Question:
If $A=\left[\begin{array}{cc}x & 1 \\ -1 & -x\end{array}\right]$ satisfies the equation $A^{2}=0$, then $x=$______
Solution:
The given matrix is $A=\left[\begin{array}{cc}x & 1 \\ -1 & -x\end{array}\right]$.
$A^{2}=0$
$\therefore\left[\begin{array}{cc}x & 1 \\ -1 & -x\end{array}\right]\left[\begin{array}{cc}x & 1 \\ -1 & -x\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}x^{2}-1 & 0 \\ 0 & x^{2}-1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow x^{2}-1=0$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
Thus, the value of $x$ is $\pm 1$.
If $A=\left[\begin{array}{cc}x & 1 \\ -1 & -x\end{array}\right]$ satisfies the equation $A^{2}=0$, then $x=\underline{\pm1}$