Class 10: Matrices – Exercise 12d – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Find } x \text{ and } y \text{, if } \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}.\begin{bmatrix} 2x \\ 1 \end{bmatrix} +2 \begin{bmatrix} -4 \\ 5 \end{bmatrix}=4 \begin{bmatrix} 2 \\ y \end{bmatrix} \hspace{1.0cm} [2003]$

Answer:

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

$\displaystyle \begin{bmatrix} 6x-2 \\ -2x+4 \end{bmatrix} + \begin{bmatrix} -8 \\ 10 \end{bmatrix}= \begin{bmatrix} 8 \\ 4y \end{bmatrix}$

Therefore

$\displaystyle 6x-10=8 \Rightarrow x = 3$

$\displaystyle \text{ and } -2x+14=4y \Rightarrow y = 2$

$\displaystyle \text{Hence } x = 3 \text{ and } y = 2.$

$\displaystyle \\$

$\displaystyle \text{Question 2: Find } x \text{ and } y \ if \ \begin{bmatrix} 3x & 8 \end{bmatrix} . \begin{bmatrix} 1 & 4 \\ 3 & 7 \end{bmatrix}-3 \begin{bmatrix} 2 & -7 \end{bmatrix} = 5 \begin{bmatrix} 3 & 2y \end{bmatrix}$

Answer:

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

$\displaystyle \begin{bmatrix} 3x+24 & 12x+56 \end{bmatrix} - \begin{bmatrix} 6 & -21 \end{bmatrix} = \begin{bmatrix} 15 & 10y \end{bmatrix}$

$\displaystyle \begin{bmatrix} 3x+18 & 12x+77 \end{bmatrix} = \begin{bmatrix} 15 & 10y \end{bmatrix}$

Therefore

$\displaystyle 3x+18 = 15 \Rightarrow x = -1$

$\displaystyle \text{ and } 12x+77=10y \Rightarrow y = 6.5$

$\displaystyle \\$

$\displaystyle \text{Question 3: If } \begin{bmatrix} x & y \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 25 \end{bmatrix} \text{ and } \begin{bmatrix} -x & y \end{bmatrix} . \begin{bmatrix} 2x \\ y \end{bmatrix}= \begin{bmatrix} -2 \end{bmatrix}$

Find the value of $\displaystyle x \text{ and } y$ if i) $\displaystyle x, y$ are whole numbers ii) $\displaystyle x, \ y$ are integers.

Answer:

$\displaystyle \begin{bmatrix} x & y \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 25 \end{bmatrix}$

$\displaystyle \Rightarrow x^2+y^2 = 25$ … … … … … … i)

$\displaystyle \begin{bmatrix} -x & y \end{bmatrix} . \begin{bmatrix} 2x \\ y \end{bmatrix}= \begin{bmatrix} -2 \end{bmatrix}$

$\displaystyle \Rightarrow -2x^2+y^2 = -2$ … … … … … … ii)

Multiplying i) by 2 and adding it to ii) we get

$\displaystyle y^2 = 16 \Rightarrow y = \pm 4$

If $\displaystyle y = 4, x = 3 \ or\ -3$

Similarly, if $\displaystyle y = -4, x = 3 \ or \ -3$

Hence if $\displaystyle x \text{ and } y$ are whole numbers, then $\displaystyle x = 3, \text{ and } y = 4$ .

If $\displaystyle x \ and\ y$ are Integers, then $\displaystyle x \pm 3 \text{ and } y = \pm 4$ .

$\displaystyle \\$

$\displaystyle \text{Question 4: Given} \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

Find: i) the order of matrix $\displaystyle X$ ii) the matrix $\displaystyle X \hspace{1.0cm} [2012]$

Answer:

$\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

$\displaystyle A_{2 \times 2} . X_{p \times q} = B_{2 \times 1}$

$\displaystyle \text{Therefore } p = 2 \text{ and } q = 1$ . Hence the order of Matrix is $\displaystyle 2 \times 1$

$\displaystyle \text{Let } X = \begin{bmatrix} x \\ y \end{bmatrix}$

Therefore

$\displaystyle \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2x+y \\ -3x+4y \end{bmatrix} = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$

Therefore

$\displaystyle 2x+y = 7$

$\displaystyle \text{ and } -3x+4y = 6$

Solving we get $\displaystyle x = 2 \text{ and } y = 3$

$\displaystyle \text{Hence } X = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 5: Evaluate: }\begin{bmatrix} \cos{45^{\circ}} & \sin{30^{\circ}} \\ \sqrt{2} \cos{0^{\circ}} & \sin{0^{\circ}} \end{bmatrix}. \begin{bmatrix} \sin{45^{\circ}} & \cos{90^{\circ}} \\ \sin{90^{\circ}} & \cot{45^{\circ}} \end{bmatrix}$

Answer:

