COMEDK · Maths · 21. Matrices
If the product of the matrix \(B=\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]\)
with a matrix \(A\) has the inverse \(C=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\), then \(A^{-1}\) equals
- A \(\left[\begin{array}{ccc}-3 & -5 & 5 \\ 0 & 9 & 14 \\ 2 & 2 & 9\end{array}\right]\)
- B \(\left[\begin{array}{ccc}-3 & 5 & 5 \\ 0 & 0 & 9 \\ 2 & 14 & 16\end{array}\right]\)
- C \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 0 & 2 \\ 2 & 14 & 6\end{array}\right]\)
- D \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{ccc}-3 & -5 & -5 \\ 0 & 9 & 2 \\ 2 & 14 & 6\end{array}\right]\)
Step-by-step Solution
Detailed explanation
\((B A)^{-1}=C\) (given) or \(A^{-1} B^{-1}=C\) or \(A^{-1}\left[\begin{array}{ccc}2 & 6 & 4 \\ 1 & 0 & 1 \\ -1 & 1 & -1\end{array}\right]^{-1}=\left[\begin{array}{ccc}-1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2\end{array}\right]\) Multiply by \(B\) on both sides, we get…
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