Class 12: Algebra of Matrices

$\displaystyle \text{Question 1: Compute the following sums:}$

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

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

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

$\displaystyle \text{(ii) }$

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

$\\$

$\displaystyle \text{Question 2: Let } A=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \text{ and } C= \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}. \text{Find each of the following:}$

$\displaystyle \text{(i) } 2A-3B$ $\displaystyle \text{(ii) } B-4C$ $\displaystyle \text{(iii) } 3A-C$ $\displaystyle \text{(iv) } 3A-2B+3C$

$\displaystyle \text{Answer:}$

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

$\displaystyle = \begin{bmatrix} 4 & 8 \\ 6 & 4 \end{bmatrix} - \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4-3 & 8-9 \\ 6+6 & 4-15 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & -1 \\ 12 & -11 \end{bmatrix}$

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

$\displaystyle = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} - \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1+8 & 3-20 \\ -2-12 & 5- 16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix}$

$\displaystyle \text{(iii) } 3A-C = 3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$

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

$\displaystyle = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}$

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

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

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

$\displaystyle = \begin{bmatrix} -2 & 21 \\ 22 & 8 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 3: Let } A=\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}, B = \begin{bmatrix} -1 & 0 & 2 \\ 3 & 4 & 1 \end{bmatrix} \text{ and } C= \begin{bmatrix} -1 & 2 & 3 \\ 2 & 1 & 0 \end{bmatrix}. \text{Find each of the following:}$

$\displaystyle \text{(i) } A+ B \text{ and } B + C$

$\displaystyle \text{(ii) } 2B+ 3A \text{ and } 3C -4B$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle A+B = \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} + \begin{bmatrix} -1 & 0 & 2 \\ 3 & 4 & 1 \end{bmatrix}$

$\displaystyle A+B \text{ is not possible because matrix } A \text{ is an order of } 2 \times 2 \text{ and Matrix } B \text{ is an order of } \\ 2 \times 3, \text{ So the Sum of the matrix is only possible when their order is same. }$

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

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

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

$\displaystyle \text{(ii) }$

$\displaystyle 2B+ 3A = 2 \begin{bmatrix} -1 & 0 & 2 \\ 3 & 4 & 1 \end{bmatrix} + 3 \begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$

$\displaystyle 2B+3A \text{ also does not exist because the order of matrix B and matrix A is different, }\\ \text{So we can not find the sum of these matrix.}$

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

$\displaystyle = \begin{bmatrix} -3 & 6 & 9 \\ 6 & 3 & 0 \end{bmatrix} - \begin{bmatrix} -4 & 0 & 8 \\ 12 & 16 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -3+4 & 6-0 & 9-8 \\ 6-12 & 3-16 & 0-4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 6 & 1 \\ -6 & -13 & -4 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 4: Let } A=\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix}, B = \begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} \text{ and } C= \begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix}. \\ \text{Compute } 2A-3B+4C.$

$\displaystyle \text{Answer:}$

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

$\displaystyle = \begin{bmatrix} -2 & 0 & 4 \\ 6 & 2 & 8 \end{bmatrix} - \begin{bmatrix} 0 & -6 & 15 \\ 3 & -9 & 3 \end{bmatrix} + \begin{bmatrix} 4 & -20 & 8 \\ 24 & 0 & -16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -2+0 + 4 & 0+6-20 & 4-15+8 \\ 6-3+24 & 2+9-0 & 8 -3-16 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 5: If } A = \text{diag}( 2 \ \ -5 \ \ 9 ), \ B=\text{diag}(1 \ \ 1 \ \ -4) \text{ and } \\ C = \text{diag}(-6 \ \ 3 \ \ 4 ) \text{ find}$

$\displaystyle \text{(i) } A-2B$

$\displaystyle \text{(ii) } B+C-2A$

$\displaystyle \text{(iii) } 2A+3B-5C$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } A = \text{diag}( 2 \ \ -5 \ \ 9 ), \ B=\text{diag}(1 \ \ 1 \ \ -4) \text{ and } C = \text{diag}(-6 \ \ 3 \ \ 4 )$

