For any square matrix $\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, where \(a\

Question

For any square matrix $\mathbf{A} = \begin{pmatrix} a & b \ c & d \end{pmatrix}$, where $a$, $b$, $c$ and $d$ are real numbers, the determinant of $\mathbf{A}$ is denoted by $|\mathbf{A}| = ad - bc.$ If $|\mathbf{A}| \neq 0$, the inverse matrix of $\mathbf{A}$ denoted by $\mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \begin{pmatrix} d & -b \ -c & a \end{pmatrix} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix},$ such that $\mathbf{AA}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}$, where $\mathbf{I}$ is the identity matrix. Find the inverse of: $\mathbf{A} = \begin{pmatrix} 2 & 1 \ 5 & -3 \end{pmatrix}$. Hence, solve the system of equations $\begin{aligned} 2x + y &= 5 \ 5x - 3y &= 7 \end{aligned}$

Accepted Answer

(i) $-\frac{1}{11}\left( \begin{matrix} -3 & -1 \ -5 & 2 \ \end{matrix} \right)$ (ii) $x=2$, $y=1$

Inverse Matrix

Inverse Matrix