Verify that (AB)^T = B^TA^T

Given:

\( A = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix} \)

To Verify:

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

Step 1: Find AB

\[ AB = \begin{bmatrix} 2 & -3 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} 2(1) + (-3)(2) & 2(0) + (-3)(-4) \\ -7(1) + 5(2) & -7(0) + 5(-4) \end{bmatrix} = \begin{bmatrix} -4 & 12 \\ 3 & -20 \end{bmatrix} \]

Step 2: Find (AB)^T

\[ (AB)^T = \begin{bmatrix} -4 & 3 \\ 12 & -20 \end{bmatrix} \]

Step 3: Find A^T and B^T

\[ A^T = \begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix}, \quad B^T = \begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix} \]

Step 4: Find B^TA^T

\[ B^T A^T = \begin{bmatrix} 1 & 2 \\ 0 & -4 \end{bmatrix} \begin{bmatrix} 2 & -7 \\ -3 & 5 \end{bmatrix} = \begin{bmatrix} 1(2) + 2(-3) & 1(-7) + 2(5) \\ 0(2) + (-4)(-3) & 0(-7) + (-4)(5) \end{bmatrix} = \begin{bmatrix} -4 & 3 \\ 12 & -20 \end{bmatrix} \]

Conclusion:

\[ (AB)^T = B^T A^T \]

Hence Verified.