Question:

If $\begin{bmatrix} 2x-1 & 3x \\ 0 & y^2 - 1 \end{bmatrix}$ is a matrix, then the value of $(x - y)$ is:

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For solving determinant problems with 2x2 matrices, factorize the expression and check for roots.
Updated On: Jun 23, 2025
  • 2 or 10
  • -2 or 10
  • 2 or -10
  • 0 or 10
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The Correct Option is C

Solution and Explanation

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.
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