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 value of \( a \) for which the system of equations has no solution, we can represent the given system of equations in matrix form and analyze the determinant of the coefficient matrix.

### Given System of Equations:
1. \( ax - y - z = a - 1 \)
2. \( x - ay - z = a - 1 \)
3. \( x - y - az = a - 1 \)

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

### Step 2: Calculate 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 \( A \) and set it equal to zero.

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

Using the determinant formula for a \( 3 \times 3 \) matrix, we have:
\[ \text{det}(A) = a \begin{vmatrix} -a & -1 \\ -1 & -a \end{vmatrix} - (-1) \begin{vmatrix} 1 & -1 \\ 1 & -a \end{vmatrix} - (-1) \begin{vmatrix} 1 & -a \\ 1 & -1 \end{vmatrix} \]

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

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

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

Putting it all together:
\[ \text{det}(A) = a(a^2 - 1) + (a - 1) + (a - 1) \]
\[ = a^3 - a + 2a - 2 = a^3 + a - 2 \]

### Step 3: Set the determinant equal to zero
For the system to have no solution, we need:
\[ a^3 + a - 2 = 0 \]

### Step 4: Solve the cubic equation
To find the roots of \( a^3 + a - 2 = 0 \), we can use the Rational Root Theorem or synthetic division. Testing \( a = 1 \):
\[ 1^3 + 1 - 2 = 0 \]
So, \( a = 1 \) is a root. Now we can factor the cubic polynomial:
\[ a^3 + a - 2 = (a - 1)(a^2 + a + 2) \]

### Step 5: Analyze the quadratic factor
The quadratic \( a^2 + a + 2 \) has no real roots (as its discriminant \( 1^2 - 4 \cdot 1 \cdot 2 = 1 - 8 < 0 \)). Thus, the only real solution is \( a = 1 \).

### Step 6: Conclusion
The system of equations has no solution if \( a = 1 \).