Given A = $\begin{bmatrix$}[r] 3 & 6 \ -2 & -8 \end{bmatrix} \text{and B} = \begin{bmatrix}[r] -2 & 16 \end{bmatrix}; find the matrix X such that XA = B.

Let order of matrix X = m × n

Given,

XA = B

$X${m \times n}\begin{bmatrix}[r] 3 & 6 \ -2 & -8 \end{bmatrix}{2 \times 2} = \begin{bmatrix}[r] -2 & 16 \end{bmatrix}_{1 \times 2}

Since, the product of matrices is possible only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ n = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ m = 1

Order of matrix X = m × n = 1 × 2.

Let matrix X = $\begin{bmatrix$}[r] a & b \end{bmatrix}

Substituting matrix in XA = B we get,

$\Rightarrow \begin{bmatrix$}[r] a & b \end{bmatrix}\begin{bmatrix}[r] 3 & 6 \ -2 & -8 \end{bmatrix} = \begin{bmatrix}[r] -2 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a \times 3 + b \times -2 & a \times 6 + b \times -8 \end{bmatrix} = \begin{bmatrix}[r] -2 & 16 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3a - 2b & 6a - 8b \end{bmatrix} = \begin{bmatrix}[r] -2 & 16 \end{bmatrix}

∴ 3a - 2b = -2 ………(1)

6a - 8b = 16 ………(2)

⇒ 3a - 2b = -2

⇒ 2(3a - 2b) = 2(-2)

⇒ 6a - 4b = -4 ………(3)

Subtracting equation (2) from (3), we get :

⇒ 6a - 4b - (6a - 8b) = -4 - 16

⇒ 6a - 6a - 4b + 8b = -20

⇒ 4b = -20

⇒ b = $\dfrac{-20}{4}$

⇒ b = -5.

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

⇒ 3a - 2(-5) = -2

⇒ 3a + 10 = -2

⇒ 3a = -2 - 10

⇒ 3a = -12

⇒ a = $\dfrac{-12}{3}$

⇒ a = -4.

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