Question

If $A^{-1} = \left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$ and $B^{-1} = \left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$, then $(AB)^{-1} =$

• A. $\left[\begin{array}{cc}2 & 7 \\ 3 & -1\end{array}\right]$
• B. $\left[\begin{array}{cc}2 & -7 \\ -3 & 11\end{array}\right]$
• C. $\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$
• D. $\left[\begin{array}{cc}2 & 3 \\ 7 & -11\end{array}\right]$

Hint:

The reversal law, $(AB)^{-1} = B^{-1}A^{-1}$ (and similarly for transposes: $(AB)^T = B^T A^T$), is a fundamental property of matrices. Multiplying them in the wrong order ($A^{-1}B^{-1}$) will lead you straight to a trap option!

Answer 1

The Correct Option is C

Step 1: Read what is asked.
We know $A^{-1}$ and $B^{-1}$ and we want $(AB)^{-1}$. We do not need to find $A$ or $B$ themselves.

Step 2: Recall the reversal rule.
The inverse of a product flips the order: \[ (AB)^{-1}=B^{-1}A^{-1}. \] So we multiply $B^{-1}$ first, then $A^{-1}$.

Step 3: Write down the product to compute.
\[ (AB)^{-1}=\left[\begin{array}{cc}1&0\\-3&1\end{array}\right]\left[\begin{array}{cc}2&-3\\-1&2\end{array}\right]. \]

Step 4: Multiply the first row.
Row 1, column 1: $(1)(2)+(0)(-1)=2$. Row 1, column 2: $(1)(-3)+(0)(2)=-3$.

Step 5: Multiply the second row.
Row 2, column 1: $(-3)(2)+(1)(-1)=-6-1=-7$. Row 2, column 2: $(-3)(-3)+(1)(2)=9+2=11$.

Step 6: Collect the result.
\[ (AB)^{-1}=\left[\begin{array}{cc}2&-3\\-7&11\end{array}\right]. \]

Step 7: Match the option.
This is option (3).
\[ \boxed{\left[\begin{array}{cc}2&-3\\-7&11\end{array}\right]} \]