If $\begin{bmatrix$}[r] a & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix}[r] 4 & 3 \ -3 & 2 \end{bmatrix} = \begin{bmatrix}[r] b & 11 \ 4 & c \end{bmatrix}, find a, b and c.

Given,

$\begin{bmatrix$}[r] a & 1 \ 1 & 0 \end{bmatrix} \begin{bmatrix}[r] 4 & 3 \ -3 & 2 \end{bmatrix} = \begin{bmatrix}[r] b & 11 \ 4 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a \times 4 + 1 \times (-3) & a \times 3 + 1 \times 2 \ 1 \times 4 + 0 \times (-3) & 1 \times 3 + 0 \times 2 \end{bmatrix} = \begin{bmatrix}[r] b & 11 \ 4 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4a - 3 & 3a + 2 \ 4 + 0 & 3 + 0 \end{bmatrix} = \begin{bmatrix}[r] b & 11 \ 4 & c \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4a - 3 & 3a + 2 \ 4 & 3 \end{bmatrix} = \begin{bmatrix}[r] b & 11 \ 4 & c \end{bmatrix} \\[1em]

By definition of equality of matrices we get,

4a - 3 = b        (…Eq 1)
3a + 2 = 11     (…Eq 2)
c = 3.

Solving (Eq 2) first,

⇒ 3a + 2 = 11
⇒ 3a = 9
⇒ a = 3.

Putting value of a in Eq 1,

⇒ 4a - 3 = b
⇒ 4(3) - 3 = b
⇒ 12 - 3 = b
⇒ b = 9

∴ a = 3, b = 9 and c = 3.

Hence, the value of a = 3, b = 9 and c = 3.