COMEDK · Maths · 21. Matrices
If \(A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right] \quad B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]\) then \((A B)^{-1}\) is equal to
- A \(\left[\begin{array}{ccc}
-2 & -2 & -3 \\
19 & 18 & 29 \\
-27 & -25 & -42
\end{array}\right]\) - B \(\left[\begin{array}{ccc}
-2 & 19 & -27 \\
-2 & 18 & -25 \\
3 & 29 & 42
\end{array}\right]\) - C \(\left[\begin{array}{lll}
2 & -19 & 27 \\
2 & -18 & 25 \\
3 & -29 & 42
\end{array}\right]\) - D \(\left[\begin{array}{lll}
-2 & 19 & -27 \\
-2 & 18 & -25 \\
-3 & 29 & -42
\end{array}\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\begin{array}{lll}
-2 & 19 & -27 \\
-2 & 18 & -25 \\
-3 & 29 & -42
\end{array}\right]\)
Step-by-step Solution
Detailed explanation
We know that \((AB)^{-1} = B^{-1} A^{-1}\). First, calculate \(A^{-1}\). The determinant of \(A\) is \(|A| = 5(3-4) - 0(2-2) + 4(4-3) = 5(-1) + 4(1) = -5 + 4 = -1\). The cofactor matrix of \(A\) is…
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