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ICSE Class X Math 2016 Paper

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Question : 2 of 57
Marks: +1, -0
Given A=[20−17]A=\begin{bmatrix}2 & 0 \\ -1 & 7\end{bmatrix} and I=[1001]I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} and A2=A^2= 9A+mI9A+mI. Find mm.
Solution:  
Given :
A=[20−17]A=\begin{bmatrix}2 & 0 \\ -1 & 7\end{bmatrix} and I=[1001]I=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
  A2=A⋅A=[20−17][20−17]\; A^2= A \cdot A =\begin{bmatrix}2 & 0 \\ -1 & 7\end{bmatrix}\begin{bmatrix}2 & 0 \\ -1 & 7\end{bmatrix}
    =[40−2−70+49]\;\; = \begin{bmatrix}4 & 0 \\ -2-7 & 0+49\end{bmatrix}
    =[40−949]\;\; = \begin{bmatrix}4 & 0 \\ -9 & 49\end{bmatrix}
  9A+mI=9[20−17]+m[1001]\; 9A + mI = 9\begin{bmatrix}2 & 0 \\ -1 & 7\end{bmatrix} + m\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
=[180−963]+[m00m]= \begin{bmatrix}18 & 0 \\ -9 & 63\end{bmatrix} + \begin{bmatrix}m & 0 \\ 0 & m\end{bmatrix}
    =[18+m0−963+m]\;\; = \begin{bmatrix}18+m & 0 \\ -9 & 63+m\end{bmatrix}
Now, According to question,
A2  =9A+mIA^2\; = 9A + mI
[40−949]  =[18+m0−963+m]\begin{bmatrix}4 & 0 \\ -9 & 49\end{bmatrix} \; = \begin{bmatrix}18+m & 0 \\ -9 & 63+m\end{bmatrix}
comparing both sides,
4  =18+m   and   49=63+m4\; = 18+m \; \text{ and } \; 49=63+m
⇒ m  =−14m\; = -14
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