The value of a for whch the sytem of equations
`x+ay+z=1`
`ax+y+z=1`
`x+y+az=1`
has no solution is

Options

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

Answer

To find the value of \( a \) for which the system of equations has no solution, we can use the concept of determinants. The system of equations is given as:

1. \( x + ay + z = 1 \)
2. \( ax + y + z = 1 \)
3. \( x + y + az = 1 \)

We can express this system in matrix form as:

\[ \begin{bmatrix} 1 & a & 1 \\ a & 1 & 1 \\ 1 & 1 & a \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \]

To determine when this system has no solution, we need to find when the determinant of the coefficient matrix is zero:

\[ D = \begin{vmatrix} 1 & a & 1 \\ a & 1 & 1 \\ 1 & 1 & a \end{vmatrix} \]

### Step 1: Calculate the Determinant

We can calculate the determinant using the formula for a 3x3 matrix:

\[ D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31}) \]

Substituting the values from our matrix:

\[ D = 1 \cdot (1 \cdot a - 1 \cdot 1) - a \cdot (a \cdot a - 1 \cdot 1) + 1 \cdot (a \cdot 1 - 1 \cdot 1) \]

### Step 2: Simplify the Determinant

Now, simplifying each term:

1. First term: \( 1 \cdot (a - 1) = a - 1 \)
2. Second term: \( -a \cdot (a^2 - 1) = -a^3 + a \)
3. Third term: \( 1 \cdot (a - 1) = a - 1 \)

Putting it all together:

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

Combining like terms:

\[ D = a - 1 - a^3 + a + a - 1 = 3a - a^3 - 2 \]

### Step 3: Set the Determinant to Zero

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

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

Rearranging gives us:

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

### Step 4: Factor the Cubic Equation

To solve this cubic equation, we can use the Rational Root Theorem or trial and error to find possible rational roots. Testing \( a = 1 \):

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

So, \( a = 1 \) is a root. We can factor \( a - 1 \) out of the cubic polynomial:

Using synthetic division or polynomial long division, we can factor:

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

### Step 5: Factor the Quadratic

Now we can factor the quadratic:

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

Thus, we have:

\[ a^3 - 3a + 2 = (a - 1)^2 (a + 2) = 0 \]

### Step 6: Find the Values of \( a \)

Setting each factor to zero gives us:

1. \( (a - 1)^2 = 0 \) → \( a = 1 \) (double root)
2. \( (a + 2) = 0 \) → \( a = -2 \)

### Conclusion

The values of \( a \) for which the system of equations has no solution are \( a = 1 \) and \( a = -2 \).