Question

If \(A = [\begin{array}{cc}2 & 3 \\ 5 & -2\end{array}]\), \(B^{-1} = \left[ \begin{array}{cc}\frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & -\frac{1}{5}\end{array} \right]\), then \((AB)^{-1} =\)

• A. \(\frac{1}{95}[\begin{array}{cc}8 & -1 \\ -1 & 12\end{array}]\)
• B. \(\frac{1}{95}[\begin{array}{cc}12 & -1 \\ -1 & 8\end{array}]\)
• C. \(\frac{1}{95}[\begin{array}{cc}-12 & 1 \\ 1 & -8\end{array}]\)
• D. \(\frac{1}{95}[\begin{array}{cc}-8 & 1 \\ 1 & -12\end{array}]\)

Hint:

The inverse of a product reverses the order: (AB)^-1 = B^-1 A^-1.

Answer 1

The Correct Option is B

Step 1: Find A inverse
\(\det A = 2(-2) - 3(5) = -19\). The adjoint of \(A\) is \(\begin{pmatrix}-2 & -3\\ -5 & 2\end{pmatrix}\). So \(A^{-1} = \frac{1}{19}\begin{pmatrix}2 & 3\\ 5 & -2\end{pmatrix}\).

Step 2: Use the reversal law
\((AB)^{-1} = B^{-1}A^{-1} = \frac15\begin{pmatrix}1 & 2\\ 2 & -1\end{pmatrix}\cdot\frac1{19}\begin{pmatrix}2 & 3\\ 5 & -2\end{pmatrix}\).

Step 3: Multiply
\[ \frac{1}{95}\begin{pmatrix}2+10 & 3-4\\ 4-5 & 6+2\end{pmatrix} = \frac{1}{95}\begin{pmatrix}12 & -1\\ -1 & 8\end{pmatrix} \]

Step 4: Compare
This matches option (B). Option (A) is the product in the wrong order, \(A^{-1}B^{-1}\), which has different entries.

Final Answer:
The inverse of AB is (1/95) times the matrix with rows (12, -1) and (-1, 8). \[ \boxed{\text{(B)}\ \frac1{95}\begin{pmatrix}12 & -1\\ -1 & 8\end{pmatrix}} \]