If A = ` ({:(1,-1,1),(2,-1,0),(1,0,0):})` find `A^(3)` , Hence find `A^(-1)` Use it to solve the following system of linear equation` x-y +z=1 ,2x -y=0 , x-4 =0 `

Answer

To solve the problem step by step, we will first find \( A^3 \) and then \( A^{-1} \). Finally, we will use \( A^{-1} \) to solve the system of linear equations.

### Step 1: Define the Matrix A
The matrix \( A \) is given as:
\[ A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \]

### Step 2: Calculate \( A^2 \)
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \]

Calculating the elements of \( A^2 \):
- First row:
- First column: \( 1 \cdot 1 + (-1) \cdot 2 + 1 \cdot 1 = 1 - 2 + 1 = 0 \)
- Second column: \( 1 \cdot (-1) + (-1) \cdot (-1) + 1 \cdot 0 = -1 + 1 + 0 = 0 \)
- Third column: \( 1 \cdot 1 + (-1) \cdot 0 + 1 \cdot 0 = 1 + 0 + 0 = 1 \)

- Second row:
- First column: \( 2 \cdot 1 + (-1) \cdot 2 + 0 \cdot 1 = 2 - 2 + 0 = 0 \)
- Second column: \( 2 \cdot (-1) + (-1) \cdot (-1) + 0 \cdot 0 = -2 + 1 + 0 = -1 \)
- Third column: \( 2 \cdot 1 + (-1) \cdot 0 + 0 \cdot 0 = 2 + 0 + 0 = 2 \)

- Third row:
- First column: \( 1 \cdot 1 + 0 \cdot 2 + 0 \cdot 1 = 1 + 0 + 0 = 1 \)
- Second column: \( 1 \cdot (-1) + 0 \cdot (-1) + 0 \cdot 0 = -1 + 0 + 0 = -1 \)
- Third column: \( 1 \cdot 1 + 0 \cdot 0 + 0 \cdot 0 = 1 + 0 + 0 = 1 \)

Thus, we have:
\[ A^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{pmatrix} \]

### Step 3: Calculate \( A^3 \)
Now, we multiply \( A^2 \) by \( A \):
\[ A^3 = A^2 \cdot A = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 1 \\ 2 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \]

Calculating the elements of \( A^3 \):
- First row:
- First column: \( 0 \cdot 1 + 0 \cdot 2 + 1 \cdot 1 = 0 + 0 + 1 = 1 \)
- Second column: \( 0 \cdot (-1) + 0 \cdot (-1) + 1 \cdot 0 = 0 + 0 + 0 = 0 \)
- Third column: \( 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 0 = 0 + 0 + 0 = 0 \)

- Second row:
- First column: \( 0 \cdot 1 + (-1) \cdot 2 + 2 \cdot 1 = 0 - 2 + 2 = 0 \)
- Second column: \( 0 \cdot (-1) + (-1) \cdot (-1) + 2 \cdot 0 = 0 + 1 + 0 = 1 \)
- Third column: \( 0 \cdot 1 + (-1) \cdot 0 + 2 \cdot 0 = 0 + 0 + 0 = 0 \)

- Third row:
- First column: \( 1 \cdot 1 + (-1) \cdot 2 + 1 \cdot 1 = 1 - 2 + 1 = 0 \)
- Second column: \( 1 \cdot (-1) + (-1) \cdot (-1) + 1 \cdot 0 = -1 + 1 + 0 = 0 \)
- Third column: \( 1 \cdot 1 + (-1) \cdot 0 + 1 \cdot 0 = 1 + 0 + 0 = 1 \)

Thus, we have:
\[ A^3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I \]

### Step 4: Find \( A^{-1} \)
Since \( A^3 = I \), we have:
\[ A^{-1} = A^2 \]

### Step 5: Calculate \( A^{-1} \)
From our previous calculation:
\[ A^{-1} = A^2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{pmatrix} \]

### Step 6: Solve the System of Equations
The system of equations is:
1. \( x - y + z = 1 \)
2. \( 2x - y = 0 \)
3. \( x - 4 = 0 \)

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

Let \( \mathbf{b} = \begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix} \).

Now, we can use \( A^{-1} \) to solve for \( \begin{pmatrix} x \\ y \\ z \end{pmatrix} \):
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \cdot \mathbf{b} \]

Calculating:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 2 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 4 \end{pmatrix} \]

Calculating the elements:
- First row: \( 0 \cdot 1 + 0 \cdot 0 + 1 \cdot 4 = 4 \)
- Second row: \( 0 \cdot 1 + (-1) \cdot 0 + 2 \cdot 4 = 8 \)
- Third row: \( 1 \cdot 1 + (-1) \cdot 0 + 1 \cdot 4 = 5 \)

Thus, we have:
\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 8 \\ 5 \end{pmatrix} \]

### Final Result
The solution to the system of equations is:
- \( x = 4 \)
- \( y = 8 \)
- \( z = 5 \)