Verify that (A – B)^T = A^T – B^T

Given:

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

To Verify:

\( (A – B)^T = A^T – B^T \)

Step 1: Find A – B

\[ A – B = \begin{bmatrix} 2-1 & -3-0 \\ -7-2 & 5-(-4) \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ -9 & 9 \end{bmatrix} \]

Step 2: Find (A – B)^T

\[ (A – B)^T = \begin{bmatrix} 1 & -9 \\ -3 & 9 \end{bmatrix} \]

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 A^T – B^T

\[ A^T – B^T = \begin{bmatrix} 2-1 & -7-2 \\ -3-0 & 5-(-4) \end{bmatrix} = \begin{bmatrix} 1 & -9 \\ -3 & 9 \end{bmatrix} \]

Conclusion:

\[ (A – B)^T = A^T – B^T \]

Hence Verified.