$\displaystyle \Rightarrow A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9 \end{bmatrix} \ \ \ \ \ \ B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix} \ \ \ \ \ C = \begin{bmatrix} -6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$\displaystyle \text{(i) }$

$\displaystyle A-2B = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -8 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 2-2 & 0 & 0 \\ 0 & -5-2 & 0 \\ 0 & 0 & 9+8 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 & 0 \\ 0 & -7 & 0 \\ 0 & 0 & 17 \end{bmatrix}$

$\displaystyle \text{(ii) }$

$\displaystyle B+C-2A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix} +\begin{bmatrix} -6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} -2 \begin{bmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix} +\begin{bmatrix} -6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 18 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1-6-4 & 0 & 0 \\ 0 & 1+3+10 & 0 \\ 0 & 0 & -4+4-18 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -9 & 0 & 0 \\ 0 & 14 & 0 \\ 0 & 0 & -18 \end{bmatrix}$

$\displaystyle \text{(iii) }$

$\displaystyle 2A+3B-5C = 2 \begin{bmatrix} 2 & 0 & 0 \\ 0 & -5 & 0 \\ 0 & 0 & 9 \end{bmatrix}+ 3 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -4 \end{bmatrix}-5\begin{bmatrix} -6 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 & 0 & 0 \\ 0 & -10 & 0 \\ 0 & 0 & 18 \end{bmatrix}+ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -12 \end{bmatrix}-\begin{bmatrix} -30 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & 20 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4+3 +30 & 0 & 0 \\ 0 & -10+3-15 & 0 \\ 0 & 0 & 18-12-20 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 37 & 0 & 0 \\ 0 & -22 & 0 \\ 0 & 0 & -14 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 6: Given the matrices}$

$\displaystyle A=\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \end{bmatrix}, B = \begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \end{bmatrix} \text{ and } C= \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \end{bmatrix}$

$\displaystyle \text{Verify that } (A+B) + C = A + (B+C)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given } A=\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \end{bmatrix}, B = \begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \end{bmatrix} \text{ and } C= \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \end{bmatrix}$

$\displaystyle \text{LHS} = (A+B) + C$

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

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

$\displaystyle = \begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 13 & 4 & 3\\ 7 & 3 & 4 \\ 11 & 7 & 15 \end{bmatrix}$

$\displaystyle \text{RHS} = (A+B) + C$

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

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

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

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

$\displaystyle = \begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \end{bmatrix}$

$\displaystyle \text{Therefore, LHS = RHS. Hence proved. }$

$\\$

$\displaystyle \text{Question 7: Find matrices } X \text{ and } Y \text{ if } X+Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} \text{ and } X-Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } X+Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} \text{ and } X-Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}$

$\displaystyle \Rightarrow 2X = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 5+ 3 & 2+ 6 \\ 0 & 9-1 \end{bmatrix}= \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}$

$\displaystyle \Rightarrow X = \frac{1}{2} \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}$

$\displaystyle \Rightarrow Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - X = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 5-4 & 2-4 \\ 0 & 9-4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 8: Find } X \text{ if } Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \text{ and } 2X+Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 2X+Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$

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

$\displaystyle \Rightarrow X = \frac{1}{2} \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 9: Find matrices } X \text{ and } Y, \text{ if } 2X-Y = \begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1 \end{bmatrix} \text{ and } \\ X+2Y = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & - 7 \end{bmatrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 2X-Y = \begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1 \end{bmatrix} \text{ and } X+2Y = \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & - 7 \end{bmatrix}$

$\displaystyle \Rightarrow 4X-2Y = 2 \begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2 \end{bmatrix}$

$\displaystyle \Rightarrow 5X = \begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 2 & 5 \\ -2 & 1 & - 7 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 12+3 & -12+2 & 0+5\\ -8-2 & 4+1 & 2-7 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 15 & -10 & 5\\ -10 & 2 & -5 \end{bmatrix}$

