If A = $\begin{bmatrix$}[r] 3 & a \ -4 & 8 \end{bmatrix}, B = \begin{bmatrix}[r] c & 4 \ -3 & 0 \end{bmatrix}, C = \begin{bmatrix}[r] -1 & 4 \ 3 & b \end{bmatrix} and 3A - 2C = 6B, find the values of a, b and c.

Given,

$\Rightarrow 3A - 2C = 6B \\[1em] \Rightarrow 3\begin{bmatrix$}[r] 3 & a \ -4 & 8 \end{bmatrix} - 2\begin{bmatrix}[r] -1 & 4 \ 3 & b \end{bmatrix} = 6\begin{bmatrix}[r] c & 4 \ -3 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 & 3a \ -12 & 24 \end{bmatrix} - \begin{bmatrix}[r] -2 & 8 \ 6 & 2b \end{bmatrix} = \begin{bmatrix}[r] 6c & 24 \ -18 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 - (-2) & 3a - 8 \ -12 - 6 & 24 - 2b \end{bmatrix} = \begin{bmatrix}[r] 6c & 24 \ -18 & 0 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 11 & 3a - 8 \ -18 & 24 - 2b \end{bmatrix} = \begin{bmatrix}[r] 6c & 24 \ -18 & 0 \end{bmatrix}

By definition of equality of matrices we get,

6c = 11
⇒ c = $\dfrac{11}{6} = 1\dfrac{5}{6}.$

3a - 8 = 24
⇒ 3a = 32
⇒ a = $\dfrac{32}{3} = 10\dfrac{2}{3}$

24 - 2b = 0
⇒ 2b = 24
⇒ b = 12.

Hence, a = $10\dfrac{2}{3}, b = 12, c = 1\dfrac{5}{6}.$