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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

  1. A \(\left[\begin{array}{ccc}
    -2 & -2 & -3 \\
    19 & 18 & 29 \\
    -27 & -25 & -42
    \end{array}\right]\)
  2. B \(\left[\begin{array}{ccc}
    -2 & 19 & -27 \\
    -2 & 18 & -25 \\
    3 & 29 & 42
    \end{array}\right]\)
  3. C \(\left[\begin{array}{lll}
    2 & -19 & 27 \\
    2 & -18 & 25 \\
    3 & -29 & 42
    \end{array}\right]\)
  4. D \(\left[\begin{array}{lll}
    -2 & 19 & -27 \\
    -2 & 18 & -25 \\
    -3 & 29 & -42
    \end{array}\right]\)
Verified Solution

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…