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Question:

If $A=\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]$, find $k$ such that $A^{2}-8 A+k i=0$

Solution:

Given: $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$

Now,

$A^{2}=A A$

$\Rightarrow A^{2}=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}1-0 & 0+0 \\ -1-7 & 0+49\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]$

$A^{2}-8 A+k I=0$

$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]-8\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]+k\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{cc}1 & 0 \\ -8 & 49\end{array}\right]-\left[\begin{array}{cc}8 & 0 \\ -8 & 56\end{array}\right]+\left[\begin{array}{ll}k & 0 \\ 0 & k\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{cc}1-8+k & 0-0+0 \\ -8+8+0 & 49-56+k\end{array}\right]=0$

$\Rightarrow\left[\begin{array}{cc}-7+k & 0 \\ 0 & -7+k\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\therefore-7+k=0$

$\Rightarrow k=7$

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