If the equations `x + ay – z=0,2x -y + az=0` and `ax + y + 2z=0` are consistent, then a is equal to

Options

A. `-2`
B. 2
C. `1+sqrt3`
D. `1-sqrt3`

Answer

To determine the value of \( a \) for which the system of equations is consistent, we need to analyze the given equations:

1. \( x + ay - z = 0 \) (Equation 1)
2. \( 2x - y + az = 0 \) (Equation 2)
3. \( ax + y + 2z = 0 \) (Equation 3)

### Step 1: Form the Coefficient Matrix and Calculate the Determinant

The coefficient matrix \( A \) of the system can be represented as:

\[ A = \begin{bmatrix} 1 & a & -1 \\ 2 & -1 & a \\ a & 1 & 2 \end{bmatrix} \]

To find the condition for consistency, we need to calculate the determinant of matrix \( A \) and set it equal to zero:

\[ \Delta = \begin{vmatrix} 1 & a & -1 \\ 2 & -1 & a \\ a & 1 & 2 \end{vmatrix} \]

### Step 2: Calculate the Determinant

Using the determinant formula for a \( 3 \times 3 \) matrix:

\[ \Delta = 1 \cdot \begin{vmatrix} -1 & a \\ 1 & 2 \end{vmatrix} - a \cdot \begin{vmatrix} 2 & a \\ a & 2 \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -1 \\ a & 1 \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:

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

2. \( \begin{vmatrix} 2 & a \\ a & 2 \end{vmatrix} = (2)(2) - (a)(a) = 4 - a^2 \)

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

Now substituting these back into the determinant formula:

\[ \Delta = 1(-2 - a) - a(4 - a^2) - 1(2 + a) \]

Expanding this gives:

\[ \Delta = -2 - a - (4a - a^3) - (2 + a) \]
\[ = -2 - a - 4a + a^3 - 2 - a \]
\[ = a^3 - 6a - 4 \]

### Step 3: Set the Determinant to Zero

For the system to be consistent, we set the determinant equal to zero:

\[ a^3 - 6a - 4 = 0 \]

### Step 4: Solve the Cubic Equation

To find the roots of the cubic equation \( a^3 - 6a - 4 = 0 \), we can use trial and error or synthetic division to find rational roots.

Testing \( a = 2 \):

\[ 2^3 - 6(2) - 4 = 8 - 12 - 4 = -8 \quad \text{(not a root)} \]

Testing \( a = -2 \):

\[ (-2)^3 - 6(-2) - 4 = -8 + 12 - 4 = 0 \quad \text{(is a root)} \]

### Step 5: Factor the Polynomial

Since \( a = -2 \) is a root, we can factor \( a + 2 \) out of \( a^3 - 6a - 4 \) using synthetic division:

\[ a^3 - 6a - 4 = (a + 2)(a^2 - 2a - 2) \]

### Step 6: Solve the Quadratic Equation

Now we solve the quadratic \( a^2 - 2a - 2 = 0 \) using the quadratic formula:

\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)} = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} \]
\[ = 1 \pm \sqrt{3} \]

### Final Values of \( a \)

Thus, the values of \( a \) for which the system is consistent are:

1. \( a = -2 \)
2. \( a = 1 + \sqrt{3} \)
3. \( a = 1 - \sqrt{3} \)

### Conclusion

The value of \( a \) that makes the system consistent is \( a = -2 \).

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