To solve the problem, we will first compute the product of matrices A and B, and then use the result to solve the system of equations.
### Step 1: Matrix Multiplication
Given:
\[ A = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]
\[ B = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]
We need to find \( AB \).
The product of two matrices \( A \) and \( B \) is calculated by taking the dot product of the rows of \( A \) with the columns of \( B \).
#### Calculation of \( AB \):
1. **First Row of A with Columns of B**:
- First element: \( (2 \cdot 1) + (2 \cdot 2) + (-4 \cdot 0) = 2 + 4 + 0 = 6 \)
- Second element: \( (2 \cdot -1) + (2 \cdot 3) + (-4 \cdot 1) = -2 + 6 - 4 = 0 \)
- Third element: \( (2 \cdot 0) + (2 \cdot 4) + (-4 \cdot 2) = 0 + 8 - 8 = 0 \)
So, the first row of \( AB \) is \( (6, 0, 0) \).
2. **Second Row of A with Columns of B**:
- First element: \( (-4 \cdot 1) + (2 \cdot 2) + (-4 \cdot 0) = -4 + 4 + 0 = 0 \)
- Second element: \( (-4 \cdot -1) + (2 \cdot 3) + (-4 \cdot 1) = 4 + 6 - 4 = 6 \)
- Third element: \( (-4 \cdot 0) + (2 \cdot 4) + (-4 \cdot 2) = 0 + 8 - 8 = 0 \)
So, the second row of \( AB \) is \( (0, 6, 0) \).
3. **Third Row of A with Columns of B**:
- First element: \( (2 \cdot 1) + (-1 \cdot 2) + (5 \cdot 0) = 2 - 2 + 0 = 0 \)
- Second element: \( (2 \cdot -1) + (-1 \cdot 3) + (5 \cdot 1) = -2 - 3 + 5 = 0 \)
- Third element: \( (2 \cdot 0) + (-1 \cdot 4) + (5 \cdot 2) = 0 - 4 + 10 = 6 \)
So, the third row of \( AB \) is \( (0, 0, 6) \).
Combining all rows, we get:
\[ AB = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} \]
### Step 2: Solve the System of Equations
The equations given are:
1. \( x - y = 3 \) (Equation 1)
2. \( 2x + 3y + 4z = 17 \) (Equation 2)
3. \( y + 2z = 7 \) (Equation 3)
We can express these equations in matrix form:
\[
\begin{pmatrix}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
=
\begin{pmatrix}
3 \\
17 \\
7
\end{pmatrix}
\]
Let \( M = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \) and \( N = \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \).
To find \( X \) (which contains \( x, y, z \)), we need to find \( M^{-1}N \).
Since we found that \( AB = 6I \), we know that \( B^{-1} = \frac{1}{6}A \).
Thus, we can write:
\[ X = B^{-1}N = \frac{1}{6}A \cdot N \]
Calculating \( A \cdot N \):
\[ A \cdot N = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \]
1. First element: \( (2 \cdot 3) + (2 \cdot 17) + (-4 \cdot 7) = 6 + 34 - 28 = 12 \)
2. Second element: \( (-4 \cdot 3) + (2 \cdot 17) + (-4 \cdot 7) = -12 + 34 - 28 = -6 \)
3. Third element: \( (2 \cdot 3) + (-1 \cdot 17) + (5 \cdot 7) = 6 - 17 + 35 = 24 \)
So, \( A \cdot N = \begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} \).
Now, we calculate \( X \):
\[ X = \frac{1}{6} \begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \]
Thus, the solution is:
\[ x = 2, \quad y = -1, \quad z = 4 \]
### Final Answer:
The solution to the equations is:
- \( x = 2 \)
- \( y = -1 \)
- \( z = 4 \)
