Question

If

\[ A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \]

then verify that \( (AB)^T = B^T A^T \).

Hint:

The property \( (AB)^T = B^T A^T \) holds for matrix multiplication and transposition.

Answer 1

Step 1: Compute \( AB \).

First, compute the product \( AB \):

\[ AB = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \]

Performing the matrix multiplication:

\[ AB = \begin{bmatrix} 2(1) + 1(0) + 3(5) & 2(-1) + 1(2) + 3(0) \\ 4(1) + 1(0) + 0(5) & 4(-1) + 1(2) + 0(0) \end{bmatrix} \]

\[ AB = \begin{bmatrix} 17 & 0 \\ 4 & -2 \end{bmatrix} \]

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

The transpose of \( AB \) is:

\[ (AB)^T = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \]

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

First compute the transposes:

\[ B^T = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix}, \quad A^T = \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix} \]

Now compute the product \( B^T A^T \):

\[ B^T A^T = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix} \]

\[ B^T A^T = \begin{bmatrix} 1(2) + 0(1) + 5(3) & 1(4) + 0(1) + 5(0) \\ -1(2) + 2(1) + 0(3) & -1(4) + 2(1) + 0(0) \end{bmatrix} \]

\[ B^T A^T = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \]

Step 4: Conclusion.

Since \[ (AB)^T = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} = B^T A^T, \] the identity \( (AB)^T = B^T A^T \) is verified.