Let A = $\begin{bmatrix} 2 & -1 \ -3 & 4 \end{bmatrix}$ and B = $\begin{bmatrix} 8 \ -17 \end{bmatrix}$. Find a matrix C such that AC = B.

A = $\begin{bmatrix} 2 & -1 \ -3 & 4 \end{bmatrix}$ and B = $\begin{bmatrix} 8 \ -17 \end{bmatrix}$

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 1

Since AC exists, we have :

Number of rows of C = Number of columns in A = 2

Number of columns of C = Number of columns in B = 1

Order of C is 2 × 1.

Let C = $\begin{bmatrix} x \ y \end{bmatrix}$

AC = B

$\Rightarrow \begin{bmatrix} 2 & -1 \ -3 & 4 \end{bmatrix} \times \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ -17 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 2x - y \ -3x + 4y \end{bmatrix} = \begin{bmatrix} 8 \ -17 \end{bmatrix}.$

∴ 2x - y = 8

⇒ y = 2x - 8 …….(1)

∴ -3x + 4y = -17

Substituting value of y from equation(1) in -3x + 4y = -17, we get:

⇒ -3x + 4(2x - 8) = -17

⇒ -3x + 8x - 32 = -17

⇒ 5x - 32 = -17

⇒ 5x = -17 + 32

⇒ 5x = 15

⇒ x = $\dfrac{15}{5}$

⇒ x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 2(3) - 8

⇒ y = 6 - 8

⇒ y = -2.

Hence, C = $\begin{bmatrix} 3 \ -2 \end{bmatrix}.$