Question:
If \( A = \begin{pmatrix} x & x-1 \\ 2x & 1 \end{pmatrix} \) and if \( \det A = -9 \), then the values of \(x\) are
If \( A = \begin{pmatrix} x & x-1 \\ 2x & 1 \end{pmatrix} \) and if \( \det A = -9 \), then the values of \(x\) are
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Always expand determinant carefully before forming quadratic equations.
Updated On: May 8, 2026
- \( \frac{3}{2}, -3 \)
- \( -\frac{2}{3}, 3 \)
- \( \frac{2}{3}, 3 \)
- \( -\frac{3}{2}, 3 \)
- \( -\frac{3}{2}, -3 \)
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The Correct Option is D
Solution and Explanation
Concept:
Determinant of \(2\times2\) matrix:
\[
\det A = ad - bc
\]
Step 1: Compute determinant
\[ \det A = x \cdot 1 - (x-1)(2x) \] \[ = x - 2x(x-1) \]
Step 2: Expand
\[ = x - 2x^2 + 2x \] \[ = 3x - 2x^2 \]
Step 3: Given condition
\[ 3x - 2x^2 = -9 \]
Step 4: Rearrangement
\[ 2x^2 - 3x - 9 = 0 \]
Step 5: Solve quadratic
\[ x = \frac{3 \pm \sqrt{9 + 72}}{4} \] \[ = \frac{3 \pm 9}{4} \] \[ x = 3,\; -\frac{3}{2} \] \[ \boxed{-\frac{3}{2}, 3} \]
Step 1: Compute determinant
\[ \det A = x \cdot 1 - (x-1)(2x) \] \[ = x - 2x(x-1) \]
Step 2: Expand
\[ = x - 2x^2 + 2x \] \[ = 3x - 2x^2 \]
Step 3: Given condition
\[ 3x - 2x^2 = -9 \]
Step 4: Rearrangement
\[ 2x^2 - 3x - 9 = 0 \]
Step 5: Solve quadratic
\[ x = \frac{3 \pm \sqrt{9 + 72}}{4} \] \[ = \frac{3 \pm 9}{4} \] \[ x = 3,\; -\frac{3}{2} \] \[ \boxed{-\frac{3}{2}, 3} \]
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