Question:
If matrix \(\begin{pmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{pmatrix}\) is non-invertible matrix and distinct values of \(x\) are \(a\) and \(b\) then \(9(a^{2}+b^{2})=\cdots \cdots \cdots\)
If matrix \(\begin{pmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{pmatrix}\) is non-invertible matrix and distinct values of \(x\) are \(a\) and \(b\) then \(9(a^{2}+b^{2})=\cdots \cdots \cdots\)
Show Hint
A matrix is non-invertible when its determinant is zero; form the quadratic in x and use sum and product of roots.
Updated On: Nov 12, 2026
- \(271\)
- \(721\)
- \(217\)
- \(-278\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
A matrix is non-invertible exactly when its determinant is zero.
Step 2: Expand the determinant
\(x(8-3x)-1(4+3x)+2(6+12)=0\)
\(8x-3x^{2}-4-3x+36=0\).
Step 3: Form the quadratic
\(-3x^{2}+5x+32=0\), that is \(3x^{2}-5x-32=0\).
So \(a+b=\dfrac{5}{3}\) and \(ab=-\dfrac{32}{3}\).
Step 4: Calculation
\(a^{2}+b^{2}=(a+b)^{2}-2ab=\dfrac{25}{9}+\dfrac{64}{3}=\dfrac{217}{9}\).
Hence \(9(a^{2}+b^{2})=217\).
Final Answer:
The value of 9(a squared plus b squared) is 217. \[ \boxed{\text{(C) }217} \]
A matrix is non-invertible exactly when its determinant is zero.
Step 2: Expand the determinant
\(x(8-3x)-1(4+3x)+2(6+12)=0\)
\(8x-3x^{2}-4-3x+36=0\).
Step 3: Form the quadratic
\(-3x^{2}+5x+32=0\), that is \(3x^{2}-5x-32=0\).
So \(a+b=\dfrac{5}{3}\) and \(ab=-\dfrac{32}{3}\).
Step 4: Calculation
\(a^{2}+b^{2}=(a+b)^{2}-2ab=\dfrac{25}{9}+\dfrac{64}{3}=\dfrac{217}{9}\).
Hence \(9(a^{2}+b^{2})=217\).
Final Answer:
The value of 9(a squared plus b squared) is 217. \[ \boxed{\text{(C) }217} \]
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