Find `A^(-1)` if `A=|(0,1,1),(1,0,1),(1,1,0)|` and show that `A^(-1)=(A^(2)-3I)/2`

Explanation

To find the inverse of the matrix \( A \) and show that \( A^{-1} = \frac{A^2 - 3I}{2} \), we will follow these steps:

### Step 1: Calculate the Determinant of \( A \)

Given the matrix:
\[ A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \]

The determinant of \( A \) can be calculated using the formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the elements are:
- \( a = 0, b = 1, c = 1 \)
- \( d = 1, e = 0, f = 1 \)
- \( g = 1, h = 1, i = 0 \)

Calculating:
\[ \text{det}(A) = 0(0 \cdot 0 - 1 \cdot 1) - 1(1 \cdot 0 - 1 \cdot 1) + 1(1 \cdot 1 - 0 \cdot 1) \]
\[ = 0 - 1(-1) + 1(1) \]
\[ = 1 + 1 = 2 \]

### Step 2: Check if the Inverse Exists

Since the determinant \( \text{det}(A) = 2 \) is non-zero, the inverse of \( A \) exists.

### Step 3: Calculate the Adjoint of \( A \)

To find the inverse, we need the adjoint of \( A \). The adjoint is the transpose of the cofactor matrix.

#### Step 3.1: Calculate the Cofactors

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = (-1)^{1+1} \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 1(0 - 1) = -1 \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = (-1)^{1+2} \cdot \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = -1(0 - 1) = 1 \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = (-1)^{1+3} \cdot \text{det}\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = 1(1 - 0) = 1 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = (-1)^{2+1} \cdot \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = -1(0 - 1) = 1 \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = (-1)^{2+2} \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 1(0 - 1) = -1 \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = (-1)^{2+3} \cdot \text{det}\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix} = -1(0 - 1) = 1 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = (-1)^{3+1} \cdot \text{det}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = 1(1 - 0) = 1 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = (-1)^{3+2} \cdot \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = -1(0 - 1) = 1 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = (-1)^{3+3} \cdot \text{det}\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = 1(0 - 1) = -1 \]

#### Step 3.2: Form the Cofactor Matrix and Adjoint

The cofactor matrix is:
\[ \text{Cof}(A) = \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \]

The adjoint of \( A \) is the transpose of the cofactor matrix:
\[ \text{adj}(A) = \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \]

### Step 4: Calculate the Inverse of \( A \)

Using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
\[ A^{-1} = \frac{1}{2} \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \]
\[ = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 5: Show that \( A^{-1} = \frac{A^2 - 3I}{2} \)

#### Step 5.1: Calculate \( A^2 \)

\[ A^2 = A \cdot A = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} \]

Calculating each element:
- First row:
- \( (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 1) = 2 \)
- \( (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) = 1 \)
- \( (0 \cdot 1 + 1 \cdot 1 + 1 \cdot 0) = 1 \)

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

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

Thus,
\[ A^2 = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} \]

#### Step 5.2: Calculate \( 3I \)

The identity matrix \( I \) is:
\[ I = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ 3I = 3 \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} \]

#### Step 5.3: Calculate \( A^2 - 3I \)

\[ A^2 - 3I = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix} - \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} = \begin{pmatrix} 2 - 3 & 1 - 0 & 1 - 0 \\ 1 - 0 & 2 - 3 & 1 - 0 \\ 1 - 0 & 1 - 0 & 2 - 3 \end{pmatrix} \]
\[ = \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} \]

#### Step 5.4: Calculate \( \frac{A^2 - 3I}{2} \)

\[ \frac{A^2 - 3I}{2} = \frac{1}{2} \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \end{pmatrix} \]

### Conclusion

Thus, we have shown that:
\[ A^{-1} = \frac{A^2 - 3I}{2} \]