If B = $\begin{bmatrix$}[r] -4 & 2 \ 5 & -1 \end{bmatrix} \text{ and C} = \begin{bmatrix}[r] 17 & -1 \ 47 & -13 \end{bmatrix}, find the matrix A such that AB = C.

Given,

AB = C

$\Rightarrow A\begin{bmatrix$}[r] -4 & 2 \ 5 & -1 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix}[r] 4 & 2 \ 5 & -1 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{A is a } 2 \times 2 \text{ matrix}.

We know that matrix A will be of order 2 × 2. Let matrix be

$\text{A} = \begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a & b \ c & d \end{bmatrix} \begin{bmatrix}[r] -4 & 2 \ 5 & -1 \end{bmatrix} = \begin{bmatrix}[r] 17 & -1 \ 47 & -13 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a \times (-4) + b \times 5 & a \times 2 + b \times (-1) \ c \times (-4) + d \times 5 & c \times 2 + d \times (-1) \end{bmatrix} = \begin{bmatrix}[r] 17 & -1 \ 47 & -13 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -4a + 5b & 2a - b \ -4c + 5d & 2c - d \end{bmatrix} = \begin{bmatrix}[r] 17 & -1 \ 47 & -13 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

-4a + 5b = 17    (…Eq 1)

2a - b = -1
⇒ b = 2a + 1     (…Eq 2)

-4c + 5d = 47    (…Eq 3)

2c - d = -13
⇒ d = 2c + 13    (…Eq 4)

Putting value of b from Eq 2 in Eq 1

⇒ -4a + 5b = 17
⇒ -4a + 5(2a + 1) = 17
⇒ -4a + 10a + 5 = 17
⇒ 6a + 5 = 17
⇒ 6a = 12
⇒ a = 2.

∴ a = 2, b = 2a + 1 = 2(2) + 1 = 5.

Putting value of d from Eq 4 in Eq 3

⇒ -4c + 5d = 47
⇒ -4c + 5(2c + 13) = 47
⇒ -4c + 10c + 65 = 47
⇒ 6c + 65 = 47
⇒ 6c = -18
⇒ c = -3.

∴ c = -3, d = 2c + 13 = 2(-3) + 13 = 7.

Since,

$\text{A} = \begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} \\[1em] \therefore \text{A} = \begin{bmatrix}[r] 2 & 5 \ -3 & 7 \end{bmatrix}

Hence, the matrix A = $\begin{bmatrix$}[r] 2 & 5 \ -3 & 7 \end{bmatrix}.