Question:

The value of \(x\) such that the matrix \[ \begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix} \] is not invertible is

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A square matrix is invertible if and only if its determinant is non-zero. Always start such problems by setting the determinant equal to zero.
Updated On: Jan 26, 2026
  • \( -\dfrac{10}{7} \)
  • \( \dfrac{7}{10} \)
  • \( -\dfrac{7}{10} \)
  • \( \dfrac{10}{7} \)
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The Correct Option is D

Solution and Explanation

Step 1: Use the condition for non-invertibility.
A matrix is not invertible if and only if its determinant is zero. Hence, \[ \det \begin{bmatrix} x & 2 & 3 \\ 4 & 5 & 6 \\ 2 & 3 & 5 \end{bmatrix} = 0 \] Step 2: Evaluate the determinant.
Expanding along the first row, \[ \det = x(5\cdot5 - 6\cdot3) - 2(4\cdot5 - 6\cdot2) + 3(4\cdot3 - 5\cdot2) \] \[ = x(25 - 18) - 2(20 - 12) + 3(12 - 10) \] \[ = 7x - 16 + 6 \] Step 3: Solve for \(x\).
\[ 7x - 10 = 0 \Rightarrow x = \frac{10}{7} \] Step 4: Final conclusion.
The matrix is not invertible when \[ x = \frac{10}{7} \]
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