Let A = $\begin{bmatrix} 3 & 2 \ -1 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}$. Find a matrix C such that AC = B.

A = $\begin{bmatrix} 3 & 2 \ -1 & 1 \end{bmatrix}$ and B = $\begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}$

Given,

AC = B

Order of A = 2 × 2

Order of AC = Order of B = 2 × 2

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 = 2

Order of C is 2 × 2.

Let C = $\begin{bmatrix} a & b\ c & d \end{bmatrix}$

AC = B

$\Rightarrow \begin{bmatrix} 3 & 2 \ -1 & 1 \end{bmatrix} \times \begin{bmatrix} a & b\ c & d \end{bmatrix} = \begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix} 3a + 2c & 3b + 2d \ -a + c & -b + d \end{bmatrix} = \begin{bmatrix} 14 & 3\ 2 & 4 \end{bmatrix}.$

Solving for a and c:

∴ -a + c = 2

⇒ c = a + 2 …(1)

∴ 3a + 2c = 14 ……(2)

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

⇒ 3a + 2(a + 2) = 14

⇒ 3a + 2a + 4 = 14

⇒ 5a = 14 - 4

⇒ 5a = 10

⇒ a = $\dfrac{10}{5}$

⇒ a = 2.

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

⇒ c = 2 + 2

⇒ c = 4.

Solving for b and d :

∴ -b + d = 4

⇒ d = b + 4 …….(3)

∴ 3b + 2d = 3 …….(4)

Substituting value of d from equation (3) in (4), we get:

⇒ 3b + 2(b + 4) = 3

⇒ 3b + 2b + 8 = 3

⇒ 5b + 8 = 3

⇒ 5b = 3 - 8

⇒ 5b = -5

⇒ b = $\dfrac{-5}{5}$

⇒ b = -1.

Substituting value of b in equation (3), we get :

⇒ d = -1 + 4

⇒ d = 3.

Hence, C = $\begin{bmatrix} 2 & -1\ 4 & 3 \end{bmatrix}.$