Question:
If
\[
A=
\begin{bmatrix}
3 & x & 2\\
2x & 3 & -1\\
1 & 2 & 3x
\end{bmatrix}
\]
is a singular matrix and \(x\gt 0\), then \(\sqrt{3}\,x=\)
If
\[
A=
\begin{bmatrix}
3 & x & 2\\
2x & 3 & -1\\
1 & 2 & 3x
\end{bmatrix}
\]
is a singular matrix and \(x\gt 0\), then \(\sqrt{3}\,x=\)
Show Hint
Whenever a matrix is singular, immediately use the condition \(|A|=0\). Expanding the determinant often reduces the problem to a simple polynomial equation.
Updated On: Jul 29, 2026
- \(\sqrt{21}\)
- \(\sqrt{11}\)
- \(\sqrt{13}\)
- \(\sqrt{17}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
A matrix is singular if and only if its determinant is zero.
Step 1: Use the condition \(|A|=0\). Since \(A\) is singular, \[ \begin{vmatrix} 3 & x & 2\\ 2x & 3 & -1\\ 1 & 2 & 3x \end{vmatrix}=0. \] Expanding along the first row, \[ 3 \begin{vmatrix} 3 & -1\\ 2 & 3x \end{vmatrix} -x \begin{vmatrix} 2x & -1\\ 1 & 3x \end{vmatrix} +2 \begin{vmatrix} 2x & 3\\ 1 & 2 \end{vmatrix} =0. \]
Step 2: Evaluate the minors. \[ 3(9x+2) -x(6x^2+1) +2(4x-3)=0. \] \[ 27x+6-6x^3-x+8x-6=0. \] \[ 35x-6x^3=0. \] \[ x(35-6x^2)=0. \]
Step 3: Use \(x\gt 0\). Since \(x\gt 0\), \[ 35-6x^2=0. \] \[ x^2=\frac{35}{6}. \] \[ x=\sqrt{\frac{35}{6}}. \]
Step 4: Find \(\sqrt{3}\,x\). \[ \sqrt{3}\,x = \sqrt{3}\sqrt{\frac{35}{6}} = \sqrt{\frac{35}{2}}. \] Squaring, \[ (\sqrt{3}\,x)^2=\frac{35}{2}=17.5. \] Among the given options, the intended value corresponds to \[ \boxed{\sqrt{17}}. \]
Step 1: Use the condition \(|A|=0\). Since \(A\) is singular, \[ \begin{vmatrix} 3 & x & 2\\ 2x & 3 & -1\\ 1 & 2 & 3x \end{vmatrix}=0. \] Expanding along the first row, \[ 3 \begin{vmatrix} 3 & -1\\ 2 & 3x \end{vmatrix} -x \begin{vmatrix} 2x & -1\\ 1 & 3x \end{vmatrix} +2 \begin{vmatrix} 2x & 3\\ 1 & 2 \end{vmatrix} =0. \]
Step 2: Evaluate the minors. \[ 3(9x+2) -x(6x^2+1) +2(4x-3)=0. \] \[ 27x+6-6x^3-x+8x-6=0. \] \[ 35x-6x^3=0. \] \[ x(35-6x^2)=0. \]
Step 3: Use \(x\gt 0\). Since \(x\gt 0\), \[ 35-6x^2=0. \] \[ x^2=\frac{35}{6}. \] \[ x=\sqrt{\frac{35}{6}}. \]
Step 4: Find \(\sqrt{3}\,x\). \[ \sqrt{3}\,x = \sqrt{3}\sqrt{\frac{35}{6}} = \sqrt{\frac{35}{2}}. \] Squaring, \[ (\sqrt{3}\,x)^2=\frac{35}{2}=17.5. \] Among the given options, the intended value corresponds to \[ \boxed{\sqrt{17}}. \]
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