Question

If $A=\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ and x² = −1, then show that (A + B)² = A² + B².

Answer 1

Given: $A=\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right],B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ and x² = −1

To show: (A + B)² = A² + B²

LHS:
$$\begin{gathered} A+B=\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right]+\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}0+0& -x+1\\ x+1& 0+0\end{array}\right] \\ =\left[\begin{array}{cc}0& -x+1\\ x+1& 0\end{array}\right] \\ {\left(A+B\right)}^2=\left[\begin{array}{cc}0& -x+1\\ x+1& 0\end{array}\right]\left[\begin{array}{cc}0& -x+1\\ x+1& 0\end{array}\right] \\ =\left[\begin{array}{cc}0+\left(1-x\right)\left(1+x\right)& 0+0\\ 0+0& \left(x+1\right)\left(1-x\right)+0\end{array}\right] \\ =\left[\begin{array}{cc}1-x^2& 0\\ 0& 1-x^2\end{array}\right]...\left(1\right) \end{gathered}$$

RHS:
$$\begin{gathered} A=\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right] \\ {A}^2=\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right]\left[\begin{array}{cc}0& -x\\ x& 0\end{array}\right] \\ =\left[\begin{array}{cc}0-x^2& 0+0\\ 0+0& -x^2+0\end{array}\right] \\ =\left[\begin{array}{cc}-x^2& 0\\ 0& -x^2\end{array}\right]...\left(2\right) \\ B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ {B}^2=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \\ =\left[\begin{array}{cc}0+1& 0+0\\ 0+0& 1+0\end{array}\right] \\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]...\left(3\right) \\ \mathrm{Adding}\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ {A}^2+B^2=\left[\begin{array}{cc}-x^2& 0\\ 0& -x^2\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\left[\begin{array}{cc}1-x^2& 0\\ 0& 1-x^2\end{array}\right]...\left(4\right) \end{gathered}$$

Comparing (1) and (4), we get

(A + B)² = A² + B²