MHT CET · Maths · Matrices
If \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 1 & 3 & 4 \\ 3 & 4 & 3\end{array}\right]\), then \(A^{-1}=\)
- A \(-\frac{1}{4}\left[\begin{array}{ccc}-7 & -6 & -1 \\ 9 & 6 & -1 \\ -5 & -2 & 1\end{array}\right]\)
- B \(\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]\)
- C \(-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & 1 \\ 9 & -1 & 1 \\ -5 & 2 & 1\end{array}\right]\)
- D \(-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\(\because A^{-1}=\frac{\operatorname{adj}(A)}{|A|}=-\frac{1}{4}\left[\begin{array}{ccc}-7 & 6 & -1 \\ 9 & -6 & -1 \\ -5 & 2 & 1\end{array}\right]\)
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