The value of k for which the set of equations `3x+ky-2z=0, x + ky + 3z = 0 and 2x+3y-4z=0` has non-trivial solution is (A) 15 (B) 16 (C) 31/2 (D) 33/2

Answer

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

\[ 3x + ky - 2z = 0, \]
\[ x + ky + 3z = 0, \]
\[ 2x + 3y - 4z = 0 \]

has a non-trivial solution, we need to set up the corresponding coefficient matrix and find its determinant. The condition for a non-trivial solution is that the determinant of the coefficient matrix must be zero.

### Step 1: Write the coefficient matrix

The coefficient matrix \( A \) for the given equations is:

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

### Step 2: Calculate the determinant of the matrix

To find the determinant of \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix:

- \( a = 3, b = k, c = -2 \)
- \( d = 1, e = k, f = 3 \)
- \( g = 2, h = 3, i = -4 \)

Now substituting these values into the determinant formula:

\[ \text{det}(A) = 3(k \cdot -4 - 3 \cdot 3) - k(1 \cdot -4 - 3 \cdot 2) - 2(1 \cdot 3 - k \cdot 2) \]

### Step 3: Simplify the determinant expression

Calculating each term:

1. \( 3(k \cdot -4 - 9) = 3(-4k - 9) = -12k - 27 \)
2. \( -k(-4 - 6) = -k(-10) = 10k \)
3. \( -2(3 - 2k) = -6 + 4k \)

Now combine these results:

\[ \text{det}(A) = -12k - 27 + 10k - 6 + 4k \]

Combine like terms:

\[ \text{det}(A) = (-12k + 10k + 4k) + (-27 - 6) = 2k - 33 \]

### Step 4: Set the determinant to zero

To find the value of \( k \) for which there is a non-trivial solution, we set the determinant to zero:

\[ 2k - 33 = 0 \]

### Step 5: Solve for \( k \)

Solving for \( k \):

\[ 2k = 33 \implies k = \frac{33}{2} \]

### Conclusion

Thus, the value of \( k \) for which the set of equations has a non-trivial solution is

\[ \boxed{\frac{33}{2}}. \]