Find a, b, c and d if $3\begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} = \begin{bmatrix}[r] 4 & a + b \ c + d & 3 \end{bmatrix} + \begin{bmatrix}[r] a & 6 \ -1 & 2d \end{bmatrix}

Given,

$3\begin{bmatrix$}[r] a & b \ c & d \end{bmatrix} = \begin{bmatrix}[r] 4 & a + b \ c + d & 3 \end{bmatrix} + \begin{bmatrix}[r] a & 6 \ -1 & 2d \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 3a & 3b \ 3c & 3d \end{bmatrix} = \begin{bmatrix}[r] 4 + a & a + b + 6 \ c + d - 1 & 3 + 2d \end{bmatrix} \\[0.5em]

By definition of equality of matrices

⇒ 3a = 4 + a     (…Eq 1)

⇒ 3b = a + b + 6     (…Eq 2)

⇒ 3c = c + d - 1     (…Eq 3)

⇒ 3d = 3 + 2d     (…Eq 4)

Solving Eq 1 we get,

⇒ 3a = 4 + a
⇒ 3a - a = 4
⇒ 2a = 4
⇒ a = 2.

Solving Eq 2 by putting value of a from Eq 1 we get,

⇒ 3b = a + b + 6
⇒ 3b = 2 + b + 6
⇒ 3b = b + 8
⇒ 3b - b = 8
⇒ 2b = 8
⇒ b = 4.

Solving Eq 4 we get,

⇒ 3d = 3 + 2d
⇒ 3d - 2d = 3
⇒ d = 3.

Solving Eq 3 by putting value of d from Eq 4 we get,

⇒ 3c = c + d - 1
⇒ 3c = c + 3 - 1
⇒ 3c - c = 2
⇒ 2c = 2
⇒ c = 1.

∴ a = 2, b = 4, c = 1 and d = 3.

Hence, the value of a = 2, b = 4, c = 1 and d = 3.