$\displaystyle \Rightarrow X = \frac{1}{5} \begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5 \end{bmatrix}= \begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 10: If } X-Y = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \text{ and } X+Y = \begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix}, \text{ Find } X \text{ and } Y.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } X-Y = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \text{ and } X+Y = \begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow 2X = \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix} = \begin{bmatrix} 1+3 & 1+5 & 1+1\\ 1-1 & 1+1 & 0+4 \\ 1+11 & 0+8 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow X = \frac{1}{2} \begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{bmatrix}= \begin{bmatrix} 2 & 3 & 1\\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{bmatrix}$

$\displaystyle \text{Now, } X+Y = \begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix}$

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

$\\$

$\displaystyle \text{Question 11: Find matrix } A, \text{ if } \begin{bmatrix} 1 & 2 & -1\\ 0 & 4 & 9 \end{bmatrix} + A = \begin{bmatrix} 9 & -1 & 4\\ -2 & 1 & 3 \end{bmatrix}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } \begin{bmatrix} 1 & 2 & -1\\ 0 & 4 & 9 \end{bmatrix} + A = \begin{bmatrix} 9 & -1 & 4\\ -2 & 1 & 3 \end{bmatrix}$

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

$\\$

$\displaystyle \text{Question 12: If } A=\begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix}, \text{ and } B = \begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix} \text{ find matrix } C \text{ such that } \\ 5A+3B + 2C \text{ is a null matrix. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 5A+3B + 2C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

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

$\displaystyle \Rightarrow 2C = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - 5\begin{bmatrix} 9 & 1 \\ 7 & 8 \end{bmatrix} - 3\begin{bmatrix} 1 & 5 \\ 7 & 12 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 45 & 5 \\ 35 & 40 \end{bmatrix} - \begin{bmatrix} 3 & 15 \\ 21 & 36 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 0-45-3 & 0-5-15 \\ 0-35-21 & 0-40-36 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} -48 & -20 \\ -56 & -76 \end{bmatrix}$

$\displaystyle \Rightarrow C= \frac{1}{2} \begin{bmatrix} -48 & -20 \\ -56 & -76 \end{bmatrix} = \begin{bmatrix} -24 & -10 \\ -28 & -38 \end{bmatrix}$

$\\$

$\displaystyle \text{Question 13: If } A=\begin{bmatrix} 2 & -2 \\ 4 & 2 \\-5 & 1 \end{bmatrix}, \text{ and } B = \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix}, \text{ find matrix } X \text{ such that } \\ 2A+3X=5B.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 2A+3X=5B$

$\displaystyle \Rightarrow 3X = 5B - 2A$

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

$\displaystyle = \begin{bmatrix} 40 & 0 \\ 20 & -10 \\ 15 & 30 \end{bmatrix} - \begin{bmatrix} 4 & -4 \\ 8 & 4 \\-10 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 40-4 & 0+4 \\ 20-8 & -10-4 \\ 15+10 & 30-2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 36 & 4 \\ 12 & -14 \\ 25 & 28 \end{bmatrix}$

$\displaystyle \Rightarrow X = \frac{1}{3} \begin{bmatrix} 36 & 4 \\ 12 & -14 \\ 25 & 28 \end{bmatrix} = \begin{bmatrix} 12 & \frac{4}{3} \\ & \\ 4 & -\frac{14}{3} \\ & \\ \frac{25}{3} & \frac{28}{3} \end{bmatrix}$

$\\$

$\displaystyle \text{Question 14: If } A=\begin{bmatrix} 1 & -3 & 2 \\ 2 & 0 & 2 \end{bmatrix}, \text{ and } B = \begin{bmatrix} 2 & -1 & - 1 \\ 1 & 0 & -1 \end{bmatrix} \text{ find matrix } C \text{ such that } \\ A+B+C \text{ is zero matrix. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } A+B+C = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

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

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

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

$\\$

$\displaystyle \text{Question 15: Find } x, y \text{ satisfying the matrix equations }$

