If `A= (1)/(9) ((-8,1,4),(4,4,7),(1,-8,4))`, prove that, `A^(-1)= A^(T)`. Hence solve the following system of equations:
`-8x + y + 4z= 0, 4x + 4y + 7z= 18, x- 8y + 4z= 9`

Answer

To solve the problem step by step, we will first prove that \( A^{-1} = A^{T} \) and then use this result to solve the system of equations.

### Step 1: Find the Transpose of Matrix \( A \)

Given:
\[ A = \frac{1}{9} \begin{pmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{pmatrix} \]

The transpose of \( A \), denoted \( A^{T} \), is obtained by swapping rows with columns:
\[ A^{T} = \frac{1}{9} \begin{pmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{pmatrix} \]

### Step 2: Multiply \( A \) and \( A^{T} \)

Next, we need to compute the product \( A \cdot A^{T} \):
\[ A \cdot A^{T} = \left( \frac{1}{9} \begin{pmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{pmatrix} \right) \cdot \left( \frac{1}{9} \begin{pmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{pmatrix} \right) \]

Calculating the product:
1. The element at (1,1):
\[ = \frac{1}{81} \left( (-8)(-8) + (1)(1) + (4)(4) \right) = \frac{1}{81} (64 + 1 + 16) = \frac{81}{81} = 1 \]

2. The element at (1,2):
\[ = \frac{1}{81} \left( (-8)(4) + (1)(4) + (4)(7) \right) = \frac{1}{81} (-32 + 4 + 28) = \frac{0}{81} = 0 \]

3. The element at (1,3):
\[ = \frac{1}{81} \left( (-8)(1) + (1)(-8) + (4)(4) \right) = \frac{1}{81} (-8 - 8 + 16) = \frac{0}{81} = 0 \]

4. The element at (2,1):
\[ = \frac{1}{81} \left( (4)(-8) + (4)(1) + (7)(4) \right) = \frac{1}{81} (-32 + 4 + 28) = \frac{0}{81} = 0 \]

5. The element at (2,2):
\[ = \frac{1}{81} \left( (4)(4) + (4)(4) + (7)(7) \right) = \frac{1}{81} (16 + 16 + 49) = \frac{81}{81} = 1 \]

6. The element at (2,3):
\[ = \frac{1}{81} \left( (4)(1) + (4)(-8) + (7)(4) \right) = \frac{1}{81} (4 - 32 + 28) = \frac{0}{81} = 0 \]

7. The element at (3,1):
\[ = \frac{1}{81} \left( (1)(-8) + (-8)(1) + (4)(4) \right) = \frac{1}{81} (-8 - 8 + 16) = \frac{0}{81} = 0 \]

8. The element at (3,2):
\[ = \frac{1}{81} \left( (1)(4) + (-8)(4) + (4)(7) \right) = \frac{1}{81} (4 - 32 + 28) = \frac{0}{81} = 0 \]

9. The element at (3,3):
\[ = \frac{1}{81} \left( (1)(1) + (-8)(-8) + (4)(4) \right) = \frac{1}{81} (1 + 64 + 16) = \frac{81}{81} = 1 \]

Thus, we have:
\[ A \cdot A^{T} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \]

### Step 3: Conclude that \( A^{-1} = A^{T} \)

Since \( A \cdot A^{T} = I \), we conclude that:
\[ A^{-1} = A^{T} \]

### Step 4: Solve the System of Equations

Now we can solve the system of equations:
\[ \begin{align*} -8x + y + 4z &= 0 \\ 4x + 4y + 7z &= 18 \\ x - 8y + 4z &= 9 \end{align*} \]

We can represent this system in matrix form as:
\[ \begin{pmatrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 18 \\ 9 \end{pmatrix} \]

Using \( A^{-1} = A^{T} \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{T} \begin{pmatrix} 0 \\ 18 \\ 9 \end{pmatrix} \]

Calculating:
\[ A^{T} = \frac{1}{9} \begin{pmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{pmatrix} \]

Now multiply \( A^{T} \) with the constants:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{9} \begin{pmatrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{pmatrix} \begin{pmatrix} 0 \\ 18 \\ 9 \end{pmatrix} \]

Calculating each component:
1. For \( x \):
\[ x = \frac{1}{9} \left( -8 \cdot 0 + 4 \cdot 18 + 1 \cdot 9 \right) = \frac{1}{9} (0 + 72 + 9) = \frac{81}{9} = 9 \]

2. For \( y \):
\[ y = \frac{1}{9} \left( 1 \cdot 0 + 4 \cdot 18 - 8 \cdot 9 \right) = \frac{1}{9} (0 + 72 - 72) = \frac{0}{9} = 0 \]

3. For \( z \):
\[ z = \frac{1}{9} \left( 4 \cdot 0 + 7 \cdot 18 + 4 \cdot 9 \right) = \frac{1}{9} (0 + 126 + 36) = \frac{162}{9} = 18 \]

Thus, the solution to the system of equations is:
\[ x = 9, \quad y = 0, \quad z = 18 \]