Question

If \( A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \) then verify that \( (AB)' = B'A' \)

Hint:

For any product of matrices \( ABC \dots Z \), the transpose is always the reverse product of transposes: \( (ABC \dots Z)' = Z' \dots C'B'A' \).

Answer 1

Step 1: Understanding the Concept:
The problem asks to verify the reversal law for transposes, which states that the transpose of the product of two matrices is equal to the product of their transposes in reverse order.
Step 2: Key Formula or Approach:
1. Find the product \( AB \).
2. Find the transpose of the product, \( (AB)' \).
3. Find the transposes of individual matrices \( A' \) and \( B' \).
4. Multiply \( B' \) and \( A' \) and compare the results.
Step 3: Detailed Explanation:
Given \( A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \).
First, compute \( AB \):
\[ AB = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix} \] \[ AB = \begin{bmatrix} (2 \times 1 + 1 \times 0 + 3 \times 5) & (2 \times -1 + 1 \times 2 + 3 \times 0) \\ (4 \times 1 + 1 \times 0 + 0 \times 5) & (4 \times -1 + 1 \times 2 + 0 \times 0) \end{bmatrix} \] \[ AB = \begin{bmatrix} 2+0+15 & -2+2+0 \\ 4+0+0 & -4+2+0 \end{bmatrix} = \begin{bmatrix} 17 & 0 \\ 4 & -2 \end{bmatrix} \] Now, find \( (AB)' \) by swapping rows and columns:
\[ (AB)' = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \] ... (i)
Next, find \( A' \) and \( B' \):
\[ A' = \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix}, \quad B' = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix} \] Compute \( B'A' \):
\[ B'A' = \begin{bmatrix} 1 & 0 & 5 \\ -1 & 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 1 & 1 \\ 3 & 0 \end{bmatrix} \] \[ B'A' = \begin{bmatrix} (1 \times 2 + 0 \times 1 + 5 \times 3) & (1 \times 4 + 0 \times 1 + 5 \times 0) \\ (-1 \times 2 + 2 \times 1 + 0 \times 3) & (-1 \times 4 + 2 \times 1 + 0 \times 0) \end{bmatrix} \] \[ B'A' = \begin{bmatrix} 2+0+15 & 4+0+0 \\ -2+2+0 & -4+2+0 \end{bmatrix} = \begin{bmatrix} 17 & 4 \\ 0 & -2 \end{bmatrix} \] ... (ii)
Step 4: Final Answer:
From (i) and (ii), we see that \( (AB)' = B'A' \).
Thus, the property is verified.