If A = $\begin{bmatrix$}[r] 2 & -1 \ -4 & 5 \end{bmatrix} \text{and B } = \begin{bmatrix}[r] 0 & -3 \ \end{bmatrix}, find matrix C such that CA = B.

Given,

CA = B.

$C\begin{bmatrix$}[r] 2 & -1 \ -4 & 5 \end{bmatrix} = \begin{bmatrix}[r] 0 & -3 \ \end{bmatrix} \\[1em] \Rightarrow C\begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix} \text{ is a } 1 \times 2 \text{ matrix, but} \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[1em] \text{Let matrix C } = \begin{bmatrix}[r] x & y \ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x & y \ \end{bmatrix} \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix} = \begin{bmatrix}[r] 0 & -3 \ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x \times 2 + y \times (-4) & x \times (-1) + y \times 5 \end{bmatrix} = \begin{bmatrix}[r] 0 & -3 \ \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x - 4y & -x + 5y \ \end{bmatrix} = \begin{bmatrix}[r] 0 & -3 \ \end{bmatrix}

By definition of equality of matrices we get,

2x - 4y = 0                             (…Eq 1)

-x + 5y = -3 or x = 5y + 3     (…Eq 2)

Putting value of x from Eq 2 in Eq 1,

⇒ 2x - 4y = 0
⇒ 2(5y + 3) - 4y = 0
⇒ 10y + 6 - 4y = 0
⇒ 6y + 6 = 0
⇒ 6y = -6
⇒ y = -1.

Now finding value of x,

⇒ x = 5y + 3 = 5(-1) + 3 = -5 + 3 = -2.

∴ x = -2, y = -1.

$\text{Since, C }= \begin{bmatrix$}[r] x & y \ \end{bmatrix} \\[0.5em] \therefore C = \begin{bmatrix}[r] -2 & -1 \ \end{bmatrix}

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