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:
Compute (AB)⁻¹ for given matrices A (2×3) and B (3×2).

Step 2: Key Formula or Approach:
Multiply matrices, then for a 2×2 matrix M, M⁻¹ = (1/|M|)·adj(M).

Step 3: Detailed Explanation:
AB = [[–5,6],[–4,5]]. Determinant = (–5)(5)–(6)(–4) = –25+24 = –1. Adjoint = [[5,–6],[4,–5]]. Inverse = (1/–1)·adj = [[–5,6],[–4,5]].

Step 4: Final Answer:
(AB)⁻¹ = [[–5,6],[–4,5]], matching option (C).