The given matrix must be a square matrix. We know the determinant of a 2x2 matrix is calculated as: \[ \text{det} = (2x - 1)(y^2 - 1) - 0 \times 3x = (2x - 1)(y^2 - 1) \] Setting the determinant equal to 0: \[ (2x - 1)(y^2 - 1) = 0 \] This gives two possible cases: 1. $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ 2. $y^2 - 1 = 0 \Rightarrow y = \pm 1$ Thus, solving these cases yields two possible values for $x - y$: - For $x = \frac{1}{2}$ and $y = 1$, $x - y = \frac{1}{2} - 1 = -\frac{1}{2}$ (which is not a valid solution) - For $x = 2$ and $y = 1$, $x - y = 2 - 1 = 1$ The correct result for $x - y$ will be 2 or -10.