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If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and $\mathbf{A}^{2}-\mathrm{kA}-\mathrm{I}_{2}=\mathrm{O}$, where $\mathrm{I}_{2}$ is the $2 \times 2$ identity matrix, then what is the value of $\mathrm{k}$ ?
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The correct answer is:
4
$\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1+4 & 2+6 \\ 2+6 & 4+9\end{array}\right]=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]$
$\mathrm{A}^{2}-\mathrm{kA}-\mathrm{I}_{2}=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]-\left[\begin{array}{cc}\mathrm{k} & 2 \mathrm{k} \\ 2 \mathrm{k} & 3 \mathrm{k}\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}5-k-1 & 8-2 k-0 \\ 8-2 k-0 & 13-3 k-1\end{array}\right]$
$=\left[\begin{array}{cc}4-k & 8-2 k \\ 8-2 k & 12-3 k\end{array}\right]$
Given, $A^{2}-k A-I_{2}=0$
$\therefore 4-k=0 \Rightarrow k=4$
$\mathrm{A}^{2}=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]=\left[\begin{array}{ll}1+4 & 2+6 \\ 2+6 & 4+9\end{array}\right]=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]$
$\mathrm{A}^{2}-\mathrm{kA}-\mathrm{I}_{2}=\left[\begin{array}{cc}5 & 8 \\ 8 & 13\end{array}\right]-\left[\begin{array}{cc}\mathrm{k} & 2 \mathrm{k} \\ 2 \mathrm{k} & 3 \mathrm{k}\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}5-k-1 & 8-2 k-0 \\ 8-2 k-0 & 13-3 k-1\end{array}\right]$
$=\left[\begin{array}{cc}4-k & 8-2 k \\ 8-2 k & 12-3 k\end{array}\right]$
Given, $A^{2}-k A-I_{2}=0$
$\therefore 4-k=0 \Rightarrow k=4$
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