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If $A=\left[\begin{array}{cc}\mathrm{k} & 2 \\ -2 & -\mathrm{k}\end{array}\right]$, then $\mathrm{A}^{-1}$ does not exists if $\mathrm{k}=$
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Verified Answer
The correct answer is:
$\pm 2$
$$
\begin{aligned}
& A=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right] \\
& \therefore|\mathrm{A}|=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right]=-\mathrm{k}^2+4
\end{aligned}
$$
When $-\mathrm{k}^2+4=0$, we get $\mathrm{k}= \pm 2$
\begin{aligned}
& A=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right] \\
& \therefore|\mathrm{A}|=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right]=-\mathrm{k}^2+4
\end{aligned}
$$
When $-\mathrm{k}^2+4=0$, we get $\mathrm{k}= \pm 2$
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