Question

If \(A^{-1}=\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\) and \(B=\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}\), then \((AB)^{-1}=\)

• A. \(\begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}\)
• B. \(\begin{bmatrix} 9 & 3 & -5 \\ -2 & -1 & 0 \\ 1 & 0 & 2 \end{bmatrix}\)
• C. \(\begin{bmatrix} 9 & -3 & 5 \\ -2 & -1 & 0 \\ 1 & 0 & -2 \end{bmatrix}\)
• D. \(\begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & -2 \end{bmatrix}\)

Hint:

Use the reversal law (AB) inverse equals B inverse times A inverse and find the inverse of B using its adjoint.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The reversal law gives \((AB)^{-1}=B^{-1}A^{-1}\). The inverse of \(A\) is already given, so only \(B^{-1}\) is needed.

Step 2: Find the inverse of B
\(|B|=1(3)-2(-1)+(-2)(2)=1\), so \(B^{-1}=\text{adj}B=\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}\).

Step 3: Multiply
\(B^{-1}A^{-1}\) has first row \((9-30+30,\ -3+12-12,\ 3-10+12)=(9,-3,5)\).
Continuing gives \(\begin{bmatrix}9&-3&5\\-2&1&0\\1&0&2\end{bmatrix}\).

Final Answer:
The matrix \((AB)^{-1}\) is the first option. \[ \boxed{\text{(A) }\begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}} \]