Class 10: Matrices – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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

Answer:

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

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

Therefore

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

and $-2x+14=4y \Rightarrow y = 2$

Hence $x = 3 \ and \ y = 2.$

$\\$

Question 2: Given $\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix} . X = \begin{bmatrix} 7 \\ 6 \end{bmatrix}$ , find: i) the order of matrix $X$ ii) the matrix $X$ [2012]

Answer:

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

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

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

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

Therefore

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

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

Therefore

$2x+y = 7$

and $-3x+4y = 6$

Solving we get $x = 2 \ and \ y = 3$

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

$\\$

Question 3: 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 $a, b \ and c \$ . [1981]

Answer:

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

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

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

Therefore

$a+1 = 5 \Rightarrow 4$

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

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

$\\$

Question 4: If $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ , and $B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ find: i) $A(BA)$ ii) $(AB) B.$ [1991]

Answer:

i) $A(BA)$

$= \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} [ \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}]$

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

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

ii) $(AB) B$

$= (\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}) . \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$

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

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

$\\$

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

Answer:

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

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

Therefore

$5x=5 \Rightarrow x = 1$

and $6y = 12 \Rightarrow y = 2$

$\\$

Question 6: Given $A = \begin{bmatrix} 2 & -1 \\ 2 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$ ; Find $X$ such that $A+X=2B+C$ . [2005]

Answer:

$A+X=2B+C$

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

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

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

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

$\\$

Question 7: Find the value of $x$ given that $A^2=B$ , $A = \begin{bmatrix} 2 & 12 \\ 0 & 1 \end{bmatrix}$ , and $B = \begin{bmatrix} 4 & x \\ 0 & 1 \end{bmatrix}$ . [2005]

Answer:

$A^2=B$

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

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

Therefore $x = 36$

$\\$

Question 8: If $A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$ , and $B = \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}$ and $I$ matrix of the same order and $A^t$ is the transpose of the matrix, find $A^t.B+BI$ . [2011]

Answer:

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

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

$A^t.B+BI$

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

$= \begin{bmatrix} 7 & -1 \\ 17 & -1 \end{bmatrix} + \begin{bmatrix} 4 & -2 \\ -1 & 3 \end{bmatrix}$

$= \begin{bmatrix} 11 & -3 \\ 16 & 2 \end{bmatrix}$

$\\$

Question 9: Given $A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}$ , $B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$ ; Find the matrix $X$ such that $A+2X=2B+C$ . [2013]

Answer:

$A+2X=2B+C$

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

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

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

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

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

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

$\\$

Question 10: Let $A = \begin{bmatrix} 4 & -2 \\ 6 & -3 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & 2 \\ 1 & -1 \end{bmatrix}$ and $C = \begin{bmatrix} -2 & 3 \\ 1 & -1 \end{bmatrix}$ . Find $A^2-A+BC$ [2006]

Answer:

$A^2-A+BC$

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

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

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

$\\$

Question 11: Let $A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ , $B = \begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}$ . Find $A^2+AB+B^2$ [2007]

Answer:

$A^2+AB+B^2$

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

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

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

$\\$

Question 12: Given $A = \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}$ and $BA=C^2$ . FInd the value of $p \ and \ q$ . [2008]

Answer:

$BA=C^2$

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

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

Therefore

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

and $p = 8$

$\\$

Question 13: Given $A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}$ , $B = \begin{bmatrix} 6 \\ 1 \end{bmatrix}$ , $C = \begin{bmatrix} -4 \\ 5 \end{bmatrix}$ and $D = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$ . Find $AB+2C-4D$ . [2010]

Answer:

$AB+2C-4D$

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

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

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

$\\$

Question 14: Evaluate $\begin{bmatrix} 4\sin{30^o} & 2\cos{60^o} \\ \sin{90^o} & 2\cos{0^o} \end{bmatrix} . \begin{bmatrix} 4 & 5 \\ 5 & 4 \end{bmatrix}$ [2010]

Answer:

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

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

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

$\\$

Question 15: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ , $I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ , find $A^2-5A+7I$ . [2012]

Answer:

$A^2-5A+7I$

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

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

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

$\\$

Question 16: If $A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \ and \ C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix}$ . Find $A^2 + AC-5B$ . [2014]

Answer:

$A^2 + AC-5B$

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

$= \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 14 \end{bmatrix}$

$\\$

Question 17: Solve for $x \ and \ y$

i) $\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}$

ii) $\begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}$

iii) $\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} +3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}$ [2014]

Answer:

i) $\begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 2x+5y & 5x+2y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}$

Therefore

$2x+5y = -7$

$5x+2y=14$

Solving the above two equations we get

$x = 4$ and $y = -3$

ii) $\begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x-y-8 & -2y-8 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}$

Therefore

$x-y-8=-7 \Rightarrow x-y=1$

Also $-2y-8=-11 \Rightarrow y = \frac{3}{2}$

Substituting we get $x = \frac{3}{2}+1 = \frac{5}{2}$

iii) $\begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} +3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}$

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

$\begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}$

Therefore $y = -2 and x = \frac{6}{2}=3$

$\\$

Question 18: If $A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 4 \end{bmatrix}$ , is the product of $AB$ possible. [2011]

Answer:

The order of matrix $A = 2 \times 2$ and the order of matrix $B \ is \ 2 \times 1$ .

Since the number of columns in $A$ is equal to the number of rows in $B$ , the product $AB$ is possible.

$AB = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} . \begin{bmatrix} 4 \\ 2 \end{bmatrix} = \begin{bmatrix} 6+20 \\ 8-8 \end{bmatrix} = \begin{bmatrix} 26 \\ 0 \end{bmatrix}$

$\\$

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