The rank of matrix `[(x, -1, 0), (0, x, -1), (-1, 0,x)]` is `2` then value of `x` is:

Options

A. 3
B. 2
C. 1
D. none of these

Answer

To find the value of \( x \) for which the rank of the matrix

\[ A = \begin{pmatrix} x & -1 & 0 \\ 0 & x & -1 \\ -1 & 0 & x \end{pmatrix} \]

is 2, we need to compute the determinant of the matrix and set it equal to zero. This is because if the rank of a matrix is less than its order (which is 3 in this case), then its determinant must be zero.

### Step-by-Step Solution:

1. **Calculate the Determinant of the Matrix:**
The determinant of matrix \( A \) can be calculated using the formula for the determinant of a 3x3 matrix:

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

For our matrix:

\[ A = \begin{pmatrix} x & -1 & 0 \\ 0 & x & -1 \\ -1 & 0 & x \end{pmatrix} \]

We can expand the determinant along the first row:

\[ \text{det}(A) = x \begin{vmatrix} x & -1 \\ 0 & x \end{vmatrix} - (-1) \begin{vmatrix} 0 & -1 \\ -1 & x \end{vmatrix} + 0 \begin{vmatrix} 0 & x \\ -1 & 0 \end{vmatrix} \]

This simplifies to:

\[ = x (x \cdot x - (-1) \cdot 0) + 1 (0 \cdot x - (-1) \cdot -1) \]

\[ = x^3 + 1 \]

2. **Set the Determinant Equal to Zero:**
To find the values of \( x \) for which the rank is 2, we set the determinant equal to zero:

\[ x^3 + 1 = 0 \]

This can be rewritten as:

\[ x^3 = -1 \]

Taking the cube root of both sides gives:

\[ x = -1 \]

3. **Verify the Result:**
We can verify that \( x = -1 \) is indeed a solution by substituting it back into the determinant equation:

\[ \text{det}(A) = (-1)^3 + 1 = -1 + 1 = 0 \]

Since the determinant is zero, the rank of the matrix is less than 3, confirming that the rank is indeed 2.

### Conclusion:
The value of \( x \) for which the rank of the matrix is 2 is:

\[ \boxed{-1} \]