For a real number a, if the system `{:[(1,a,a^2),(a,1,a),(a^2,a,1)][(x),(y),(z)]=[(1),(-1),(1)]:}`
of the linear equations, has infinitely many solutions, then `1+a+a^2=`

Options

A. 1
B. 0
C. -1
D. 2

Explanation

To solve the problem, we need to analyze the given system of equations represented by the matrix and find the value of \(1 + a + a^2\) under the condition that the system has infinitely many solutions.

### Step-by-Step Solution:

1. **Identify the Matrix and the System of Equations**:
The system of equations is represented by the matrix:
\[ \begin{bmatrix} 1 & a & a^2 \\ a & 1 & a \\ a^2 & a & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \]
For the system to have infinitely many solutions, the determinant of the coefficient matrix must be zero.

2. **Calculate the Determinant**:
The determinant of the matrix can be calculated using the formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ \text{det}(A) = 1 \cdot (1 \cdot 1 - a \cdot a) - a \cdot (a \cdot 1 - a^2 \cdot 1) + a^2 \cdot (a \cdot a - 1 \cdot 1) \]
Simplifying this gives:
\[ = 1(1 - a^2) - a(a - a^2) + a^2(a^2 - 1) \]
\[ = 1 - a^2 - a^2 + a^3 + a^4 - a^2 \]
\[ = 1 - 3a^2 + a^3 + a^4 \]

3. **Set the Determinant to Zero**:
For the system to have infinitely many solutions, we set the determinant equal to zero:
\[ 1 - 3a^2 + a^3 + a^4 = 0 \]
Rearranging gives:
\[ a^4 + a^3 - 3a^2 + 1 = 0 \]

4. **Factor the Polynomial**:
We can try to factor this polynomial. By testing possible rational roots, we find that \(a = 1\) and \(a = -1\) are roots:
\[ (a^2 - 1)^2 = 0 \]
This gives us:
\[ a^2 - 1 = 0 \implies a = \pm 1 \]

5. **Check Each Root**:
- If \(a = 1\):
The system becomes:
\[ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \]
This represents the same plane, leading to no solutions.
- If \(a = -1\):
The system becomes:
\[ \begin{bmatrix} 1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \]
This represents different planes, leading to infinitely many solutions.

6. **Calculate \(1 + a + a^2\)**:
Now substituting \(a = -1\):
\[ 1 + a + a^2 = 1 - 1 + 1 = 1 \]

### Final Answer:
Thus, the value of \(1 + a + a^2\) is:
\[ \boxed{1} \]