MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{cc}\lambda & i \\ i & -\lambda\end{array}\right]\) and \(A^{-1}\) does not exist, then \(\lambda=(\) where \(I=\sqrt{-1})\)
- A \(\pm 2\)
- B \(\pm 1\)
- C 0
- D \(\pm 3\)
Answer & Solution
Correct Answer
(B) \(\pm 1\)
Step-by-step Solution
Detailed explanation
\(A=\left[\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right] \Rightarrow|\mathrm{A}|=\left|\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right|\)
Since \(\mathrm{A}^{-1}\) does not exist, we write \(|\mathrm{A}|=0\)
\(\therefore\left(-\lambda^2\right)-\left(\mathrm{i}^2\right)=0 \quad \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\)
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right] \Rightarrow|\mathrm{A}|=\left|\begin{array}{cc}
\lambda & \mathrm{i} \\
\mathrm{i} & -\lambda
\end{array}\right|\)
Since \(\mathrm{A}^{-1}\) does not exist, we write \(|\mathrm{A}|=0\)
\(\therefore\left(-\lambda^2\right)-\left(\mathrm{i}^2\right)=0 \quad \Rightarrow \lambda^2=1 \Rightarrow \lambda= \pm 1\)
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