To show that \((AB)' = B'A'\), we will follow these steps:
### Step 1: Define the matrices A and B
Given:
\[
A = \begin{pmatrix} 1 & 0 & -3 \\ 2 & 3 & 4 \\ -4 & 5 & -2 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & 0 & -1 \\ 2 & 5 & -4 \\ 4 & -1 & 2 \end{pmatrix}
\]
### Step 2: Calculate the product \(AB\)
To find \(AB\), we will multiply matrix \(A\) by matrix \(B\).
1. **First Row, First Column**:
\[
(1 \cdot 3) + (0 \cdot 2) + (-3 \cdot 4) = 3 + 0 - 12 = -9
\]
2. **First Row, Second Column**:
\[
(1 \cdot 0) + (0 \cdot 5) + (-3 \cdot -1) = 0 + 0 + 3 = 3
\]
3. **First Row, Third Column**:
\[
(1 \cdot -1) + (0 \cdot -4) + (-3 \cdot 2) = -1 + 0 - 6 = -7
\]
4. **Second Row, First Column**:
\[
(2 \cdot 3) + (3 \cdot 2) + (4 \cdot 4) = 6 + 6 + 16 = 28
\]
5. **Second Row, Second Column**:
\[
(2 \cdot 0) + (3 \cdot 5) + (4 \cdot -1) = 0 + 15 - 4 = 11
\]
6. **Second Row, Third Column**:
\[
(2 \cdot -1) + (3 \cdot -4) + (4 \cdot 2) = -2 - 12 + 8 = -6
\]
7. **Third Row, First Column**:
\[
(-4 \cdot 3) + (5 \cdot 2) + (-2 \cdot 4) = -12 + 10 - 8 = -10
\]
8. **Third Row, Second Column**:
\[
(-4 \cdot 0) + (5 \cdot 5) + (-2 \cdot -1) = 0 + 25 + 2 = 27
\]
9. **Third Row, Third Column**:
\[
(-4 \cdot -1) + (5 \cdot -4) + (-2 \cdot 2) = 4 - 20 - 4 = -20
\]
Thus, the product \(AB\) is:
\[
AB = \begin{pmatrix} -9 & 3 & -7 \\ 28 & 11 & -6 \\ -10 & 27 & -20 \end{pmatrix}
\]
### Step 3: Calculate the transpose of \(AB\)
To find \((AB)'\), we will transpose the matrix \(AB\):
\[
(AB)' = \begin{pmatrix} -9 & 28 & -10 \\ 3 & 11 & 27 \\ -7 & -6 & -20 \end{pmatrix}
\]
### Step 4: Calculate the transpose of B and A
1. **Transpose of B**:
\[
B' = \begin{pmatrix} 3 & 2 & 4 \\ 0 & 5 & -1 \\ -1 & -4 & 2 \end{pmatrix}
\]
2. **Transpose of A**:
\[
A' = \begin{pmatrix} 1 & 2 & -4 \\ 0 & 3 & 5 \\ -3 & 4 & -2 \end{pmatrix}
\]
### Step 5: Calculate the product \(B'A'\)
Now we will multiply \(B'\) and \(A'\):
1. **First Row, First Column**:
\[
(3 \cdot 1) + (2 \cdot 0) + (4 \cdot -3) = 3 + 0 - 12 = -9
\]
2. **First Row, Second Column**:
\[
(3 \cdot 2) + (2 \cdot 3) + (4 \cdot 4) = 6 + 6 + 16 = 28
\]
3. **First Row, Third Column**:
\[
(3 \cdot -4) + (2 \cdot 5) + (4 \cdot -2) = -12 + 10 - 8 = -10
\]
4. **Second Row, First Column**:
\[
(0 \cdot 1) + (5 \cdot 0) + (-1 \cdot -3) = 0 + 0 + 3 = 3
\]
5. **Second Row, Second Column**:
\[
(0 \cdot 2) + (5 \cdot 3) + (-1 \cdot 4) = 0 + 15 - 4 = 11
\]
6. **Second Row, Third Column**:
\[
(0 \cdot -4) + (5 \cdot 5) + (-1 \cdot -2) = 0 + 25 + 2 = 27
\]
7. **Third Row, First Column**:
\[
(-1 \cdot 1) + (-4 \cdot 0) + (2 \cdot -3) = -1 + 0 - 6 = -7
\]
8. **Third Row, Second Column**:
\[
(-1 \cdot 2) + (-4 \cdot 3) + (2 \cdot 4) = -2 - 12 + 8 = -6
\]
9. **Third Row, Third Column**:
\[
(-1 \cdot -4) + (-4 \cdot 5) + (2 \cdot -2) = 4 - 20 - 4 = -20
\]
Thus, the product \(B'A'\) is:
\[
B'A' = \begin{pmatrix} -9 & 28 & -10 \\ 3 & 11 & 27 \\ -7 & -6 & -20 \end{pmatrix}
\]
### Step 6: Compare \((AB)'\) and \(B'A'\)
We have:
\[
(AB)' = \begin{pmatrix} -9 & 28 & -10 \\ 3 & 11 & 27 \\ -7 & -6 & -20 \end{pmatrix}
\]
\[
B'A' = \begin{pmatrix} -9 & 28 & -10 \\ 3 & 11 & 27 \\ -7 & -6 & -20 \end{pmatrix}
\]
Since \((AB)' = B'A'\), we have shown that the statement is true.
