Question
If $A=\left[\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right]$, find $\mathrm{k}$ so that $\mathrm{A}^2-8 \mathrm{~A}-\mathrm{kl}=\mathrm{O}$, where $\mathrm{I}$ is a unit matrix and $\mathrm{O}$ is a null
matrix of order 2.
matrix of order 2.
$\begin{aligned} \therefore\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right]-8\left[\begin{array}{rr}1 & 0 \\ -1 & 7\end{array}\right] & -\mathrm{k}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]\end{aligned}$
$\begin{array}{ll}\therefore & {\left[\begin{array}{rr}1+0 & 0+0 \\ -1-7 & 0+49\end{array}\right]-\left[\begin{array}{rr}8 & 0 \\ -8 & 56\end{array}\right]-\left[\begin{array}{ll}\mathrm{k} & 0 \\ 0 & \mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]} \\ \therefore & {\left[\begin{array}{rc}1 & 0 \\ -8 & 49\end{array}\right]-\left[\begin{array}{rr}8 & 0 \\ -8 & 56\end{array}\right]-\left[\begin{array}{ll}\mathrm{k} & 0 \\ 0 & \mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]} \\ \therefore & {\left[\begin{array}{rc}1-8-\mathrm{k} & 0-0-0 \\ -8+8-0 & 49-56-\mathrm{k}\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]}\end{array}$
∴ by equality of matrices, we get 1 – 8 – k = 0 ∴ k = -7
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