Class 10: Matrices – Exercise 12d – 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: Find $x \ and \ y$ 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:

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

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

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

Therefore

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

and $12x+77=10y \Rightarrow y = 6.5$

$\\$

Question 3: If $\begin{bmatrix} x & y \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix}= \begin{bmatrix} 25 \end{bmatrix}$ and $\begin{bmatrix} -x & y \end{bmatrix} . \begin{bmatrix} 2x \\ y \end{bmatrix}= \begin{bmatrix} -2 \end{bmatrix}$ find the value of $x \ and \ y$ if i) $x, y$ are whole numbers ii) $x, \ y$ are integers.

Answer:

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

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

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

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

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

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

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

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

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

If $x \ and\ y$ are Integers, then $x \pm 3 \ and \ y = \pm 4$ .

$\\$

Question 4: 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 5: Evaluate: $\begin{bmatrix} \cos{45^o} & \sin{30^o} \\ \sqrt{2} \cos{0^o} & \sin{0^o} \end{bmatrix}. \begin{bmatrix} \sin{45^o} & \cos{90^o} \\ \sin{90^o} & \cot{45^o} \end{bmatrix}$

Answer:

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

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

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

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

$\\$

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

Answer:

$3A \times M = 2B$

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

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

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

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

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

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

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

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

$\\$

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 $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 8: 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 9: 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 10: If the matrix $X = \begin{bmatrix} -3 & 4 \\ 2 & -3 \end{bmatrix}. \begin{bmatrix} 2 \\ -2 \end{bmatrix}$ and $2X-3Y=\begin{bmatrix} 10 \\ -8 \end{bmatrix}$ . Find the matrix $X \ and \ Y$ .

Answer:

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

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

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

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

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

$\\$

Question 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: If $A = \begin{bmatrix} 3 & a \\ -4 & 8 \end{bmatrix}$ , $B = \begin{bmatrix} c & 4 \\ -3 & 0 \end{bmatrix}$ and $C = \begin{bmatrix} -1 & 4 \\ 3 & b \end{bmatrix}$ and $3A-2C=6B$ , find the value of $a, b, \ and \ c$ .

Answer:

$3A-2C=6B$

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

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

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

Therefore

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

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

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

$\\$

Question 18: 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 19: Given $A = \begin{bmatrix} 3 & -2 \\ -1 & 4 \end{bmatrix}$ , $B = \begin{bmatrix} 6 \\ 1 \end{bmatrix}$ , $C = \begin{bmatrix} -4 \\ 5 \end{bmatrix}$ 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 20: 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 21: 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:

$A^2-5A+7I$

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

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

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

$\\$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 10: Matrices – Exercise 12d

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