Solution

The system of equations can be written in the form \(AX = B\), where \(A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}\), \(X = \begin{bmatrix} x \\ y \end{bmatrix}\) and \(B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}\). Now, \(|A| = 4 - 15 = -11 \neq 0\). Hence, A is nonsingular matrix and so a unique solution exists. \(A^{-1} = -\frac{1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\) \(X = A^{-1}B = -\frac{1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 7 \end{bmatrix}\) \(\begin{bmatrix} x \\ y \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} 2-35 \\ -3+14 \end{bmatrix} = -\frac{1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}\) Hence \(x=3, y=-1\).