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 over the set of rational is

Options

A. `33//2`
B. `31//2`
C. 16
D. 15

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. A non-trivial solution exists if the determinant of the coefficient matrix is 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 \( |A| \), we can use the formula for the determinant of a 3x3 matrix:

\[ |A| = a(ei - fh) - b(di - fg) + c(dh - eg) \]

where the matrix is:

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

For our matrix:

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

### Step 3: Calculate the 2x2 determinants

Calculating the first determinant:

\[ \begin{vmatrix} k & 3 \\ 3 & -4 \end{vmatrix} = k(-4) - 3(3) = -4k - 9 \]

Calculating the second determinant:

\[ \begin{vmatrix} 1 & 3 \\ 2 & -4 \end{vmatrix} = 1(-4) - 3(2) = -4 - 6 = -10 \]

Calculating the third determinant:

\[ \begin{vmatrix} 1 & k \\ 2 & 3 \end{vmatrix} = 1(3) - k(2) = 3 - 2k \]

### Step 4: Substitute back into the determinant formula

Now substituting back into the determinant formula:

\[ |A| = 3(-4k - 9) - k(-10) - 2(3 - 2k) \]

Expanding this gives:

\[ |A| = -12k - 27 + 10k - 6 + 4k \]

Combining like terms:

\[ |A| = (-12k + 10k + 4k) + (-27 - 6) = 2k - 33 \]

### Step 5: Set the determinant to zero

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

\[ 2k - 33 = 0 \]

### Step 6: Solve for \( k \)

Solving for \( k \):

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

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

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