If A = $\begin{bmatrix$}[r] 0 & 2 \ 5 & -2 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix} and I is a unit matrix of order 2 × 2, find :

(i) AB

(ii) BA

(iii) AI

(iv) IB

(v) A²

(vi) B²A

(i) Substituting value in AB,

$\Rightarrow AB = \begin{bmatrix$}[r] 0 & 2 \ 5 & -2 \end{bmatrix}\begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 \times 1 + 2 \times 3 & 0 \times (-1) + 2 \times 2 \ 5 \times 1 + (-2) \times 3 & 5 \times (-1) + (-2) \times 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 + 6 & 0 + 4 \ 5 - 6 & -5 - 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6 & 4 \ -1 & -9 \end{bmatrix}.

Hence, AB = $\begin{bmatrix$}[r] 6 & 4 \ -1 & -9 \end{bmatrix}.

(ii) Substituting value in BA,

$\Rightarrow BA = \begin{bmatrix$}[r] 1 & -1 \ 3 & 2 \end{bmatrix}\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 0 + (-1) \times 5 & 1 \times 2 + (-1) \times (-2) \ 3 \times 0 + 2 \times 5 & 3 \times 2 + 2 \times (-2) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 - 5 & 2 + 2 \ 0 + 10 & 6 - 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -5 & 4 \ 10 & 2 \end{bmatrix}.

Hence, BA = $\begin{bmatrix$}[r] -5 & 4 \ 10 & 2 \end{bmatrix}.

(iii) Substituting value in AI,

$\Rightarrow AI = \begin{bmatrix$}[r] 0 & 2 \ 5 & -2 \end{bmatrix}\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 \times 1 + 2 \times 0 & 0 \times 0 + 2 \times 1 \ 5 \times 1 + (-2) \times 0 & 5 \times 0 + (-2) \times 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} = A

Hence, AI = matrix A.

(iv) Substituting value in IB,

$\Rightarrow IB = \begin{bmatrix$}[r] 1 & 0 \ 0 & 1 \end{bmatrix}\begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 0 \times 3 & 1 \times (-1) + 0 \times 2 \ 0 \times 1 + 1 \times 3 & 0 \times (-1) + 1 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 0 & -1 + 0 \ 0 + 3 & 0 + 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix} = B

Hence, IB = matrix B

(v) Substituting value in A²,

$\Rightarrow A^2 = \begin{bmatrix$}[r] 0 & 2 \ 5 & -2 \end{bmatrix}\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 \times 0 + 2 \times 5 & 0 \times 2 + 2 \times -2 \ 5 \times 0 + (-2) \times 5 & 5 \times 2 + (-2) \times (-2) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 + 10 & 0 - 4 \ 0 - 10 & 10 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 10 & -4 \ -10 & 14 \end{bmatrix}.

Hence, $A^2 = \begin{bmatrix$}[r] 10 & -4 \ -10 & 14 \end{bmatrix}.

(vi) Substituting value in B²A,

$\Rightarrow B^2A = \begin{bmatrix$}[r] 1 & -1 \ 3 & 2 \end{bmatrix}\begin{bmatrix}[r] 1 & -1 \ 3 & 2 \end{bmatrix}\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + (-1) \times 3 & 1 \times -1 + (-1) \times 2 \ 3 \times 1 + 2 \times 3 & 3 \times (-1) + 2 \times 2 \end{bmatrix}\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 & -3 \ 9 & 1 \end{bmatrix}\begin{bmatrix}[r] 0 & 2 \ 5 & -2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -2 \times 0 + (-3) \times 5 & (-2) \times 2 + (-3) \times (-2) \ 9 \times 0 + 1 \times 5 & 9 \times 2 + 1 \times (-2) \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 0 - 15 & -4 + 6 \ 0 + 5 & 18 - 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] -15 & 2 \ 5 & 16 \end{bmatrix}

Hence, $B^2A = \begin{bmatrix$}[r] -15 & 2 \ 5 & 16 \end{bmatrix}.