If `4x-ay+3z=0, x+2y+az=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 solve the problem, we need to determine the values of \( a \) for which the system of equations has a non-trivial solution. The equations given are:

1. \( 4x - ay + 3z = 0 \) (Equation 1)
2. \( x + 2y + az = 0 \) (Equation 2)
3. \( ax + 2z = 0 \) (Equation 3)

For the system to have a non-trivial solution, the determinant of the coefficients must be equal to zero.

### Step 1: Form the Coefficient Matrix

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

\[ A = \begin{bmatrix} 4 & -a & 3 \\ 1 & 2 & a \\ a & 0 & 2 \end{bmatrix} \]

### Step 2: Calculate the Determinant

We need to calculate the determinant of the matrix \( A \):

\[ \text{det}(A) = \begin{vmatrix} 4 & -a & 3 \\ 1 & 2 & a \\ a & 0 & 2 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix, we expand it:

\[ \text{det}(A) = 4 \begin{vmatrix} 2 & a \\ 0 & 2 \end{vmatrix} - (-a) \begin{vmatrix} 1 & a \\ a & 2 \end{vmatrix} + 3 \begin{vmatrix} 1 & 2 \\ a & 0 \end{vmatrix} \]

Calculating the smaller determinants:

1. \( \begin{vmatrix} 2 & a \\ 0 & 2 \end{vmatrix} = 2 \cdot 2 - 0 \cdot a = 4 \)

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

3. \( \begin{vmatrix} 1 & 2 \\ a & 0 \end{vmatrix} = 1 \cdot 0 - 2 \cdot a = -2a \)

Substituting these back into the determinant:

\[ \text{det}(A) = 4 \cdot 4 + a(2 - a^2) + 3(-2a) \]

This simplifies to:

\[ \text{det}(A) = 16 + 2a - a^3 - 6a = -a^3 - 4a + 16 \]

### Step 3: Set the Determinant to Zero

For a non-trivial solution, we set the determinant to zero:

\[ -a^3 - 4a + 16 = 0 \]

This can be rearranged to:

\[ a^3 + 4a - 16 = 0 \]

### Step 4: Find the Roots

To find the real values of \( a \), we can use the Rational Root Theorem or trial and error. 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 the cubic polynomial.

Using synthetic division or polynomial long division, we divide \( a^3 + 4a - 16 \) by \( a - 2 \):

\[ a^3 + 0a^2 + 4a - 16 = (a - 2)(a^2 + 2a + 8) \]

### Step 5: Analyze the Quadratic Factor

Now we need to analyze the quadratic \( a^2 + 2a + 8 \):

The discriminant \( D \) of this quadratic is:

\[ D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 8 = 4 - 32 = -28 \]

Since the discriminant is negative, \( a^2 + 2a + 8 \) has no real roots.

### Conclusion

The only real solution for \( a \) is \( a = 2 \). Therefore, the number of real values of \( a \) is:

\[ \boxed{1} \]