MHT CET · Maths · Matrices
If \((B A)^{-1}=C\) where \(B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]\) and \(C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\), then \(A^{-1}\) is given by
- A \(\left[\begin{array}{ccc}-3 & -3 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
- B \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
- C \(\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 6\end{array}\right]\)
- D \(\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
Answer & Solution
Correct Answer
(B) \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\((B A)^{-1}=C\)
\(\begin{array}{l}\Rightarrow A^{-1} B^{-1}=C \\ \Rightarrow A^{-1} B^{-1} B=C B \\ \Rightarrow A^{-1}=C B\end{array}\)
\(\Rightarrow A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]=\) \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
\(\begin{array}{l}\Rightarrow A^{-1} B^{-1}=C \\ \Rightarrow A^{-1} B^{-1} B=C B \\ \Rightarrow A^{-1}=C B\end{array}\)
\(\Rightarrow A^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]=\) \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
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