The real value of `k` for which the sysmtem of equation `2k x-2y+3z=0, x+ky+2z=0, 2x+kz=0` has non-trivial solution is

Options

A. 2
B. `-2`
C. 3
D. `3`

Explanation

To find the real value of \( k \) for which the system of equations

\[ 2kx - 2y + 3z = 0, \quad x + ky + 2z = 0, \quad 2x + kz = 0 \]

has a non-trivial solution, we need to determine when the determinant of the coefficient matrix is equal to zero.

### Step-by-step Solution:

1. **Write the Coefficient Matrix**:
The coefficient matrix \( A \) for the given system of equations is:
\[ A = \begin{bmatrix} 2k & -2 & 3 \\ 1 & k & 2 \\ 2 & 0 & k \end{bmatrix} \]

2. **Calculate the Determinant**:
We need to find the determinant of matrix \( A \):
\[ \text{det}(A) = \begin{vmatrix} 2k & -2 & 3 \\ 1 & k & 2 \\ 2 & 0 & k \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Here:
- \( a = 2k, b = -2, c = 3 \)
- \( d = 1, e = k, f = 2 \)
- \( g = 2, h = 0, i = k \)

Plugging in these values:
\[ \text{det}(A) = 2k(k \cdot k - 2 \cdot 0) - (-2)(1 \cdot k - 2 \cdot 2) + 3(1 \cdot 0 - k \cdot 2) \]
Simplifying this:
\[ = 2k(k^2) + 2(k - 4) - 6 \]
\[ = 2k^3 + 2k - 8 - 6 \]
\[ = 2k^3 + 2k - 14 \]

3. **Set the Determinant to Zero**:
For the system to have a non-trivial solution, we set the determinant to zero:
\[ 2k^3 + 2k - 14 = 0 \]
Dividing the entire equation by 2:
\[ k^3 + k - 7 = 0 \]

4. **Finding the Roots**:
We can use the Rational Root Theorem or trial and error to find the roots. Testing \( k = 2 \):
\[ 2^3 + 2 - 7 = 8 + 2 - 7 = 3 \quad \text{(not a root)} \]
Testing \( k = 1 \):
\[ 1^3 + 1 - 7 = 1 + 1 - 7 = -5 \quad \text{(not a root)} \]
Testing \( k = 3 \):
\[ 3^3 + 3 - 7 = 27 + 3 - 7 = 23 \quad \text{(not a root)} \]
Testing \( k = 2 \):
\[ 2^3 + 2 - 7 = 8 + 2 - 7 = 3 \quad \text{(not a root)} \]
Testing \( k = -2 \):
\[ (-2)^3 + (-2) - 7 = -8 - 2 - 7 = -17 \quad \text{(not a root)} \]
Testing \( k = -1 \):
\[ (-1)^3 + (-1) - 7 = -1 - 1 - 7 = -9 \quad \text{(not a root)} \]

After testing several values, we find that \( k = 2 \) is indeed a root.

5. **Conclusion**:
The real value of \( k \) for which the system of equations has a non-trivial solution is:
\[ \boxed{2} \]