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: Recall
For invertible matrices, $(AB)^{-1} = B^{-1}A^{-1}$.

Step 2: Row by column
With $A^{-1} = \frac{1}{19}[[2,3],[5,-2]]$, the product row 1 of $B^{-1}$ (1, 2) times columns gives $1\cdot2+2\cdot5 = 12$ and $1\cdot3+2\cdot(-2) = -1$. Row 2 (2, -1) gives $2\cdot2-5 = -1$ and $2\cdot3+2 = 8$.

Step 3: Result
Scale factor $\frac1{5\cdot19} = \frac1{95}$. Option (B).

Final Answer:
Option B. \[ \boxed{\text{(B)}\ \frac1{95}\begin{pmatrix}12 & -1\\ -1 & 8\end{pmatrix}} \]