If the system of linear equations x+ky+3z=0 3x+ky-2z=0 2x+4y-3z=0 has a non-zero solution (x,y,z) then `(xz)/(y^2)` is equal to

Options

A. 30
B. -10
C. 10
D. -30

Explanation

To solve the given system of linear equations and find the value of \(\frac{xz}{y^2}\) when there is a non-zero solution, we can follow these steps:

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

### Step 2: Formulate the determinant
For the system to have a non-zero solution, the determinant of the coefficients must be zero. The determinant \(\Delta\) is given by:

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

### Step 3: Calculate the determinant
We will expand the determinant:

\[ \Delta = 1 \cdot \begin{vmatrix} k & -2 \\ 4 & -3 \end{vmatrix} - k \cdot \begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} + 3 \cdot \begin{vmatrix} 3 & k \\ 2 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

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

2. \(\begin{vmatrix} 3 & -2 \\ 2 & -3 \end{vmatrix} = 3(-3) - (-2)(2) = -9 + 4 = -5\)

3. \(\begin{vmatrix} 3 & k \\ 2 & 4 \end{vmatrix} = 3(4) - k(2) = 12 - 2k\)

Putting it all together:

\[ \Delta = 1(-3k + 8) - k(-5) + 3(12 - 2k) \]
\[ = -3k + 8 + 5k + 36 - 6k \]
\[ = (-3k + 5k - 6k) + (8 + 36) = -4k + 44 \]

### Step 4: Set the determinant to zero
For a non-zero solution, we set \(\Delta = 0\):

\[ -4k + 44 = 0 \]
\[ 4k = 44 \implies k = 11 \]

### Step 5: Substitute \(k\) back into the equations
Now substituting \(k = 11\) into the equations:

1. \(x + 11y + 3z = 0\) (Equation 1)
2. \(3x + 11y - 2z = 0\) (Equation 2)
3. \(2x + 4y - 3z = 0\) (Equation 3)

### Step 6: Solve the equations
Using Equation 1 and Equation 2:

From Equation 1:
\[ x = -11y - 3z \]

Substituting into Equation 2:
\[ 3(-11y - 3z) + 11y - 2z = 0 \]
\[ -33y - 9z + 11y - 2z = 0 \]
\[ -22y - 11z = 0 \implies 2y + z = 0 \implies z = -2y \]

Now substituting \(z = -2y\) into Equation 1:
\[ x + 11y + 3(-2y) = 0 \]
\[ x + 11y - 6y = 0 \implies x + 5y = 0 \implies x = -5y \]

### Step 7: Find \(\frac{xz}{y^2}\)
Now we have:
- \(x = -5y\)
- \(z = -2y\)

Calculating \(\frac{xz}{y^2}\):
\[ \frac{xz}{y^2} = \frac{(-5y)(-2y)}{y^2} = \frac{10y^2}{y^2} = 10 \]

### Final Answer
Thus, the value of \(\frac{xz}{y^2}\) is \(10\).

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