Question
If $A=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$, show that $(A+B)(A-B) \neq A^2-B^2$.
$(A+B) \cdot(A-B) \neq A^2-B^2$
i.e, to prove that $A(A-B)+B(A-B) \neq A^2-B^2$
i.e, to prove that $A^2-A B+B A-B^2 \neq A^2-B^2$
i.e, to prove that $A B \neq B A$.
$\begin{aligned} A B & =\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0+1 & 0+0 \\ 0+0 & -1+0\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]\end{aligned}$
$\begin{aligned} \mathbf{B A} & =\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0-1 & 0-0 \\ 0+0 & 1+0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\end{aligned}$
∴ AB ≠ BA
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