Question

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

• A. $\begin{bmatrix} 5 & -6 \\ -4 & 5 \end{bmatrix}$
• B. $\begin{bmatrix} 5 & 6 \\ 4 & 5 \end{bmatrix}$
• C. $\begin{bmatrix} -5 & 6 \\ -4 & 5 \end{bmatrix}$
• D. $\begin{bmatrix} -5 & -6 \\ -4 & -5 \end{bmatrix}$

Hint:

Notice that the determinant is $-1$. Multiplying a matrix by $-1$ flips the signs of all its components. Since the original product matrix $AB$ was $\begin{bmatrix} -5 & 6 \\ -4 & 5 \end{bmatrix}$, its inverse turns out to be identical to itself! This rare property means $AB$ is an involutory matrix.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a $2 \times 3$ matrix $A$ and a $3 \times 2$ matrix $B$. We need to compute their product matrix $AB$, verify its invertibility, and find its inverse matrix $(AB)^{-1}$.

Step 2: Key Formula or Approach:
First, calculate the product matrix $AB$ via standard row-column dot products. For any $2 \times 2$ matrix $M = \begin{bmatrix} w & x \\ y & z \end{bmatrix}$, its inverse is given by the standard formula: $$M^{-1} = \frac{1}{|M|} \cdot \text{adj } M = \frac{1}{wz - xy} \begin{bmatrix} z & -x \\ -y & w \end{bmatrix}$$

Step 3: Detailed Explanation:
Let's perform the matrix multiplication to find $AB$: $$AB = \begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 1 \\ 0 & 2 \end{bmatrix}$$ $$AB = \begin{bmatrix} (1)(1) + (2)(-3) + (1)(0) & (1)(2) + (2)(1) + (1)(2) \\ (-1)(1) + (1)(-3) + (3)(0) & (-1)(2) + (1)(1) + (3)(2) \end{bmatrix}$$ $$AB = \begin{bmatrix} 1 - 6 + 0 & 2 + 2 + 2 \\ -1 - 3 + 0 & -2 + 1 + 6 \end{bmatrix} = \begin{bmatrix} -5 & 6 \\ -4 & 5 \end{bmatrix}$$ Now, compute the determinant of this product matrix $|AB|$: $$|AB| = \begin{vmatrix} -5 & 6 \\ -4 & 5 \end{vmatrix} = (-5)(5) - (6)(-4) = -25 - (-24) = -25 + 24 = -1$$ Next, find the adjoint matrix $\text{adj}(AB)$ by swapping diagonal elements and reversing the signs of the off-diagonal elements: $$\text{adj}(AB) = \begin{bmatrix} 5 & -6 \\ 4 & -5 \end{bmatrix}$$ Finally, calculate the inverse matrix $(AB)^{-1}$: $$(AB)^{-1} = \frac{1}{|AB|} \cdot \text{adj}(AB) = \frac{1}{-1} \begin{bmatrix} 5 & -6 \\ 4 & -5 \end{bmatrix} = \begin{bmatrix} -5 & 6 \\ -4 & 5 \end{bmatrix}$$

Step 4: Final Answer:
The inverse matrix $(AB)^{-1}$ is $\begin{bmatrix} -5 & 6 \\ -4 & 5 \end{bmatrix}$, which corresponds to option (C).