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

$\displaystyle \text{(ii) } \begin{bmatrix} x & y+2 & z-3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$

$\displaystyle \text{(iii) } x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = 0$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Given, } \begin{bmatrix} x-y & 2 & - 2 \\ 4 & x & 6 \end{bmatrix} + \begin{bmatrix} 3 & -2 & 2 \\ 1 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 5 & 2x+y & 5 \end{bmatrix}$

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

$\displaystyle \Rightarrow \begin{bmatrix} x-y+3 & 0 & 0 \\ 5 & x & 5 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 5 & 2x+y & 5 \end{bmatrix}$

$\displaystyle \Rightarrow x-y+3 = 6 \Rightarrow x-y = 3 \ldots \ldots (i)$

$\displaystyle \Rightarrow x = 2x+y \Rightarrow x = - y \ldots \ldots (ii)$

$\displaystyle \text{From (i) and (ii) we get }$

$\displaystyle -2y = 3 \Rightarrow y = -\frac{3}{2}$

$\displaystyle \Rightarrow x = \frac{3}{2}$

$\displaystyle \text{Hence, } x = \frac{3}{2} \text{ and } y = -\frac{3}{2}$

$\displaystyle \text{(ii) }$

$\displaystyle \text{Given, } \begin{bmatrix} x & y+2 & z-3 \end{bmatrix} + \begin{bmatrix} y & 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} x+y & y+2+4 & z-3+5 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} x+y & y+6 & z+2 \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12 \end{bmatrix}$

$\displaystyle \Rightarrow y+6 = 9 \Rightarrow y = 3$

$\displaystyle \Rightarrow z+2 = 12 \Rightarrow z = 10$

$\displaystyle \Rightarrow x+y = 4 \Rightarrow x + 3 = 4 \Rightarrow x = 1$

$\displaystyle \text{Therefore, } x = 1, y = 3, \text{ and } z = 10$

$\displaystyle \text{(iii) }$

$\displaystyle \text{Given, } x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y \begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = 0$

$\displaystyle \Rightarrow \begin{bmatrix} 2x \\ x \end{bmatrix} + \begin{bmatrix} 3y \\ 5y \end{bmatrix} = \begin{bmatrix} 8 \\ 11 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x+ 3y \\ x+5y \end{bmatrix} = \begin{bmatrix} 8 \\ -11 \end{bmatrix}$

$\displaystyle \Rightarrow 2x+3y = 8 \ldots \ldots (i)$

$\displaystyle \Rightarrow x+5y = 11 \ldots \ldots (ii)$

$\displaystyle \text{Solving (i) and (ii) we get } x = 1 \text{ and } y = 2$

$\\$

$\displaystyle \text{Question 16: If } 2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix}+ \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}, \text{Find } x \text{ and } y$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix}+ \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 6 & 8 \\ 10 & 2x \end{bmatrix}+ \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 6+1 & 8+y \\ 10 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 7 & 8+y \\ 10 & 2x+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}$

$\displaystyle \Rightarrow 8+y = 0 \Rightarrow y = -8$

$\displaystyle \Rightarrow 2x+1 = 5 \Rightarrow x = 2$

$\displaystyle \text{Therefore we get } x = 2 \text{ and } y = -8$

$\\$

$\displaystyle \text{Question 17: Find the value of } \lambda \text{ a non-zero scalar, if } \\ \\ \lambda \begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \end{bmatrix} + 2\begin{bmatrix} 1 & 2 & 3 \\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{bmatrix}.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given} \lambda \begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \end{bmatrix} + 2\begin{bmatrix} 1 & 2 & 3 \\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} \lambda & 0 & 2\lambda \\ 3\lambda & 4\lambda & 5\lambda \end{bmatrix} + \begin{bmatrix} 2 & 4 & 6 \\ -2 & -6 & 4 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} \lambda & 0 & 2\lambda \\ 3\lambda & 4\lambda & 5\lambda \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10 \\ 4 & 2 & 14 \end{bmatrix} - \begin{bmatrix} 2 & 4 & 6 \\ -2 & -6 & 4 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} \lambda & 0 & 2\lambda \\ 3\lambda & 4\lambda & 5\lambda \end{bmatrix} = \begin{bmatrix} 4-2 & 4-4 & 10-6 \\ 4+2 & 2+6 & 14-4 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} \lambda & 0 & 2\lambda \\ 3\lambda & 4\lambda & 5\lambda \end{bmatrix} = \begin{bmatrix} 2 & 0 & 4 \\ 6 & 8 & 10 \end{bmatrix}$

