If `4x-ay+3z=0, x+2y+ax=0`
and `ax+2z=0` have a non - trivial solution, then the number of real value(s) of a is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To find the number of real values of \( a \) for which the equations \( 4x - ay + 3z = 0 \), \( x + 2y + ax = 0 \), and \( ax + 2z = 0 \) have a non-trivial solution, we need to set up the corresponding coefficient matrix and find its determinant.

### Step-by-Step Solution:

1. **Set up the coefficient matrix**:
The equations can be rewritten in matrix form as:
\[ \begin{bmatrix} 4 & -a & 3 \\ 1 & 2 & a \\ a & 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 \]

2. **Calculate the determinant**:
We need to find the determinant of the matrix:
\[ D = \begin{vmatrix} 4 & -a & 3 \\ 1 & 2 & a \\ a & 0 & 2 \end{vmatrix} \]
Using the determinant formula for a 3x3 matrix, we can expand it:
\[ D = 4 \begin{vmatrix} 2 & a \\ 0 & 2 \end{vmatrix} - (-a) \begin{vmatrix} 1 & a \\ a & 2 \end{vmatrix} + 3 \begin{vmatrix} 1 & 2 \\ a & 0 \end{vmatrix} \]

3. **Calculate the minors**:
- The first minor:
\[ \begin{vmatrix} 2 & a \\ 0 & 2 \end{vmatrix} = (2)(2) - (0)(a) = 4 \]
- The second minor:
\[ \begin{vmatrix} 1 & a \\ a & 2 \end{vmatrix} = (1)(2) - (a)(a) = 2 - a^2 \]
- The third minor:
\[ \begin{vmatrix} 1 & 2 \\ a & 0 \end{vmatrix} = (1)(0) - (2)(a) = -2a \]

4. **Substituting back into the determinant**:
Now substituting the minors back into the determinant:
\[ D = 4(4) + a(2 - a^2) + 3(-2a) \]
Simplifying:
\[ D = 16 + 2a - a^3 - 6a = -a^3 - 4a + 16 \]

5. **Set the determinant to zero**:
For a non-trivial solution, we set the determinant equal to zero:
\[ -a^3 - 4a + 16 = 0 \quad \Rightarrow \quad a^3 + 4a - 16 = 0 \]

6. **Finding the roots**:
We can use the Rational Root Theorem to test for possible rational roots. Testing \( a = 2 \):
\[ 2^3 + 4(2) - 16 = 8 + 8 - 16 = 0 \]
Thus, \( a = 2 \) is a root. We can factor \( a - 2 \) out of \( a^3 + 4a - 16 \) using polynomial long division.

7. **Perform polynomial long division**:
Dividing \( a^3 + 4a - 16 \) by \( a - 2 \) gives:
\[ a^3 + 4a - 16 = (a - 2)(a^2 + 2a + 8) \]

8. **Finding the discriminant of the quadratic**:
Now, we need to check the quadratic \( a^2 + 2a + 8 \):
\[ D = b^2 - 4ac = 2^2 - 4(1)(8) = 4 - 32 = -28 \]
Since the discriminant is negative, this quadratic has no real roots.

9. **Conclusion**:
The only real solution for \( a \) is \( a = 2 \). Thus, the number of real values of \( a \) is:
\[ \boxed{1} \]