Solve the matrix equation $\begin{bmatrix$}[r] 4 \ 1 \end{bmatrix} X = \begin{bmatrix}[r] -4 & 8 \ -1 & 2 \end{bmatrix} .

Since,

$\begin{bmatrix$}[r] 4 \ 1 \end{bmatrix} X = \begin{bmatrix}[r] -4 & 8 \ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 \ 1 \end{bmatrix} X \text{ is a } 2 \times 2 \text{ matrix, but} \begin{bmatrix}[r] 4 \ 1 \end{bmatrix} \text{ is a } 2 \times 1 \text{ matrix}. \\[1em] \Rightarrow \text{X is a } 1 \times 2 \text{ matrix}.

We know that X matrix will be of order 1 × 2. So, let matrix X be $\begin{bmatrix$}[r] x & y \end{bmatrix}.

Given,

$\begin{bmatrix$}[r] 4 \ 1 \end{bmatrix} X = \begin{bmatrix}[r] -4 & 8 \ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 \ 1 \end{bmatrix} \begin{bmatrix}[r] x & y \end{bmatrix} = \begin{bmatrix}[r] -4 & 8 \ -1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4x & 4y \ x & y \end{bmatrix} = \begin{bmatrix}[r] -4 & 8 \ -1 & 2 \end{bmatrix} \\[1em]

From definition of equality of matrices we get,

⇒ x = -1 and y = 2.

$\text{Since, X }= \begin{bmatrix$}[r] x & y \ \end{bmatrix} \\[1em] \therefore \text{X } = \begin{bmatrix}[r] -1 & 2 \ \end{bmatrix}

Hence, the matrix $\text{X} = \begin{bmatrix$}[r] -1 & 2 \end{bmatrix}.