$\displaystyle \begin{bmatrix} \cos{45^{\circ}} & \sin{30^{\circ}} \\ \sqrt{2} \cos{0^{\circ}} & \sin{0^{\circ}} \end{bmatrix}. \begin{bmatrix} \sin{45^{\circ}} & \cos{90^{\circ}} \\ \sin{90^{\circ}} & \cot{45^{\circ}} \end{bmatrix}$

$\displaystyle = \begin{bmatrix} \frac{1}{\sqrt{2}} &\frac{1}{2} \\ \sqrt{2} & 0 \end{bmatrix}. \begin{bmatrix} \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}+\frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} \\ 1 & 0 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 6: If } A = \begin{bmatrix} 0 & -1 \\ 4 & -3 \end{bmatrix} , B = \begin{bmatrix} -5 \\ 6 \end{bmatrix} \text{ and } 3A \times M = 2B$

Find matrix $\displaystyle M$ .

Answer:

$\displaystyle 3A \times M = 2B$

$\displaystyle A_{2 \times 2} . M_{p \times q} = B_{2 \times 1}$

$\displaystyle \text{Therefore } p = 2 \text{ and } q = 1$ . Hence the order of Matrix is $\displaystyle 2 \times 1$

$\displaystyle \text{Let } M = \begin{bmatrix} x \\ y \end{bmatrix}$

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

$\displaystyle \begin{bmatrix} -3y \\ 12x-9y \end{bmatrix} = \begin{bmatrix} -10 \\ 12 \end{bmatrix}$

$\displaystyle \text{Therefore } -3y = -10 \Rightarrow y = \frac{10}{3}$

$\displaystyle \text{ and } 12x-9y = 12 \Rightarrow x = \frac{7}{2}$

$\displaystyle \text{Hence } M = \begin{bmatrix} \frac{7}{2} \\ \frac{10}{3} \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 7: If } \begin{bmatrix} a & 3 \\ 4 & 1 \end{bmatrix}+\begin{bmatrix} 2 & b \\ 1 & -2 \end{bmatrix}-\begin{bmatrix} 1 & 1 \\ -2 & c \end{bmatrix}=\begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}$

Find the value of $\displaystyle a, b \text{ and } c \hspace{1.0cm} [ 1981]$

Answer:

$\displaystyle \begin{bmatrix} a & 3 \\ 4 & 1 \end{bmatrix}+\begin{bmatrix} 2 & b \\ 1 & -2 \end{bmatrix}-\begin{bmatrix} 1 & 1 \\ -2 & c \end{bmatrix}=\begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}$

$\displaystyle \begin{bmatrix} a+2-1 & 3+b-1 \\ 4+1+2 & 1-2-c \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}$

$\displaystyle \begin{bmatrix} a+1 & b+2 \\ 7 & -c-1 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 7 & 3 \end{bmatrix}$

Therefore

$\displaystyle a+1 = 5 \Rightarrow 4$

$\displaystyle b+2 = 0 \Rightarrow b = -2$

$\displaystyle -c-1=3 \Rightarrow c = -4$

$\displaystyle \\$

$\displaystyle \text{Question 8: If } A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} , \text{ and } B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

Find: i) $\displaystyle A(BA)$ ii) $\displaystyle (AB) B. \hspace{1.0cm} [1991]$

Answer:

i) $\displaystyle A(BA) = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \Bigg( \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \Bigg) =\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}. \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix}$

ii) $\displaystyle (AB) B = \Bigg( \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \Bigg) . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 9: Find } x \text{ and } y \text{, if } \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix} [1992, 2013]$

Answer:

$\displaystyle \begin{bmatrix} x & 3x \\ y & 4y \end{bmatrix}. \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 12 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 2x + 3x \\ 2y+4y \end{bmatrix}. = \begin{bmatrix} 5 \\ 12 \end{bmatrix}$

Therefore

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

$\displaystyle \text{ and } 6y = 12 \Rightarrow y = 2$

$\displaystyle \\$

$\displaystyle \text{Question 10: If the matrix } X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix}. \begin{bmatrix} 2 \\ -2 \end{bmatrix} \text{ and } 2X-3Y=\begin{bmatrix} 10 \\ -8 \end{bmatrix}$ .

Find the matrix $\displaystyle X \text{ and } Y$ .

Answer:

$\displaystyle X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix}. \begin{bmatrix} 2 \\ -2 \end{bmatrix}$

$\displaystyle X = \begin{bmatrix} -14 \\ 10 \end{bmatrix}$

$\displaystyle 2X-3Y=\begin{bmatrix} 10 \\ -8 \end{bmatrix}$

$\displaystyle 2\begin{bmatrix} -14 \\ 10 \end{bmatrix}-3Y=\begin{bmatrix} 10 \\ -8 \end{bmatrix}$

$\displaystyle Y = \frac{1}{3} \begin{bmatrix} -38 \\ 28 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 11: Given } A= \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix} , B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$

Find $\displaystyle X$ such that $\displaystyle A+X=2B+C$ . [2005]

Answer:

