Let $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}, B = \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$. Then compute $AB$. Hence, solve the following system of equations: $2x + y = 4, 3x + 2y = 1$.

$x = 7, y = -10$

The correct answer is Option (3) → $x = 7, y = -10$ ##

$A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$

and $B = \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$

then $AB = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 4 & -6 \\ -2 & 4 \end{bmatrix}$

$AB = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = 2I$

$\Rightarrow A \left( \frac{1}{2}B \right) = I$

On multiplying by $A^{-1}$

$A^{-1} = \frac{1}{2}B$

$A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$

The given system of equations are equivalent to $A'X = C$

where $X = \begin{bmatrix} x \\ y \end{bmatrix}$

and $C = \begin{bmatrix} 4 \\ 1 \end{bmatrix}, A' = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$

$X = (A')^{-1}C = (A^{-1})'C$

$\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -10 \end{bmatrix}$

$∴x = 7, y = -10$