If A = $\begin{bmatrix$}[r] -1 & 3 \ 2 & 0 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & -2 \ 0 & 3 \end{bmatrix}, C = \begin{bmatrix}[r] 1 & -4 \end{bmatrix} \text{ and } D = \begin{bmatrix}[r] 4 \ 1 \end{bmatrix}

(a) Is the product AC possible? Justify your answer.

(b) Find the matrix X, such that X = AB + B² − DC

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2

The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.

Hence, product AC is not possible.

(b) Given,

$\Rightarrow X = AB + B^2 - DC \\[1em] \Rightarrow X = \begin{bmatrix$}[r] -1 & 3 \ 2 & 0 \end{bmatrix}\begin{bmatrix}[r] 1 & -2 \ 0 & 3 \end{bmatrix} + \begin{bmatrix}[r] 1 & -2 \ 0 & 3 \end{bmatrix}\begin{bmatrix}[r] 1 & -2 \ 0 & 3 \end{bmatrix} - \begin{bmatrix}[r] 4 \ 1 \end{bmatrix}\begin{bmatrix}[r] 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix} + \begin{bmatrix}[r] 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix} - \begin{bmatrix}[r] 4 \times 1 & 4 \times -4 \ 1 \times 1 & 1 \times -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -1 + 0 & 2 + 9 \ 2 + 0 & -4 + 0 \end{bmatrix} + \begin{bmatrix}[r] 1 + 0 & -2 - 6 \ 0 + 0 & 0 + 9 \end{bmatrix} - \begin{bmatrix}[r] 4 & -16 \ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -1 & 11 \ 2 & -4 \end{bmatrix} + \begin{bmatrix}[r] 1 & -8 \ 0 & 9 \end{bmatrix} - \begin{bmatrix}[r] 4 & -16 \ 1 & -4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -1 + 1 - 4 & 11 + (-8) - (-16) \ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -4 & 11 - 8 + 16 \ 1 & -4 + 9 + 4 \end{bmatrix} \\[1em] \Rightarrow X = \begin{bmatrix}[r] -4 & 19 \ 1 & 9 \end{bmatrix}

Hence, X = $\begin{bmatrix$}[r] -4 & 19 \ 1 & 9 \end{bmatrix}.