Let A be a matrix such that $\begin{bmatrix} 5 & -2 \ 1 & 3 \end{bmatrix}$ × A = $\begin{bmatrix} 4 \ 11 \end{bmatrix}$.

(i) Write the order of A.

(ii) Find A.

(i) Let B = $\begin{bmatrix} 5 & -2 \ 1 & 3 \end{bmatrix}$ Then, BA = $\begin{bmatrix} 4 \ 11 \end{bmatrix}$

Order of B = 2 × 2

Order of BA = 2 × 1

Since BA exists, we have:

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

Number of columns of A = Number of columns in BA = 1

Order of A is 2 × 1.

$B${2 \times 2} \times A{2 \times 1} = BA_{2 \times 1}

Hence, order of A is 2 × 1.

(ii) Let A = $\begin{bmatrix} x \ y \end{bmatrix}$.

Then,

$\Rightarrow \begin{bmatrix} 5 & -2 \ 1 & 3 \end{bmatrix} \times \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 4 \ 11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 5x - 2y \ x + 3y \end{bmatrix} = \begin{bmatrix} 4 \ 11 \end{bmatrix}.$

Solving for x and y:

∴ 5x - 2y = 4… (1)

∴ x + 3y = 11

⇒ x = 11 - 3y….(2)

Substituting value of d from equation (2) in 5x - 2y = 4, we get:

⇒ 5(11 - 3y) - 2y = 4

⇒ 55 - 15y - 2y = 4

⇒ 55 - 17y = 4

⇒ 17y = 55 - 4

⇒ 17y = 51

⇒ y = $\dfrac{51}{17}$

⇒ y = 3

Substituting value of y in equation (2), we get:

⇒ x = 11 - 3(3)

⇒ x = 11 - 9

⇒ x = 2

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