Given $\begin{bmatrix} 4 & 2 \ -1 & 1 \end{bmatrix}$ 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) Let N = $\begin{bmatrix} 4 & 2 \ -1 & 1 \end{bmatrix}$

Given, NM = 6I

NM = $\begin{bmatrix} 6 & 0 \ 0 & 6 \end{bmatrix}$

Order of N = 2 × 2

Order of NM = 2 × 2

Since NM exists, we have:

Number of rows of M = Number of columns in N = 2

Number of columns of M = Number of columns in NM = 2

Order of M is 2 × 2.

$M${2 \times 2} \times N{2 \times 2} = MN_{2 \times 2}

Hence, order of M is 2 × 2.

(ii) Let M = $\begin{bmatrix} a & b \ c & d \end{bmatrix}$.

Then,

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

Solving for a and c :

∴ -a + c = 0

⇒ c = a….(1)

∴ 4a + 2c = 6

⇒ 2(2a + c = 3)

⇒ 2a + c = 3…(2)

Substituting value of c from equation (1) in 2a + c = 3, we get:

⇒ 2a + a = 3

⇒ 3a = 3

⇒ a = $\dfrac{3}{3}$

⇒ a = 1

Hence, a = c = 1

Solving for b and d:

∴ -b + d = 6

⇒ d = b + 6…(3)

∴ 4b + 2d = 0

⇒ 2(2b + d) = 0

⇒ 2b + d = 0…(4)

Substituting value of d from equation (3) in 2b + d = 0, we get:

⇒ 2b + b + 6 = 0

⇒ 3b + 6 = 0

⇒ 3b = -6

⇒ b = $\dfrac{-6}{3}$

⇒ b = -2

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

⇒ d = -2 + 6

⇒ d = 4

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