$\displaystyle \Rightarrow \lambda = 2$

$\\$

$\displaystyle \text{Question 18: }$

$\displaystyle \text{(i) Find a matrix } A \text{ such that } 2A+B + X = O, \text{ where } \\ \\ A=\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}, B = \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} {\hspace{7.0cm} \text{[CBSE 2000]} }$

$\displaystyle \text{(ii) If } A=\begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix}, \text{ and } B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} \text{ then find the matrix } X \text{ of order } \\ \\ 3 \times 2 \text{ such that } 2A+3X=5B.$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Given} 2A+B + X = O$

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

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

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

$\displaystyle \Rightarrow \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} + X = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

$\displaystyle \Rightarrow X = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix}$

$\displaystyle \Rightarrow X = \begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}$

$\displaystyle \text{(ii) }$

$\displaystyle \text{Given} 2A+3X=5B$

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

$\displaystyle \Rightarrow \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}+3X=\begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix}$

$\displaystyle \Rightarrow 3X=\begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}$

$\displaystyle \Rightarrow 3X=\begin{bmatrix} 10-16 & -10 \\ 20-8 & 10+4 \\ -25-6 & 5-12 \end{bmatrix}$

$\displaystyle \Rightarrow 3X=\begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}$

$\displaystyle \Rightarrow X=\begin{bmatrix} -2 & -\frac{10}{3} \\ & \\ 4 & \frac{14}{3} \\ & \\ -\frac{31}{3} & -\frac{7}{3} \end{bmatrix}$

$\\$

$\displaystyle \text{Question 19: Find } x, y , z, \text{ and } t \text{ if }$

$\displaystyle \text{(i) } 3\begin{bmatrix} x & y \\ z & t \end{bmatrix}= \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix}$

$\displaystyle \text{(ii) } 2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix}+ \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ 15 & 14 \end{bmatrix} {\hspace{5.0cm} \text{[CBSE 2002, 2012]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{Given } 3\begin{bmatrix} x & y \\ z & t \end{bmatrix}= \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix}$

$\displaystyle \text{Given } \begin{bmatrix} 3x & 3y \\ 3z & 3t \end{bmatrix}= \begin{bmatrix} x+4 & 6+x+y \\ -1+z+t & 2t+3 \end{bmatrix}$

$\displaystyle \Rightarrow 3t = 2t + 3 \ \Rightarrow t = 3$

$\displaystyle \Rightarrow 3z = -1+z+t \ \Rightarrow 2z = - 1 + 3 \ \Rightarrow 2z = 2 \ \Rightarrow z = 1$

$\displaystyle \Rightarrow 3x = x + 4 \ \Rightarrow 2x = 4 \ \Rightarrow x = 2$

$\displaystyle \Rightarrow 3y = 6 + x + y \ \Rightarrow 2y = 6 + 2 \ \Rightarrow 2y = 8 \ \Rightarrow y = 4$

$\displaystyle \text{Hence, } x = 2, \ y = 4, \ z= 1 \text{ and } t = 3$

$\displaystyle \text{(ii) }$

$\displaystyle \text{Given } 2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix}+ \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ 15 & 14 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ 15 & 14 \end{bmatrix} - \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} = \begin{bmatrix} 7-3 & 14-4 \\ 15-1 & 14-2 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 2x & 10 \\ 14 & 2y-6 \end{bmatrix} = \begin{bmatrix} 4 & 10 \\ 14 & 12 \end{bmatrix}$

