{Suppose their monthly expenditures are } 5y \text{ and } 7y \text{ respectively.}$
$\displaystyle \text{Since each saves Rs } 15000 \text{ per month,}$
$\displaystyle \text{Monthly saving of Aryan: } 3x - 5y = 15000$
$\displaystyle \text{Monthly saving of Babban: } 4x - 7y = 15000$
$\displaystyle \text{The above system of equations can be written in matrix form as follows:}$
$\displaystyle \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}$
$\displaystyle \text{or, } AX = B \text{ where}$
$\displaystyle A = \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix}$
$\displaystyle X = \begin{bmatrix} x \\ y \end{bmatrix}$
$\displaystyle B = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}$
$\displaystyle \text{Now,}$
$\displaystyle |A| = \begin{vmatrix} 3 & -5 \\ 4 & -7 \end{vmatrix} = -21 - (-20) = -1$
$\displaystyle \text{Adj } A =$
$\displaystyle \begin{bmatrix} -7 & -4 \\ 5 & 3 \end{bmatrix}^{T} = \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix}$
$\displaystyle \text{So,}$
$\displaystyle A^{-1} = \frac{1}{|A|}\,\text{adj}\,A = -1 \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}$
$\displaystyle \therefore X = A^{-1}B$
$\displaystyle \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 105000 - 75000 \\ 60000 - 45000 \end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 30000 \\ 15000 \end{bmatrix}$
$\displaystyle \Rightarrow x = 30{,}000 \text{ and } y = 15{,}000$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Monthly income of Aryan } = 3 \times \text{Rs } 30{,}000 = \text{Rs } 90{,}000$
$\displaystyle \text{Monthly income of Babban } = 4 \times \text{Rs } 30{,}000 = \text{Rs } 1{,}20{,}000$
$\displaystyle \text{From this problem, we are encouraged to understand the power of savings.}$
$\displaystyle \text{We should save a certain part of our monthly income for the future.}$

$\\$

Question 79: A trust invested some money in two types of bonds. The first bond pays 10% interest and the second bond pays 12% interest. The trust received Rs. 2800 as interest. However, if the trust had interchanged the money invested in the two bonds, it would have received Rs. 100 less as interest. Using the matrix method, find the amount invested by the trust. [CBSE 2016]

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let Rs } x \text{ be invested in the first bond and Rs } y \text{ be invested in the second bond.}$
$\displaystyle \text{Let } A \text{ be the investment matrix and } B \text{ be the interest per rupee matrix. Then,}$
$\displaystyle A = [\, x \;\; y \,] \text{ and } B = \begin{bmatrix} \frac{10}{100} \\ \frac{12}{100} \end{bmatrix}$
$\displaystyle \text{Total annual interest } = AB = [\, x \;\; y \,] \begin{bmatrix} \frac{10}{100} \\ \frac{12}{100} \end{bmatrix} = \frac{10x}{100} + \frac{12y}{100}$
$\displaystyle \therefore \frac{10x}{100} + \frac{12y}{100} = 2800$
$\displaystyle \Rightarrow 10x + 12y = 280000 \qquad (1)$
$\displaystyle \text{If the rates of interest had been interchanged, then the total interest earned} \\ \text{is Rs 100 less than the previous interest}$
$\displaystyle \therefore \frac{12x}{100} + \frac{10y}{100} = 2700$
$\displaystyle \Rightarrow 12x + 10y = 270000 \qquad (2)$
$\displaystyle \text{The system of equations (1) and (2) can be expressed as}$
$\displaystyle \text{Let Rs } x \text{ be invested in the first bond and Rs } y \text{ be invested in the second bond.}$
$\displaystyle \text{Let } A \text{ be the investment matrix and } B \text{ be the interest per rupee matrix.}$
$\displaystyle A = [\, x \;\; y \,], \quad B = \begin{bmatrix} \frac{10}{100} \\ \frac{12}{100} \end{bmatrix}$
$\displaystyle \text{Total annual interest } = AB = [\, x \;\; y \,] \begin{bmatrix} \frac{10}{100} \\ \frac{12}{100} \end{bmatrix} = \frac{10x}{100} + \frac{12y}{100}$
$\displaystyle \therefore \frac{10x}{100} + \frac{12y}{100} = 2800$
$\displaystyle \Rightarrow 10x + 12y = 280000 \qquad (1)$
$\displaystyle \text{If the rates of interest are interchanged, the total interest is Rs } 100 \text{ less.}$
$\displaystyle \therefore \frac{12x}{100} + \frac{10y}{100} = 2700$
$\displaystyle \Rightarrow 12x + 10y = 270000 \qquad (2)$
$\displaystyle \text{The system of equations (1) and (2) can be expressed as } PX = Q$
$\displaystyle P = \begin{bmatrix} 10 & 12 \\ 12 & 10 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \end{bmatrix}, \quad Q = \begin{bmatrix} 280000 \\ 270000 \end{bmatrix}$
$\displaystyle |P| = \begin{vmatrix} 10 & 12 \\ 12 & 10 \end{vmatrix} = 100 - 144 = -44 \neq 0$
$\displaystyle \text{Thus, } P \text{ is invertible.}$
$\displaystyle \therefore X = P^{-1}Q$
$\displaystyle \Rightarrow X = \frac{1}{|P|} \, \text{adj}\,P \, Q$
$\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{-44} \begin{bmatrix} 10 & -12 \\ -12 & 10 \end{bmatrix} \begin{bmatrix} 280000 \\ 270000 \end{bmatrix}$
$\displaystyle \Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 10000 \\ 15000 \end{bmatrix}$
$\displaystyle \therefore x = 10000 \text{ and } y = 15000$
$\displaystyle \text{Hence, Rs } 10{,}000 \text{ is invested in the first bond and Rs } 15{,}000 \text{ in the second bond.}$