Question:
If \(B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}\) is the inverse of a \(3 \times 3\) matrix A, then the sum of all values of \(\alpha\) for which det(A) + 1 = 0, is :
If \(B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}\) is the inverse of a \(3 \times 3\) matrix A, then the sum of all values of \(\alpha\) for which det(A) + 1 = 0, is :
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Since \(B=A^{-1}\), \(\det A=1/\det B\). Set \(\det A=-1\), i.e. \(\det B=-1\), and use the sum of roots of the resulting quadratic in \(\alpha\).
Updated On: Nov 5, 2026
- \(2\)
- \(-1\)
- \(0\)
- \(1\)
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The Correct Option is D
Solution and Explanation
Expand \(\det B\) along the first row: \(5(2\cdot(-1)-1\cdot3)-2\alpha(0\cdot(-1)-1\cdot\alpha)+1(0\cdot3-2\alpha)\)
\(=5(-5)+2\alpha^2-2\alpha=2\alpha^2-2\alpha-25\).
\(\det A=\frac{1}{\det B}\). Condition \(\det A+1=0\Rightarrow\det A=-1\Rightarrow\det B=-1\).
\(2\alpha^2-2\alpha-25=-1\Rightarrow2\alpha^2-2\alpha-24=0\Rightarrow\alpha^2-\alpha-12=0\Rightarrow\alpha=4,-3\).
Sum of values \(=4+(-3)=1\).
Answer: (4) \(1\)
\(=5(-5)+2\alpha^2-2\alpha=2\alpha^2-2\alpha-25\).
\(\det A=\frac{1}{\det B}\). Condition \(\det A+1=0\Rightarrow\det A=-1\Rightarrow\det B=-1\).
\(2\alpha^2-2\alpha-25=-1\Rightarrow2\alpha^2-2\alpha-24=0\Rightarrow\alpha^2-\alpha-12=0\Rightarrow\alpha=4,-3\).
Sum of values \(=4+(-3)=1\).
Answer: (4) \(1\)
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