If A = $\begin{bmatrix$}[r] 3 & 2 \ 0 & 5 \end{bmatrix} \text{ and B } = \begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix}, find each of the following and state if they are equal :

(i) (A + B)(A - B)

(ii) A² - B²

(i) We need to find the value of (A + B)(A - B)

$(A + B)(A - B) = \Big(\begin{bmatrix$}[r] 3 & 2 \ 0 & 5 \end{bmatrix} + \begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix}\Big)\Big(\begin{bmatrix}[r] 3 & 2 \ 0 & 5 \end{bmatrix} - \begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix}\Big) \\[0.5em] = \begin{bmatrix}[r] 3 + 1 & 2 + 0 \ 0 + 1 & 5 + 2 \end{bmatrix} \begin{bmatrix}[r] 3 - 1 & 2 - 0 \ 0 - 1 & 5 - 2 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 4 & 2 \ 1 & 7 \end{bmatrix} \begin{bmatrix}[r] 2 & 2 \ -1 & 3 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 4 \times 2 + 2 \times (-1) & 4 \times 2 + 2 \times 3 \ 1 \times 2 + 7 \times (-1) & 1 \times 2 + 7 \times 3 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 8 - 2 & 8 + 6 \ 2 - 7 & 2 + 21 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 6 & 14 \ -5 & 23 \end{bmatrix}.

Hence, the value of (A + B)(A - B) = $\begin{bmatrix$}[r] 6 & 14 \ -5 & 23 \end{bmatrix}.

(ii) We need to find the value of A² - B²

$A^2 - B^2 = \begin{bmatrix$}[r] 3 & 2 \ 0 & 5 \end{bmatrix} \begin{bmatrix}[r] 3 & 2 \ 0 & 5 \end{bmatrix} - \begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix} \begin{bmatrix}[r] 1 & 0 \ 1 & 2 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 3 \times 3 + 2 \times 0 & 3 \times 2 + 2 \times 5 \ 0 \times 3 + 5 \times 0 & 0 \times 2 + 5 \times 5 \end{bmatrix} - \begin{bmatrix}[r] 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 2 \ 1 \times 1 + 2 \times 1 & 1 \times 0 + 2 \times 2 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 9 & 16 \ 0 & 25 \end{bmatrix} - \begin{bmatrix}[r] 1 & 0 \ 3 & 4 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 9 - 1 & 16 - 0 \ 0 - 3 & 25 - 4 \end{bmatrix} \\[0.5em] = \begin{bmatrix}[r] 8 & 16 \ -3 & 21 \end{bmatrix} .

Hence, the value of $A^2 - B^2 = \begin{bmatrix$}[r] 8 & 16 \ -3 & 21 \end{bmatrix} \text{ and (A + B)(A - B) } \neq A^2 - B^2.