The system of equations `ax-y- z=a-1 ,x-ay-z=a-1,x-y-az=a-1 ` has no solution if `a` is:

Options

A. either -2 or 1
B. -2
C. 1
D. not(-2)

Answer

To determine the values of \( a \) for which the system of equations has no solution, we need to analyze the given equations and find the determinant of the coefficient matrix. Here are the steps to solve the problem:

### Step 1: Write the system of equations in matrix form

The given equations are:
1. \( ax - y - z = a - 1 \)
2. \( x - ay - z = a - 1 \)
3. \( x - y - az = a - 1 \)

We can express this system in the form \( A \mathbf{x} = \mathbf{b} \), where:
\[ A = \begin{bmatrix} a & -1 & -1 \\ 1 & -a & -1 \\ 1 & -1 & -a \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} a - 1 \\ a - 1 \\ a - 1 \end{bmatrix} \]

### Step 2: Find the determinant of matrix \( A \)

To find the values of \( a \) for which the system has no solution, we need to calculate the determinant of matrix \( A \) and set it to zero.

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

### Step 3: Calculate the determinant using cofactor expansion

We can expand the determinant along the first row:

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

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -a & -1 \\ -1 & -a \end{vmatrix} = a^2 - 1 \)

2. \( \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} = -a + 1 \)

3. \( \begin{vmatrix} 1 & -a \\ 1 & -1 \end{vmatrix} = -1 + a \)

Substituting these back into the determinant:

\[ \text{det}(A) = a(a^2 - 1) + (a - 1) + (a - 1) \]
\[ = a^3 - a + 2a - 2 \]
\[ = a^3 + a - 2 \]

### Step 4: Set the determinant to zero

For the system to have no solution, we set the determinant equal to zero:

\[ a^3 + a - 2 = 0 \]

### Step 5: Find the roots of the polynomial

To find the roots, we can use the Rational Root Theorem or trial and error. Testing \( a = 1 \):

\[ 1^3 + 1 - 2 = 0 \]

Thus, \( a - 1 \) is a factor. We can perform polynomial long division of \( a^3 + a - 2 \) by \( a - 1 \):

After dividing, we find:

\[ a^3 + a - 2 = (a - 1)(a^2 + a + 2) \]

### Step 6: Solve for the other factors

The quadratic \( a^2 + a + 2 \) has no real roots (discriminant \( 1^2 - 4 \cdot 1 \cdot 2 < 0 \)). Therefore, the only real solution is:

\[ a = 1 \]

### Conclusion

The system of equations has no solution if \( a = 1 \).