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: Understanding the Question:
We are given two matrices $A$ and $B$ of dimension $2 \times 2$. We need to compute the composite matrix configuration $(B^{-1}A^{-1})^{-1}$.

Step 2: Key Formula or Approach:
According to the reversal property of matrix inversion, the inverse of a product of matrices is the product of their individual inverses in reverse order: $$(X \cdot Y)^{-1} = Y^{-1} \cdot X^{-1}$$ Applying this principle to our target expression with $X = B^{-1}$ and $Y = A^{-1}$: $$(B^{-1}A^{-1})^{-1} = (A^{-1})^{-1} \cdot (B^{-1})^{-1} = A \cdot B$$ This simplification bypasses calculating separate matrix inverses entirely and reduces the task to simple matrix multiplication!

Step 3: Detailed Explanation:
Let's compute the matrix product $A \cdot B$ directly using standard row-by-column multiplication rules: $$A \cdot B = \begin{bmatrix} 2 & -2 \\ 2 & -3 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$ Let's evaluate each entry systematically:

• Row 1, Column 1: $(2)(0) + (-2)(1) = 0 - 2 = -2$

• Row 1, Column 2: $(2)(-1) + (-2)(0) = -2 + 0 = -2$

• Row 2, Column 1: $(2)(0) + (-3)(1) = 0 - 3 = -3$

• Row 2, Column 2: $(2)(-1) + (-3)(0) = -2 + 0 = -2$
Assembling these entries together gives the product matrix: $$A \cdot B = \begin{bmatrix} -2 & -2 \\ -3 & -2 \end{bmatrix}$$ This matches option (A).

Step 4: Final Answer:
The value of the expression is $\begin{bmatrix} -2 & -2 \\ -3 & -2 \end{bmatrix}$, which corresponds to option (A).