Verify that (AB)T = BTAT
Given:
\[ A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \]
To Verify:
\[ (AB)^T = B^T A^T \]
Step 1: Find AB
\[ AB = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} = \begin{bmatrix} (2\cdot1 + 1\cdot0 + 3\cdot5) & (2\cdot(-1) + 1\cdot2 + 3\cdot0) \\ (4\cdot1 + 1\cdot0 + 0\cdot5) & (4\cdot(-1) + 1\cdot2 + 0\cdot0) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 17 & 0 \\ 4 & -2 \end{bmatrix} \]
Step 2: Find (AB)T
\[ (AB)^T = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \]
Step 3: Find AT and BT
\[ A^T = \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix}, \quad B^T = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix} \]
Step 4: Find BTAT
\[ B^T A^T = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} (1\cdot2 + 0\cdot1 + 5\cdot3) & (1\cdot4 + 0\cdot1 + 5\cdot0) \\ (-1\cdot2 + 2\cdot1 + 0\cdot3) & (-1\cdot4 + 2\cdot1 + 0\cdot0) \end{bmatrix} \]
\[ B^T A^T = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \]
Conclusion:
\[ (AB)^T = B^T A^T \]
Hence Verified.