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 problem step by step, we will first calculate the product \( BA \) and then use it to solve the system of equations.

### Step 1: Define the matrices \( A \) and \( B \)

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

### Step 2: Calculate the product \( BA \)

To find \( BA \), we multiply matrix \( B \) by matrix \( A \):

\[ BA = B \cdot A = \begin{pmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{pmatrix} \cdot \begin{pmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{pmatrix} \]

Calculating each element of the resulting matrix:

- First row:
- \( (1 \cdot 2) + (-1 \cdot -4) + (0 \cdot 2) = 2 + 4 + 0 = 6 \)
- \( (1 \cdot 2) + (-1 \cdot 2) + (0 \cdot -1) = 2 - 2 + 0 = 0 \)
- \( (1 \cdot -4) + (-1 \cdot -4) + (0 \cdot 5) = -4 + 4 + 0 = 0 \)

- Second row:
- \( (2 \cdot 2) + (3 \cdot -4) + (4 \cdot 2) = 4 - 12 + 8 = 0 \)
- \( (2 \cdot 2) + (3 \cdot 2) + (4 \cdot -1) = 4 + 6 - 4 = 6 \)
- \( (2 \cdot -4) + (3 \cdot -4) + (4 \cdot 5) = -8 - 12 + 20 = 0 \)

- Third row:
- \( (0 \cdot 2) + (1 \cdot -4) + (2 \cdot 2) = 0 - 4 + 4 = 0 \)
- \( (0 \cdot 2) + (1 \cdot 2) + (2 \cdot -1) = 0 + 2 - 2 = 0 \)
- \( (0 \cdot -4) + (1 \cdot -4) + (2 \cdot 5) = 0 - 4 + 10 = 6 \)

Putting it all together, we have:
\[ BA = \begin{pmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{pmatrix} = 6I \]
where \( I \) is the identity matrix.

### Step 3: Finding the inverse of \( B \)

Since \( BA = 6I \), we can find the inverse of \( B \):
\[ B^{-1} = \frac{1}{6} A \]

### Step 4: Set up the system of equations

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

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

### Step 5: Solve for \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)

Using the inverse of \( B \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = B^{-1} \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} = \frac{1}{6} A \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} \]

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

Calculating each element:
- First row: \( 2 \cdot 3 + 2 \cdot 17 - 4 \cdot 7 = 6 + 34 - 28 = 12 \)
- Second row: \( -4 \cdot 3 + 2 \cdot 17 - 4 \cdot 7 = -12 + 34 - 28 = -6 \)
- Third row: \( 2 \cdot 3 - 1 \cdot 17 + 5 \cdot 7 = 6 - 17 + 35 = 24 \)

Thus:
\[ A \begin{pmatrix} 3 \\ 17 \\ 7 \end{pmatrix} = \begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} \]

Now, multiplying by \( \frac{1}{6} \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{6} \begin{pmatrix} 12 \\ -6 \\ 24 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} \]

### Final Solution

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