The number of integral value(s) of k such that the system of equations `kz-2y-z=x, ky-z=z+3x` and `2x+kz=2y-z` has non - trivial solution, is/are

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to determine the integral values of \( k \) such that the system of equations has a non-trivial solution. The equations given are:

1. \( kz - 2y - z = x \)
2. \( ky - z = z + 3x \)
3. \( 2x + kz = 2y - z \)

### Step 1: Rewrite the equations
We can rewrite the equations in the form \( ax + by + cz = 0 \).

1. From \( kz - 2y - z = x \), we can rearrange it to:
\[ x - kz + 2y + z = 0 \quad \text{(Equation 1)} \]

2. From \( ky - z = z + 3x \), rearranging gives:
\[ 3x - ky + 2z = 0 \quad \text{(Equation 2)} \]

3. From \( 2x + kz = 2y - z \), rearranging gives:
\[ 2x - 2y + kz + z = 0 \quad \text{(Equation 3)} \]

### Step 2: Formulate the coefficient matrix
The system can be represented in matrix form \( A \mathbf{v} = 0 \), where \( A \) is the coefficient matrix and \( \mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \).

The coefficient matrix \( A \) is:
\[ A = \begin{pmatrix} 1 & 2 & -k + 1 \\ 3 & -k & 2 \\ 2 & -2 & k + 1 \end{pmatrix} \]

### Step 3: Find the determinant of the matrix
For the system to have a non-trivial solution, the determinant of the matrix \( A \) must be zero.

Calculating the determinant:
\[ \text{det}(A) = \begin{vmatrix} 1 & 2 & -k + 1 \\ 3 & -k & 2 \\ 2 & -2 & k + 1 \end{vmatrix} \]

Using the determinant formula for a \( 3 \times 3 \) matrix:
\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -k & 2 \\ -2 & k + 1 \end{vmatrix} - 2 \cdot \begin{vmatrix} 3 & 2 \\ 2 & k + 1 \end{vmatrix} + (-k + 1) \cdot \begin{vmatrix} 3 & -k \\ 2 & -2 \end{vmatrix} \]

Calculating each of the \( 2 \times 2 \) determinants:
1. \( \begin{vmatrix} -k & 2 \\ -2 & k + 1 \end{vmatrix} = -k(k + 1) - (-4) = -k^2 - k + 4 \)

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

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

Putting it all together:
\[ \text{det}(A) = 1(-k^2 - k + 4) - 2(3k - 1) + (-k + 1)(2k - 6) \]
Expanding this gives:
\[ -k^2 - k + 4 - 6k + 2 + (-2k^2 + 6k + 2k - 6) = -k^2 - k + 4 - 6k - 2k^2 + 8k - 6 \]
Combining like terms:
\[ -k^2 - 2k^2 + (-k - 6k + 8k) + (4 - 6) = -3k^2 + k - 2 \]

### Step 4: Set the determinant to zero
Setting the determinant equal to zero for non-trivial solutions:
\[ -3k^2 + k - 2 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = -3, b = 1, c = -2 \):
\[ k = \frac{-1 \pm \sqrt{1^2 - 4(-3)(-2)}}{2(-3)} = \frac{-1 \pm \sqrt{1 - 24}}{-6} = \frac{-1 \pm \sqrt{-23}}{-6} \]
Since the discriminant is negative, there are no real solutions for \( k \).

### Conclusion
Since we are looking for integral values of \( k \) and there are no real solutions, the number of integral values of \( k \) such that the system has a non-trivial solution is **0**.