Given that `A=[{:(1,-1,0),(2,3,4),(0,1,2):}]` and `B=[{:(2,2,-4),(-4,2,-4),(2,-1,5):}]`. Find `AB`.
Use this to solve that following system of equations :
`x-y=3`, `2x+3y+4z=17`, `y+2z=7`.

Answer

To solve the problem step by step, we need to follow these procedures:

### Step 1: Define Matrices A and B
Given:
\[ A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \]
\[ B = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]

### Step 2: Calculate the Product AB
To find \( AB \), we multiply matrix \( A \) by matrix \( B \).

The element in the \( i^{th} \) row and \( j^{th} \) column of the product \( AB \) is calculated as follows:
\[ (AB)_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj} \]

#### Calculation of each element:
1. **First Row, First Column**:
\[ (AB)_{11} = 1 \cdot 2 + (-1) \cdot (-4) + 0 \cdot 2 = 2 + 4 + 0 = 6 \]

2. **First Row, Second Column**:
\[ (AB)_{12} = 1 \cdot 2 + (-1) \cdot 2 + 0 \cdot (-1) = 2 - 2 + 0 = 0 \]

3. **First Row, Third Column**:
\[ (AB)_{13} = 1 \cdot (-4) + (-1) \cdot (-4) + 0 \cdot 5 = -4 + 4 + 0 = 0 \]

4. **Second Row, First Column**:
\[ (AB)_{21} = 2 \cdot 2 + 3 \cdot (-4) + 4 \cdot 2 = 4 - 12 + 8 = 0 \]

5. **Second Row, Second Column**:
\[ (AB)_{22} = 2 \cdot 2 + 3 \cdot 2 + 4 \cdot (-1) = 4 + 6 - 4 = 6 \]

6. **Second Row, Third Column**:
\[ (AB)_{23} = 2 \cdot (-4) + 3 \cdot (-4) + 4 \cdot 5 = -8 - 12 + 20 = 0 \]

7. **Third Row, First Column**:
\[ (AB)_{31} = 0 \cdot 2 + 1 \cdot (-4) + 2 \cdot 2 = 0 - 4 + 4 = 0 \]

8. **Third Row, Second Column**:
\[ (AB)_{32} = 0 \cdot 2 + 1 \cdot 2 + 2 \cdot (-1) = 0 + 2 - 2 = 0 \]

9. **Third Row, Third Column**:
\[ (AB)_{33} = 0 \cdot (-4) + 1 \cdot (-4) + 2 \cdot 5 = 0 - 4 + 10 = 6 \]

### Result of AB:
Combining all the calculated elements, we have:
\[ AB = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I \]
where \( I \) is the identity matrix.

### Step 3: Solve the System of Equations
The given system of equations is:
1. \( x - y = 3 \) (Equation 1)
2. \( 2x + 3y + 4z = 17 \) (Equation 2)
3. \( y + 2z = 7 \) (Equation 3)

We can express this in matrix form \( AX = B \), where:
\[ A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \]

### Step 4: Find the Inverse of Matrix A
Since \( AB = 6I \), we can find \( A^{-1} \) as follows:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Given \( \text{det}(A) = 6 \), we have:
\[ A^{-1} = \frac{1}{6} \cdot \text{adj}(A) \]

### Step 5: Calculate \( A^{-1}B \)
To find \( X \):
\[ X = A^{-1}B \]
Substituting \( A^{-1} \):
\[ X = \frac{1}{6} \cdot \text{adj}(A) \cdot B \]
Calculating \( \text{adj}(A) \) from the previous steps, we have:
\[ \text{adj}(A) = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]

Now, calculate \( X \):
\[ X = \frac{1}{6} \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \]

Calculating each element of \( X \):
1. **First Element**:
\[ x = \frac{1}{6} (2 \cdot 3 + 2 \cdot 17 - 4 \cdot 7) = \frac{1}{6} (6 + 34 - 28) = \frac{12}{6} = 2 \]

2. **Second Element**:
\[ y = \frac{1}{6} (-4 \cdot 3 + 2 \cdot 17 - 4 \cdot 7) = \frac{1}{6} (-12 + 34 - 28) = \frac{-6}{6} = -1 \]

3. **Third Element**:
\[ z = \frac{1}{6} (2 \cdot 3 - 1 \cdot 17 + 5 \cdot 7) = \frac{1}{6} (6 - 17 + 35) = \frac{24}{6} = 4 \]

### Final Solution:
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = -1, \quad z = 4 \]