Given $\begin{bmatrix$}[r] 4 & 2 \ -1 & 1 \end{bmatrix} M = 6I, where M is a matrix and I is the unit matrix of order 2 × 2.

(i) State the order of matrix M.

(ii) Find the matrix M.

(i) Given,

$\begin{bmatrix$}[r] 4 & 2 \ -1 & 1 \end{bmatrix} M = 6I \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 2 \ -1 & 1 \end{bmatrix} M \text{ is a } 2 \times 2 \text{ matrix, and} \begin{bmatrix}[r] 4 & 2 \ -1 & 1 \end{bmatrix} \text{ is a } 2 \times 2 \text{ matrix}. \\[1em] \therefore \text{M is a } 2 \times 2 \text{ matrix}. \\[1em]

Hence, the matrix M is of order 2 × 2.

(ii) Let matrix M be $\begin{bmatrix$}[r] a & b \ c & d \end{bmatrix}.

Given,

$\begin{bmatrix$}[r] 4 & 2 \ -1 & 1 \end{bmatrix} M = 6I \\[1em] \Rightarrow \begin{bmatrix}[r] 4 & 2 \ -1 & 1 \end{bmatrix} \begin{bmatrix}[r] a & b \ c & d \end{bmatrix} = 6\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 \times a + 2 \times c & 4 \times b + 2 \times d \ (-1) \times a + 1 \times c & (-1) \times b + 1 \times d \end{bmatrix} = \begin{bmatrix}[r] 6 & 0 \ 0 & 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4a + 2c & 4b + 2d \ -a + c & -b + d \end{bmatrix} = \begin{bmatrix}[r] 6 & 0 \ 0 & 6 \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

4a + 2c = 6     (…Eq 1)

4b + 2d = 0
⇒ d = -2b       (…Eq 2)

-a + c = 0
⇒ a = c           (…Eq 3)

-b + d = 6       (…Eq 4)

Putting value of a from Eq 3 in Eq 1

⇒ 4a + 2c = 6
⇒ 4c + 2c = 6
⇒ 6c = 6
⇒ c = 1.

∴ c = 1 and a = c = 1.

Putting value of d from Eq 2 in Eq 4

⇒ -b + d = 6
⇒ -b + (-2b) = 6
⇒ -3b = 6
⇒ b = -2.

∴ b = -2 and d = -2b = 4.

Since,

$\text{M} = \begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} \\[1em] \therefore \text{M} = \begin{bmatrix}[r] 1 & -2 \ 1 & 4 \end{bmatrix}

Hence, the matrix $\text{M} = \begin{bmatrix$}[r] 1 & -2 \ 1 & 4 \end{bmatrix}.