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Given A = [20−17]\begin{bmatrix} 2 & 0 \ -1 & 7 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} and A2 = 9A + mI. Find m.

Matrices

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Answer

Given,

A = [20−17]\begin{bmatrix} 2 & 0 \ -1 & 7 \end{bmatrix} and I = [1001]\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}

⇒A2=[20−17]×[20−17]=[(2)(2)+(0)(−1)(2)(0)+(0)(7)(−1)(2)+(7)(−1)(−1)(0)+(7)(7)]=[4+00+0−2−70+49]=[40−949].\Rightarrow A^2 = \begin{bmatrix} 2 & 0 \ -1 & 7 \end{bmatrix} \times \begin{bmatrix} 2 & 0 \ -1 & 7 \end{bmatrix} \\[1em] = \begin{bmatrix} (2)(2) + (0)(-1) & (2)(0) + (0)(7) \ (-1)(2) + (7)(-1) & (-1)(0) + (7)(7) \end{bmatrix} \\[1em] = \begin{bmatrix} 4 + 0 & 0 + 0 \ -2 - 7 & 0 + 49 \end{bmatrix} \\[1em] = \begin{bmatrix} 4 & 0 \ -9 & 49 \end{bmatrix}.

Solving for A2 = 9A + mI:

⇒[40−949]=9[20−17]+m[1001]⇒[40−949]=[180−963]+m[1001]⇒[40−949]−[180−963]=[m00m]⇒[4−180−0−9−(−9)49−63]=[m00m]⇒[−1400−14]=[m00m].\Rightarrow \begin{bmatrix} 4 & 0 \ -9 & 49 \end{bmatrix} = 9\begin{bmatrix} 2 & 0 \ -1 & 7 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 0 \ -9 & 49 \end{bmatrix} = \begin{bmatrix} 18 & 0 \ -9 & 63 \end{bmatrix} + m \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 & 0 \ -9 & 49 \end{bmatrix} - \begin{bmatrix} 18 & 0 \ -9 & 63 \end{bmatrix} = \begin{bmatrix} m & 0 \ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 4 - 18 & 0 - 0 \ -9 - (-9) & 49 - 63 \end{bmatrix} = \begin{bmatrix} m & 0 \ 0 & m \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} -14 & 0 \ 0 & -14 \end{bmatrix} = \begin{bmatrix} m & 0 \ 0 & m \end{bmatrix}.

∴ m = -14.

Hence, m = -14.

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