If show that

Solution:

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

So, $(A+B)=\left[\begin{array}{cc}0+0 & 1-1 \\ 1+1 & 1+0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right]$

And, $(A-B)=\left[\begin{array}{ll}0-0 & 1+1 \\ 1-1 & 1-0\end{array}\right]=\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]$

$(A+B) \cdot(A-B)=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 4+1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right]$ $\ldots$ (i)

Also, $\quad A^{2}=A \cdot A$

$=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right] \cdot\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}0+1 & 0+1 \\ 0+1 & 1+1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]$

And, $B^{2}=B \cdot B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}0-1 & 0+0 \\ 0+0 & -1+0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$

Therefore, $A^{2}-B^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ -0 & -1\end{array}\right]=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$ $\ldots$ (ii)

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

So, $(A+B)=\left[\begin{array}{ll}0+0 & 1-1 \\ 1+1 & 1+0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right]$

And, $(A-B)=\left[\begin{array}{ll}0-0 & 1+1 \\ 1-1 & 1-0\end{array}\right]=\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]$

$(A+B) \cdot(A-B)=\left[\begin{array}{ll}0 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}0+0 & 0+0 \\ 0+0 & 4+1\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 5\end{array}\right]$ $\ldots$ (i)

Also, $\quad A^{2}=A \cdot A$

$=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right] \cdot\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{ll}0+1 & 0+1 \\ 0+1 & 1+1\end{array}\right]=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]$

And, $B^{2}=B \cdot B=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]=\left[\begin{array}{cc}0-1 & 0+0 \\ 0+0 & -1+0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$

Therefore,  $A^{2}-B^{2}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ -0 & -1\end{array}\right]=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$ ........(ii)

Hence, from (i) and (ii),

(A + B) (A – B) ≠ A2 – B2