Find the values of a, b, c and d if $\begin{bmatrix$}[r] a + b & 3 \ 5 + c & ab \end{bmatrix} = \begin{bmatrix}[r] 6 & d \ -1 & 8 \end{bmatrix} .

Given $\begin{bmatrix$}[r] a + b & 3 \ 5 + c & ab \end{bmatrix} = \begin{bmatrix}[r] 6 & d \ -1 & 8 \end{bmatrix} .

By definition of equality of matrices, we get

d = 3,
5 + c = -1 or c = -6,
a + b = 6       […Eq 1],
ab = 8            […Eq 2].

Putting value of b from Eq 1 in Eq 2,

$a + b = 6 \text{ or } b = 6 - a \\[1em] \Rightarrow a(6 - a) = 8 \\[0.5em] \Rightarrow 6a - a^2 = 8 \\[0.5em] \Rightarrow a^2 - 6a + 8 = 0 \\[0.5em] \Rightarrow a^2 - 4a - 2a + 8 = 0 \\[0.5em] \Rightarrow a(a - 4) - 2(a - 4) = 0 \\[0.5em] \Rightarrow (a - 2)(a - 4) = 0 \\[0.5em] \Rightarrow a = 2 \text{ or } a = 4.$

Now, finding value of b = 6 - a,

if,  a = 2, b = 6 - 2 = 4.

or, a = 4, b = 6 - 4 = 2.

∴ a = 2 or 4, b = 4 or 2, c = -6 and d = 3.

Hence, the values are a = 2, b = 4, c = -6, d = 3 OR a = 4, b = 2, c = -6, d = 3.