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: $2x + y = 4, 3x + 2y = 1$.
Answer & explanation
Correct answer: option 3
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 $C = \begin{bmatrix} 4 \\ 1 \end{bmatrix}, A' = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}$
$X = (A')^{-1}C$
$\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 7 \\ -10 \end{bmatrix}$
$∴x = 7, y = -10$