If A = $\begin{bmatrix$}[r] 3 & -4 \ -1 & 2 \end{bmatrix}, find the matrix B such that BA = I, where I is the unity matrix of order 2.

Given,

BA = I

$\text{B}\begin{bmatrix$}[r] 3 & -4 \ -1 & 2 \end{bmatrix} = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \text{B}\begin{bmatrix}[r] 3 & -4 \ -1 & 2 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix}[r] 3 & -4 \ -1 & 2 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{B is a } 2 \times 2 \text{ matrix}. \\[1em] \text{I } = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}

We know that B will be of order 2 × 2. So, let

$\text{B} = \begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a & b \ c & d \end{bmatrix} \begin{bmatrix}[r] 3 & -4 \ -1 & 2 \end{bmatrix} = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a \times 3 + b \times (-1) & a \times (-4) + b \times 2 \ c \times 3 + d \times (-1) & c \times (-4) + d \times 2 \end{bmatrix} = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3a - b & -4a + 2b \ 3c - d & -4c + 2d \end{bmatrix} = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

3a - b = 1          (…Eq 1)

-4a + 2b = 0
⇒ 4a = 2b
⇒ b = 2a           (…Eq 2)

3c - d = 0
⇒ d = 3c           (…Eq 3)

-4c + 2d = 1     (…Eq 4)

Putting value of b from Eq 2 in Eq 1

⇒ 3a - b = 1
⇒ 3a - 2a = 1
⇒ a = 1

∴ a = 1, b = 2a = 2.

Putting value of d from Eq 3 in Eq 4

⇒ -4c + 2d = 1
⇒ -4c + 2(3c) = 1
⇒ -4c + 6c = 1
⇒ 2c = 1
⇒ c = $\dfrac{1}{2}$

∴ c = $\dfrac{1}{2}$, d = 3c = $\dfrac{3}{2}$.

$\text{Since, B }= \begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} \\[1em] \therefore \text{B} = \begin{bmatrix}[r] 1 & 2 \ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix}

Hence, the matrix B = $\begin{bmatrix$}[r] 1 & 2 \ \dfrac{1}{2} & \dfrac{3}{2} \end{bmatrix}.