Let A be a matrix such that A × $\begin{bmatrix} 3 & 2 \ -1 & 5 \end{bmatrix}$ = $\begin{bmatrix} 9 & -11 \end{bmatrix}$.

(i) Write the order of A.

(ii) Find A.

(i) Let B = $\begin{bmatrix} 3 & 2 \ -1 & 5 \end{bmatrix}$ Then, AB = $\begin{bmatrix} 9 & -11 \end{bmatrix}$

Order of B = 2 × 2

Order of AB = 1 × 2

Since AB exists, we have:

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

Number of rows of A = Number of rows in AB = 1

Order of A is 1 × 2.

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

Hence, order of A is 1 × 2.

(ii) Let A = $\begin{bmatrix} a & b \end{bmatrix}$.

Then,

$\Rightarrow \begin{bmatrix} a & b \end{bmatrix} \times \begin{bmatrix} 3 & 2 \ -1 & 5 \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a - b & 2a + 5b \end{bmatrix} = \begin{bmatrix} 9 & -11 \end{bmatrix}.$

Solving for a and b:

∴ 3a - b = 9

⇒ b = 3a - 9 ….(1)

∴ 2a + 5b = -11 …(2)

Substituting value of b from equation (1) in (2), we get :

⇒ 2a + 5(3a - 9) = -11

⇒ 2a + 15a - 45 = -11

⇒ 17a = -11 + 45

⇒ 17a = 34

⇒ a = $\dfrac{34}{17}$

⇒ a = 2.

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

⇒ b = 3(2) - 9

⇒ b = 6 - 9

⇒ b = -3.

$\therefore A = \begin{bmatrix} 2 & -3 \end{bmatrix}$.

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