If the system of equations
`x-ky+3z=0,`
`2x+ky-2z=0` and `3x-4y+2z=0` has non - trivial solutions, then the value of `(10y)/(x)` is equal to

Options

A. 3
B. `-(15)/(2)`
C. `(5)/(7)`
D. `-(5)/(7)`

Answer

To solve the given system of equations for non-trivial solutions and find the value of \(\frac{10y}{x}\), we will follow these steps:

### Step 1: Write the system of equations
The given equations are:
1. \(x - ky + 3z = 0\)
2. \(2x + ky - 2z = 0\)
3. \(3x - 4y + 2z = 0\)

### Step 2: Form the coefficient matrix
The coefficient matrix \(A\) for the system can be written as:
\[ A = \begin{bmatrix} 1 & -k & 3 \\ 2 & k & -2 \\ 3 & -4 & 2 \end{bmatrix} \]

### Step 3: Set the determinant to zero
For the system to have non-trivial solutions, the determinant of the coefficient matrix must be zero:
\[ \text{det}(A) = 0 \]

### Step 4: Calculate the determinant
We can calculate the determinant using the formula for a \(3 \times 3\) matrix:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} k & -2 \\ -4 & 2 \end{vmatrix} - (-k) \cdot \begin{vmatrix} 2 & -2 \\ 3 & 2 \end{vmatrix} + 3 \cdot \begin{vmatrix} 2 & k \\ 3 & -4 \end{vmatrix} \]

Calculating the minors:
1. \(\begin{vmatrix} k & -2 \\ -4 & 2 \end{vmatrix} = k \cdot 2 - (-2)(-4) = 2k - 8\)

2. \(\begin{vmatrix} 2 & -2 \\ 3 & 2 \end{vmatrix} = 2 \cdot 2 - (-2)(3) = 4 + 6 = 10\)

3. \(\begin{vmatrix} 2 & k \\ 3 & -4 \end{vmatrix} = 2 \cdot (-4) - k \cdot 3 = -8 - 3k\)

Putting it all together:
\[ \text{det}(A) = 1(2k - 8) + k(10) + 3(-8 - 3k) \]
\[ = 2k - 8 + 10k - 24 - 9k \]
\[ = (2k + 10k - 9k) - 32 = 3k - 32 \]

Setting the determinant to zero:
\[ 3k - 32 = 0 \implies k = \frac{32}{3} \]

### Step 5: Substitute \(k\) back into the equations
Now, we substitute \(k = \frac{32}{3}\) into the first equation:
\[ x - \frac{32}{3}y + 3z = 0 \implies 3x - 32y + 9z = 0 \quad \text{(Multiplying by 3)} \]

### Step 6: Express \(z\) in terms of \(x\) and \(y\)
From the first equation:
\[ 3x - 32y + 9z = 0 \implies 9z = -3x + 32y \implies z = \frac{-3x + 32y}{9} \]

### Step 7: Substitute \(z\) into the second equation
Substituting \(z\) into the second equation:
\[ 2x + \frac{32}{3}y - 2\left(\frac{-3x + 32y}{9}\right) = 0 \]
Multiplying through by 9 to eliminate the fraction:
\[ 18x + 96y + 6x - 64y = 0 \]
\[ 24x + 32y = 0 \implies 3x + 4y = 0 \implies y = -\frac{3}{4}x \]

### Step 8: Find \(\frac{10y}{x}\)
Substituting \(y = -\frac{3}{4}x\) into \(\frac{10y}{x}\):
\[ \frac{10y}{x} = \frac{10\left(-\frac{3}{4}x\right)}{x} = -\frac{30}{4} = -\frac{15}{2} \]

### Final Answer
Thus, the value of \(\frac{10y}{x}\) is \(-\frac{15}{2}\).

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