Find a and b if $\begin{bmatrix$}[r] a - b & b - 4 \ b + 4 & a - 2 \end{bmatrix} \begin{bmatrix}[r] 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix}[r] -2 & -2 \ 14 & 0 \end{bmatrix}.

Given,

$\begin{bmatrix$}[r] a - b & b - 4 \ b + 4 & a - 2 \end{bmatrix} \begin{bmatrix}[r] 2 & 0 \ 0 & 2 \end{bmatrix} = \begin{bmatrix}[r] -2 & -2 \ 14 & 0 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] (a - b) \times 2 + (b - 4) \times 0 & (a - b) \times 0 + (b - 4) \times 2 \ (b + 4) \times 2 + (a - 2) \times 0 & (b + 4) \times 0 + (a - 2) \times 2 \end{bmatrix} = \begin{bmatrix}[r] -2 & -2 \ 14 & 0 \end{bmatrix} \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2a - 2b & 2b - 8 \ 2b + 8 & 2a - 4 \end{bmatrix} = \begin{bmatrix}[r] -2 & -2 \ 14 & 0 \end{bmatrix} \\[0.5em]

By definition of equality of matrices we get,

⇒ 2a - 4 = 0 or 2a = 4 or a = 2

⇒ 2b - 8 = -2 or 2b = -2 + 8 = 6 or b = 3

⇒ 2a - 2b = -2     (Eq 1)

Checking whether a = 2 and b = 3 satisfies Eq 1,

⇒ 2a - 2b = -2

L.H.S. = 2a - 2b = 2(2) - 2(3) = 4 - 6 = -2 = R.H.S.

∴ a = 2 and b = 3.

Hence, the values are a = 2 and b = 3.