$\displaystyle \Rightarrow 2x = 4 \ \Rightarrow x = 2$

$\displaystyle \Rightarrow 2y-6 = 12 \ \Rightarrow y = 9$

$\displaystyle \text{Hence, } x = 2, \ y = 9$

$\\$

$\displaystyle \text{Question 20: If } X \text{ and } Y \text{ are } 2 \times 2 \text{ matrix, then solve the following matrix equations} \\ \text{for } X \text{ and } Y.$

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

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } \\ \\ 2X+3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \ldots \ldots (i) \\ \\ \ \ \ 3X+2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} \ldots \ldots (ii)$

$\displaystyle \text{Multiplying (i) by 3 and (ii) by 2 and subtracting (ii) from (i) we get }$

$\displaystyle \Rightarrow 9Y - 4Y = 3 \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - 2 \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}$

$\displaystyle \Rightarrow 5Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} - \begin{bmatrix} -4 & 4 \\ 2 & -10 \end{bmatrix}$

$\displaystyle \Rightarrow 5Y = \begin{bmatrix} 6+4 & 9-4 \\ 12-2 & 0+10 \end{bmatrix}$

$\displaystyle \Rightarrow 5Y = \begin{bmatrix} 10 & 5 \\ 10 & 10 \end{bmatrix}$

$\displaystyle \Rightarrow Y = \frac{1}{5} \begin{bmatrix} 10 & 5 \\ 10 & 10 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}$

$\displaystyle \text{Substituting in (i) we get}$

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

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

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

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

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

$\\$

Question 21: In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind in all the colleges.

$\displaystyle \text{Answer:}$

$\displaystyle \text{Number of posts in any college is given by: } X = \begin{bmatrix} 15 \\ 6 \\ 1 \\ 1 \end{bmatrix}$

$\displaystyle \text{Total number of posts in each kind of college is } = 30X = 30 \begin{bmatrix} 15 \\ 6 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 450 \\ 180 \\ 30 \\ 30 \end{bmatrix}$

$\\$

Question 22: 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 saves Rs. 15,000 per month, find their monthly incomes using matrix method. This problem reflects which value? [CBSE 2016]

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let monthly incomes of Aryan and Babban be Rs. } 3x \text{ and Rs. } 4x \text{ while their monthly}\\ \text{expenditure be Rs. } 5y \text{ and Rs. } 7y \text{ respectively. }$

$\displaystyle \text{Given, }$

$\displaystyle 3x - 5y = 15000 \ldots \ldots (i)$

$\displaystyle 4x - 7y = 15000 \ldots \ldots (ii)$

$\displaystyle \text{Above system of linear equations (i) and (ii) may be written in matrix form as follows } \\ AX = B \text{ where}$

$\displaystyle A = \begin{bmatrix} 3 & -5 \\ 4 & -7 \end{bmatrix}, \ \ X = \begin{bmatrix} x \\ y \end{bmatrix} \text{ and } Y = \begin{bmatrix} 15000 \\ 15000 \end{bmatrix}$

$\displaystyle \because AX = B \Rightarrow X = A^{-1} B \ldots \ldots (iii)$

$\displaystyle \text{Now, } |A| = - 21 + 20 = -1$

$\displaystyle \text{adj } A = \begin{bmatrix} -7 & -4 \\ 5 & 3 \end{bmatrix}^T = \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix}$

$\displaystyle \therefore A^{-1} = \frac{1}{-1} \begin{bmatrix} -7 & 5 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix}$

$\displaystyle X = \begin{bmatrix} 7 & -5 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 15000 \\ 15000 \end{bmatrix} = \begin{bmatrix} 105000 -75000 \\ 60000 -45000 \end{bmatrix} = \begin{bmatrix} 30000 \\ 15000 \end{bmatrix}$

$\displaystyle x=30000, \ \ y = 15000$

$\displaystyle \text{Hence monthly income of Aryan = Rs. } 90,000 \\ \text{and monthly income of Babban = Rs. } 1,20,000$

$\\$

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Class 12: Algebra of Matrices – Exercise 5.2

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