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If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right], I=\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$ and $A^{2}=8 A+k I$, then the value of $K$ is
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1052 Upvotes
Verified Answer
The correct answer is:
$-7$
$$
A^{2}=A \times A=\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-8 & 49
\end{array}\right]
$$
Given $A^{2}=8 A+K I$
$$
\begin{aligned}
&=8\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]+\mathrm{k}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
8 & 0 \\
-8 & 56
\end{array}\right]+\left[\begin{array}{cc}
\mathrm{k} & 0 \\
0 & \mathrm{k}
\end{array}\right] \\
\therefore \quad\left[\begin{array}{cc}
1 & 0 \\
-8 & 49
\end{array}\right] &=\left[\begin{array}{cc}
8+\mathrm{k} & 0 \\
-8 & 56+\mathrm{k}
\end{array}\right] \Rightarrow 8+\mathrm{k}=1 \Rightarrow \mathrm{k}=-7
\end{aligned}
$$
A^{2}=A \times A=\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-8 & 49
\end{array}\right]
$$
Given $A^{2}=8 A+K I$
$$
\begin{aligned}
&=8\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]+\mathrm{k}\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
8 & 0 \\
-8 & 56
\end{array}\right]+\left[\begin{array}{cc}
\mathrm{k} & 0 \\
0 & \mathrm{k}
\end{array}\right] \\
\therefore \quad\left[\begin{array}{cc}
1 & 0 \\
-8 & 49
\end{array}\right] &=\left[\begin{array}{cc}
8+\mathrm{k} & 0 \\
-8 & 56+\mathrm{k}
\end{array}\right] \Rightarrow 8+\mathrm{k}=1 \Rightarrow \mathrm{k}=-7
\end{aligned}
$$
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