Verify that (AB)T = BTAT
Given:
\[ A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} \]
To Verify:
\[ (AB)^T = B^T A^T \]
Step 1: Find AB
\[ AB = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 5 \end{bmatrix} = \begin{bmatrix} (1\cdot1 + 3\cdot2) & (1\cdot4 + 3\cdot5) \\ (2\cdot1 + 4\cdot2) & (2\cdot4 + 4\cdot5) \end{bmatrix} \]
\[ AB = \begin{bmatrix} 7 & 19 \\ 10 & 28 \end{bmatrix} \]
Step 2: Find (AB)T
\[ (AB)^T = \begin{bmatrix} 7 & 10 \\ 19 & 28 \end{bmatrix} \]
Step 3: Find AT and BT
\[ A^T = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B^T = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \]
Step 4: Find BTAT
\[ B^T A^T = \begin{bmatrix} 1 & 2 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} (1\cdot1 + 2\cdot3) & (1\cdot2 + 2\cdot4) \\ (4\cdot1 + 5\cdot3) & (4\cdot2 + 5\cdot4) \end{bmatrix} \]
\[ B^T A^T = \begin{bmatrix} 7 & 10 \\ 19 & 28 \end{bmatrix} \]
Conclusion:
\[ (AB)^T = B^T A^T \]
Hence Verified.