If A = $\begin{bmatrix$}[r] 2 & 1 \ 1 & 3 \end{bmatrix} \text{ and } B = \begin{bmatrix}[r] 3 \ -11 \end{bmatrix} find the matrix X such that AX = B.

Given,

$\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix$}[r] 2 & 1 \ 1 & 3 \end{bmatrix}X = \begin{bmatrix}[r] 3 \ -11 \end{bmatrix}

X will be a matrix of order 2 × 1. So, let X = $\begin{bmatrix$}[r] a \ b \end{bmatrix}.

$\Rightarrow \begin{bmatrix$}[r] 2 & 1 \ 1 & 3 \end{bmatrix}\begin{bmatrix}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 3 \ -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 \times a + 1 \times b \ 1 \times a + 3 \times b \end{bmatrix} = \begin{bmatrix}[r] 3 \ -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2a + b \ a + 3b \end{bmatrix} = \begin{bmatrix}[r] 3 \ -11 \end{bmatrix}

By definition of equality of matrices we get,

2a + b = 3
⇒ b = 3 - 2a …….(i)

a + 3b = -11

Substituting value of b from (i) in above equation we get,

⇒ a + 3(3 - 2a) = -11
⇒ a + 9 - 6a = -11
⇒ -5a = -11 - 9
⇒ -5a = -20
⇒ a = 4.

b = 3 - 2a = 3 - 2(4) = 3 - 8 = -5.

Hence, X = $\begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 4 \ -5 \end{bmatrix}.