If `A=|(2,2,-4),(-4,2,-4),(2,-1,5)|` and `B=|(1,-1,0),(2,3,4),(0,1,2)|` then find BA and use ths to sovle the system of equations `y+2z=7, x-y=3` and `2x+3y+4z=17`.

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Calculate the product \( BA \)

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

We need to compute \( BA \):
\[ BA = B \cdot A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]

Calculating each element of the resulting matrix:

1. **First row, first column**:
\[ 1 \cdot 2 + (-1) \cdot (-4) + 0 \cdot 2 = 2 + 4 + 0 = 6 \]

2. **First row, second column**:
\[ 1 \cdot 2 + (-1) \cdot 2 + 0 \cdot (-1) = 2 - 2 + 0 = 0 \]

3. **First row, third column**:
\[ 1 \cdot (-4) + (-1) \cdot (-4) + 0 \cdot 5 = -4 + 4 + 0 = 0 \]

4. **Second row, first column**:
\[ 2 \cdot 2 + 3 \cdot (-4) + 4 \cdot 2 = 4 - 12 + 8 = 0 \]

5. **Second row, second column**:
\[ 2 \cdot 2 + 3 \cdot 2 + 4 \cdot (-1) = 4 + 6 - 4 = 6 \]

6. **Second row, third column**:
\[ 2 \cdot (-4) + 3 \cdot (-4) + 4 \cdot 5 = -8 - 12 + 20 = 0 \]

7. **Third row, first column**:
\[ 0 \cdot 2 + 1 \cdot (-4) + 2 \cdot 2 = 0 - 4 + 4 = 0 \]

8. **Third row, second column**:
\[ 0 \cdot 2 + 1 \cdot 2 + 2 \cdot (-1) = 0 + 2 - 2 = 0 \]

9. **Third row, third column**:
\[ 0 \cdot (-4) + 1 \cdot (-4) + 2 \cdot 5 = 0 - 4 + 10 = 6 \]

Putting it all together, we have:
\[ BA = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I \]

### Step 2: Use the result to solve the system of equations

The system of equations is:
1. \( y + 2z = 7 \)
2. \( x - y = 3 \)
3. \( 2x + 3y + 4z = 17 \)

We can express this in matrix form as:
\[ \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 2 \\ 2 & 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 7 \\ 17 \end{pmatrix} \]

Let \( B \) be the coefficient matrix and \( X \) be the variable matrix:
\[ B = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 2 \\ 2 & 3 & 4 \end{pmatrix}, \quad Z = \begin{pmatrix} -3 \\ 7 \\ 17 \end{pmatrix} \]

From the earlier calculation, we know that:
\[ BA = 6I \implies B^{-1} = \frac{1}{6}A \]

Thus, we can find \( X \):
\[ X = B^{-1}Z = \frac{1}{6}A \cdot Z \]

Calculating \( A \cdot Z \):
\[ A \cdot Z = \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \begin{pmatrix} -3 \\ 7 \\ 17 \end{pmatrix} \]

Calculating each element:
1. First row:
\[ 2 \cdot (-3) + 2 \cdot 7 + (-4) \cdot 17 = -6 + 14 - 68 = -60 \]
2. Second row:
\[ -4 \cdot (-3) + 2 \cdot 7 + (-4) \cdot 17 = 12 + 14 - 68 = -42 \]
3. Third row:
\[ 2 \cdot (-3) + (-1) \cdot 7 + 5 \cdot 17 = -6 - 7 + 85 = 72 \]

Thus:
\[ A \cdot Z = \begin{pmatrix} -60 \\ -42 \\ 72 \end{pmatrix} \]

Now, multiplying by \( \frac{1}{6} \):
\[ X = \frac{1}{6} \begin{pmatrix} -60 \\ -42 \\ 72 \end{pmatrix} = \begin{pmatrix} -10 \\ -7 \\ 12 \end{pmatrix} \]

### Final Result:
The solution to the system of equations is:
\[ x = -10, \quad y = -7, \quad z = 12 \]