If `A=[(1,-2,0),(2,1,3),(0,-2,1)] and B=[(7,2,-6),(-2,1,-3),(-4,2,5)]` , find AB
Hence , solve the system of equation
x-2y=10, 2x+y+3z=8 and -2y+z=7.

Answer

To solve the problem step by step, we will first find the product of matrices A and B, and then solve the given system of equations.

### Step 1: Calculate the product of matrices A and B

Given:
\[ A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{pmatrix} \]
\[ B = \begin{pmatrix} 7 & 2 & -6 \\ -2 & 1 & -3 \\ -4 & 2 & 5 \end{pmatrix} \]

To find \( AB \), we will multiply each row of A by each column of B.

1. **First row of A with each column of B:**
- First column: \( 1 \cdot 7 + (-2) \cdot (-2) + 0 \cdot (-4) = 7 + 4 + 0 = 11 \)
- Second column: \( 1 \cdot 2 + (-2) \cdot 1 + 0 \cdot 2 = 2 - 2 + 0 = 0 \)
- Third column: \( 1 \cdot (-6) + (-2) \cdot (-3) + 0 \cdot 5 = -6 + 6 + 0 = 0 \)

So, the first row of \( AB \) is \( (11, 0, 0) \).

2. **Second row of A with each column of B:**
- First column: \( 2 \cdot 7 + 1 \cdot (-2) + 3 \cdot (-4) = 14 - 2 - 12 = 0 \)
- Second column: \( 2 \cdot 2 + 1 \cdot 1 + 3 \cdot 2 = 4 + 1 + 6 = 11 \)
- Third column: \( 2 \cdot (-6) + 1 \cdot (-3) + 3 \cdot 5 = -12 - 3 + 15 = 0 \)

So, the second row of \( AB \) is \( (0, 11, 0) \).

3. **Third row of A with each column of B:**
- First column: \( 0 \cdot 7 + (-2) \cdot (-2) + 1 \cdot (-4) = 0 + 4 - 4 = 0 \)
- Second column: \( 0 \cdot 2 + (-2) \cdot 1 + 1 \cdot 2 = 0 - 2 + 2 = 0 \)
- Third column: \( 0 \cdot (-6) + (-2) \cdot (-3) + 1 \cdot 5 = 0 + 6 + 5 = 11 \)

So, the third row of \( AB \) is \( (0, 0, 11) \).

Thus, the product \( AB \) is:
\[ AB = \begin{pmatrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{pmatrix} = 11I \]
where \( I \) is the identity matrix.

### Step 2: Solve the system of equations

The system of equations is:
1. \( x - 2y = 10 \) (Equation 1)
2. \( 2x + y + 3z = 8 \) (Equation 2)
3. \( -2y + z = 7 \) (Equation 3)

We can express this in matrix form \( AX = B \), where:
\[ A = \begin{pmatrix} 1 & -2 & 0 \\ 2 & 1 & 3 \\ 0 & -2 & 1 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

To solve for \( X \), we can use the inverse of matrix \( A \):
\[ X = A^{-1}B \]

From the previous calculation, we found that:
\[ AB = 11I \]
Thus, we can find \( A^{-1} \):
\[ A^{-1} = \frac{1}{11}B \]

Now, we can calculate \( X \):
\[ X = A^{-1}B = \frac{1}{11}B \]

Calculating \( B \):
\[ B = \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} \]

Now, we multiply:
\[ X = \frac{1}{11} \begin{pmatrix} 10 \\ 8 \\ 7 \end{pmatrix} = \begin{pmatrix} \frac{10}{11} \\ \frac{8}{11} \\ \frac{7}{11} \end{pmatrix} \]

### Final Solution

Thus, the solution to the system of equations is:
\[ x = \frac{10}{11}, \quad y = \frac{8}{11}, \quad z = \frac{7}{11} \]