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If $A=\left[\begin{array}{cc}\lambda & i \\ i & -\lambda\end{array}\right]$ and $A^{-1}$ does not exist, then $\lambda=($ where $I=\sqrt{-1})$
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Verified Answer
The correct answer is:
$\pm 1$
$A=\left[\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right] \Rightarrow|\mathrm{A}|=\left|\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right|$
Since $\mathrm{A}^{-1}$ does not exist, we write $|\mathrm{A}|=0$
$\therefore\left(-\lambda^2\right)-\left(\mathrm{i}^2\right)=0 \quad \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1$
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right] \Rightarrow|\mathrm{A}|=\left|\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right|$
Since $\mathrm{A}^{-1}$ does not exist, we write $|\mathrm{A}|=0$
$\therefore\left(-\lambda^2\right)-\left(\mathrm{i}^2\right)=0 \quad \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1$
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