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: Define the Matrix \( A \)
Given:
\[
A = \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\]
### Step 2: Calculate the Determinant of \( A \)
The determinant of a 3x3 matrix \( A = \begin{pmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{pmatrix} \) is calculated as:
\[
\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
\]
For our matrix \( A \):
\[
\text{det}(A) = 0 \cdot (0 \cdot 0 - 1 \cdot 1) - 1 \cdot (1 \cdot 0 - 1 \cdot 1) + 1 \cdot (1 \cdot 1 - 0 \cdot 1)
\]
Calculating each term:
\[
= 0 - 1 \cdot (-1) + 1 \cdot 1 = 1 + 1 = 2
\]
### Step 3: Calculate the Cofactor Matrix
To find the cofactor matrix, we calculate the cofactor for each element of \( A \).
1. **Cofactor \( C_{11} \)**:
\[
C_{11} = (-1)^{1+1} \cdot \text{det}\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} = 1 \cdot (0 \cdot 0 - 1 \cdot 1) = -1
\]
2. **Cofactor \( C_{12} \)**:
\[
C_{12} = (-1)^{1+2} \cdot \text{det}\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix} = -1 \cdot (1 \cdot 0 - 1 \cdot 1) = 1
\]
3. **Cofactor \( C_{13} \)**:
\[
C_{13} = (-1)^{1+3} \cdot \text{det}\begin{pmatrix}
1 & 0 \\
1 & 1
\end{pmatrix} = 1 \cdot (1 \cdot 1 - 0 \cdot 1) = 1
\]
Continuing this process for all elements, we find the cofactor matrix:
\[
\text{Cofactor}(A) = \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix}
\]
### Step 4: Calculate the Adjoint of \( A \)
The adjoint of \( A \) is the transpose of the cofactor matrix:
\[
\text{adj}(A) = \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix}^T = \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix}
\]
### Step 5: Calculate the Inverse of \( A \)
Using the formula for the inverse:
\[
A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} = \frac{1}{2} \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix}
\]
Thus,
\[
A^{-1} = \begin{pmatrix}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & -\frac{1}{2}
\end{pmatrix}
\]
### Step 6: Calculate \( A^2 \)
Now we compute \( A^2 \):
\[
A^2 = A \cdot A = \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix} \cdot \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}
\]
Calculating the product:
- First row: \( (0, 1, 1) \cdot (0, 1, 1) = 2 \), \( (0, 1, 1) \cdot (1, 0, 1) = 1 \), \( (0, 1, 1) \cdot (1, 1, 0) = 1 \)
- Second row: \( (1, 0, 1) \cdot (0, 1, 1) = 1 \), \( (1, 0, 1) \cdot (1, 0, 1) = 2 \), \( (1, 0, 1) \cdot (1, 1, 0) = 1 \)
- Third row: \( (1, 1, 0) \cdot (0, 1, 1) = 1 \), \( (1, 1, 0) \cdot (1, 0, 1) = 1 \), \( (1, 1, 0) \cdot (1, 1, 0) = 2 \)
Thus,
\[
A^2 = \begin{pmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{pmatrix}
\]
### Step 7: Calculate \( A^2 - 3I \)
Where \( I \) is the identity matrix:
\[
3I = \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix}
\]
Now,
\[
A^2 - 3I = \begin{pmatrix}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{pmatrix} - \begin{pmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{pmatrix} = \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix}
\]
### Step 8: Calculate \( \frac{A^2 - 3I}{2} \)
Finally,
\[
\frac{A^2 - 3I}{2} = \frac{1}{2} \begin{pmatrix}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{pmatrix} = \begin{pmatrix}
-\frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & -\frac{1}{2}
\end{pmatrix}
\]
### Conclusion
Thus, we have shown that:
\[
A^{-1} = \frac{A^2 - 3I}{2}
\]
