JEE Mains · Maths · STD 12 - 3 and 4 . metrices and determinant
If \(A=\left[\begin{array}{cc}1 & 5 \\ \lambda & 10\end{array}\right], A ^{-1}=\alpha A+\beta I\) and \(\alpha+\beta=-2\) then \(4 \alpha^2+\beta^2+\lambda^2\) is equal to:
- A \(12\)
- B \(10\)
- C \(19\)
- D \(14\)
Answer & Solution
Correct Answer
(D) \(14\)
Step-by-step Solution
Detailed explanation
\(|A-x I|=0 \Rightarrow\left|\begin{array}{cc}1-x & 5 \\ \lambda & 10-x\end{array}\right|=0\) \(\Rightarrow x^2-11 x+10-5 \lambda=0\) \(\Rightarrow(10-5 \lambda) A^{-1}=-A+11 I\)…
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