MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{cc}\mathrm{k} & 2 \\ -2 & -\mathrm{k}\end{array}\right]\), then \(\mathrm{A}^{-1}\) does not exists if \(\mathrm{k}=\)
- A \(3\)
- B \(\pm 2\)
- C \(0\)
- D \(\pm 1\)
Answer & Solution
Correct Answer
(B) \(\pm 2\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& A=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right] \\
& \therefore|\mathrm{A}|=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right]=-\mathrm{k}^2+4
\end{aligned}
\)
When \(-\mathrm{k}^2+4=0\), we get \(\mathrm{k}= \pm 2\)
\begin{aligned}
& A=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right] \\
& \therefore|\mathrm{A}|=\left[\begin{array}{cc}
\mathrm{k} & 2 \\
-2 & -\mathrm{k}
\end{array}\right]=-\mathrm{k}^2+4
\end{aligned}
\)
When \(-\mathrm{k}^2+4=0\), we get \(\mathrm{k}= \pm 2\)
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