To find the inverse of a matrix \( A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \), we need to use the formula for the inverse of a 3x3 matrix.

The formula for the inverse of a 3x3 matrix \( A \) is:

\(A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)\)

where \(\text{adj}(A)\) is the adjugate of matrix \( A \) and \(\text{det}(A)\) is the determinant of matrix \( A \).

Step 1: Find the Determinant of Matrix \( A \)

Calculate the determinant of matrix \( A \) using the formula:

\(\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\)

For matrix \( A \),

\(a = 1, b = 0, c = 1, d = -1, e = 1, f = 1, g = 0, h = 1, i = 0\)

Substitute these values into the determinant formula:

\(\text{det}(A) = 1(1 \cdot 0 - 1 \cdot 1) - 0(-1 \cdot 0 - 1 \cdot 1) + 1(-1 \cdot 1 - 1 \cdot 0)\)

\( = 1(0 - 1) + 0 + 1(-1) \)

\( = -1 - 1 = -2 \)

Step 2: Find the Adjugate of Matrix \( A \)

The adjugate is the transpose of the cofactor matrix. Calculate the cofactor matrix by taking the determinant of minors:

\(\text{adj}(A) = \begin{bmatrix} \text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\ \text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\ \text{C}_{31} & \text{C}_{32} & \text{C}_{33} \end{bmatrix}\)

Where \(\text{C}_{ij} = (-1)^{i+j} M_{ij}\) and \(M_{ij}\) is the minor of the element \( a_{ij} \).

After computing, the adjugate matrix for \( A \) is:

\(\begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{bmatrix}\)

Step 3: Calculate the Inverse

Substitute the determinant and adjugate into the inverse formula:

\(A^{-1} = \frac{1}{-2} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{bmatrix}\)

This simplifies to:

\(\frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{bmatrix}\)

Conclusion

The correct answer is:

\(\frac{1}{2} \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 2 \\ 1 & 1 & -1 \end{bmatrix}\)