$\displaystyle A+X=2B+C$

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

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

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

$\displaystyle X = \begin{bmatrix} -7 & 5 \\ 6 & 2 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 12: Find the value of } x \text{ given that } A^2=B , A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} , \text{ and } \\ \\ B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix} \hspace{1.0cm} [2005]$

Answer:

$\displaystyle A^2=B$

$\displaystyle \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}$

$\displaystyle \text{Therefore } x = 36$

$\displaystyle \\$

$\displaystyle \text{Question 13: If } A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} , \text{ and } B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} \text{ and } I \text{ matrix of the same order } \text{ and } A^t$ is the transpose of the matrix, find $\displaystyle A^t.B+BI$ . [2011]

Answer:

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

$\displaystyle A^t.B+BI$

$\displaystyle = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} . \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}+ \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 14: Given } A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}$ , $\displaystyle B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}$ $\displaystyle \text{ and } C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$

Find the matrix $\displaystyle X$ such that $\displaystyle A+2X=2B+C [2013]$

Answer:

$\displaystyle A+2X=2B+C$

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

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

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

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

$\displaystyle X= \frac{1}{2} \begin{bmatrix} -4 & 10 \\ 6 & 2 \end{bmatrix}$

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

$\displaystyle \\$

$\displaystyle \text{Question 15: Let } A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix} , B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix} , \text{ and } C = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}$

$\displaystyle \text{Find } A^2-A+BC \hspace{1.0cm} [2006]$

Answer:

$\displaystyle A^2-A+BC$

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

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

$\displaystyle = \begin{bmatrix} 2 & -2 \\ -1 & 11 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 16: Let } A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix} , B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$

$\displaystyle \text{Find } A^2+AB+B^2 \hspace{1.0cm} [2007]$

Answer:

$\displaystyle A^2+AB+B^2$

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

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

$\displaystyle =\begin{bmatrix} 4 & 9 \\ 5 & 4 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 17: If } A = \begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix} , B = \begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}, \text{ and } C = \begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix} , \text{ and } \\ \\ 3A-2C=6B \text{ Find the value of } a, b, \text{ and } c$

Answer:

$\displaystyle 3A-2C=6B$

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

$\displaystyle \begin{bmatrix} 9 & 3a \\ -12 & 24 \end{bmatrix} -\begin{bmatrix} -2 & 8 \\ 6 & 2b \end{bmatrix}=\begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 11 & 3a-8 \\ -18 & 24-2b \end{bmatrix} = \begin{bmatrix} 6c & 24 \\ -18 & 0 \end{bmatrix}$

Therefore

$\displaystyle 6c=11 \Rightarrow c = \frac{11}{6}$

$\displaystyle 3a-8 = 24 \Rightarrow a = \frac{32}{3}$

$\displaystyle 24-2b=0 \Rightarrow b = 12$

$\displaystyle \\$

$\displaystyle \text{Question 18: Given } A = \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} , B = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} , \text{ and } C = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} , \text{ and } \\ \\ BA=C^2 \text{ Find the value of } p \text{ and } q \hspace{1.0cm} [2008]$

Answer:

$\displaystyle BA=C^2$

$\displaystyle \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}.\begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}=\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} .\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}$

$\displaystyle \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix}$

Therefore

$\displaystyle -2q = -8 \Rightarrow q = 4$

$\displaystyle \text{ and } p = 8$

$\displaystyle \\$

$\displaystyle \text{Question 19: Given } A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix} , B = \begin{bmatrix} 6 \\ 1 \end{bmatrix} , C = \begin{bmatrix} -4 \\ 5 \end{bmatrix} , \text{ and } D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$

$\displaystyle \text{Find } AB+2C-4D \hspace{1.0cm} [2010]$

Answer:

$\displaystyle AB+2C-4D$

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

$\displaystyle = \begin{bmatrix} 16 \\ -2 \end{bmatrix}+\begin{bmatrix} -8 \\ 10 \end{bmatrix}- \begin{bmatrix} 8 \\ 8 \end{bmatrix}$

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

$\displaystyle \\$

$\displaystyle \text{Question 20: Evaluate } \begin{bmatrix} 4\sin{30^{\circ}} & 2\cos{60^{\circ}} \\ \sin{90^{\circ}} & 2\cos{0^{\circ}} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix} \hspace{1.0cm} [2010]$

Answer:

$\displaystyle \begin{bmatrix} 4\sin{30^{\circ}} & 2\cos{60^{\circ}} \\ \sin{90^{\circ}} & 2\cos{0^{\circ}} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}$

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

$\displaystyle = \begin{bmatrix} 13 & 14 \\ 14 & 13 \end{bmatrix}$

$\displaystyle \\$

$\displaystyle \text{Question 21: If } A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} , I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \text{ Find } A^2-5A+7I \hspace{1.0cm} [2012]$

Answer:

$\displaystyle A^2-5A+7I$

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

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

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

$\displaystyle \\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 10: Matrices – Exercise 12d

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