Question

Verify \( (AB)^T = B^T A^T \).

Given:

\[ A = \begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix} \]

Hint:

The transpose of a product of two matrices \( AB \) is the product of their transposes in reverse order: \( (AB)^T = B^T A^T \).

Answer 1

Step 1: Compute \( AB \).

First, multiply the matrices \( A \) and \( B \):

\[ AB = \begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} 4 & 2 \\ 3 & 1 \end{bmatrix} \]

Performing the matrix multiplication:

\[ AB = \begin{bmatrix} 1\cdot4 + 2\cdot3 & 1\cdot2 + 2\cdot1 \\ 3\cdot4 + (-4)\cdot3 & 3\cdot2 + (-4)\cdot1 \end{bmatrix} = \begin{bmatrix} 10 & 4 \\ 0 & 2 \end{bmatrix} \]

Step 2: Compute \( (AB)^T \).

\[ (AB)^T = \begin{bmatrix} 10 & 0 \\ 4 & 2 \end{bmatrix} \]

Step 3: Compute \( A^T \) and \( B^T \).

\[ A^T = \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix}, \quad B^T = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \]

Step 4: Compute \( B^T A^T \).

\[ B^T A^T = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & -4 \end{bmatrix} \]

\[ B^T A^T = \begin{bmatrix} 4\cdot1 + 3\cdot2 & 4\cdot3 + 3\cdot(-4) \\ 2\cdot1 + 1\cdot2 & 2\cdot3 + 1\cdot(-4) \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 4 & 2 \end{bmatrix} \]

Step 5: Conclusion.

Since \[ (AB)^T = \begin{bmatrix} 10 & 0 \\ 4 & 2 \end{bmatrix} = B^T A^T, \] the identity \( (AB)^T = B^T A^T \) is verified.