Question

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

• A. $\begin{bmatrix} -2 & -2 \\ -3 & -2 \end{bmatrix}$
• B. $\begin{bmatrix} 2 & 2 \\ -2 & -3 \end{bmatrix}$
• C. $\begin{bmatrix} 3 & -2 \\ 2 & 2 \end{bmatrix}$
• D. $\begin{bmatrix} 1 & -1 \\ -2 & 3 \end{bmatrix}$

Hint:

Always apply the reversal property $(XY)^{-1} = Y^{-1}X^{-1}$ to clean up matrix operations early. Turning a difficult double-inverse question into a single matrix product saves immense time and keeps your calculations free from messy fractional determinants.

Answer 1

The Correct Option is A

Step 1: Read the target.
We need $(B^{-1}A^{-1})^{-1}$ for the given $2\times 2$ matrices $A$ and $B$.
Step 2: Apply the reversal law.
Taking the inverse of a product reverses the order: $(B^{-1}A^{-1})^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB$.
Step 3: Note the simplification.
So we only need the product $AB$, with no inverse to compute at all.
Step 4: Multiply the matrices.
$AB = \begin{bmatrix} 2 & -2 \\ 2 & -3 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Step 5: Compute each entry.
Top row: $(2\cdot0 + (-2)\cdot1,\ 2\cdot(-1) + (-2)\cdot0) = (-2, -2)$. Bottom row: $(2\cdot0 + (-3)\cdot1,\ 2\cdot(-1) + (-3)\cdot0) = (-3, -2)$.
Step 6: Assemble.
The result is $\begin{bmatrix} -2 & -2 \\ -3 & -2 \end{bmatrix}$, obtained without ever inverting a matrix.
\[ \boxed{\begin{bmatrix} -2 & -2 \\ -3 & -2 \end{bmatrix}} \]