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 find the number of integral values of \( k \) such that the system of equations has a non-trivial solution, we can analyze the given equations. The equations are:

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

We can rewrite these equations in the standard form \( Ax = 0 \), where \( A \) is the coefficient matrix and \( x \) is the vector of variables \( (x, y, z) \).

### Step 1: Write the equations in matrix form

The equations can be rearranged as follows:

1. \( -x + 2y + (k-1)z = 0 \)
2. \( -3x + ky + z = 0 \)
3. \( 2x - 2y + (k+1)z = 0 \)

This gives us the coefficient matrix \( A \):

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

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

For the system to have a non-trivial solution, the determinant of the matrix \( A \) must be zero. We can calculate the determinant \( |A| \):

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

Using the determinant formula for a 3x3 matrix, we expand it:

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

Calculating each of these 2x2 determinants:

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

2. \( \begin{vmatrix} -3 & 1 \\ 2 & k+1 \end{vmatrix} = -3(k+1) - 2 = -3k - 3 - 2 = -3k - 5 \)

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

Substituting these back into the determinant expression:

\[ |A| = -1(k^2 + k + 2) - 2(-3k - 5) + (k-1)(6 - 2k) \]

Expanding this gives:

\[ |A| = -k^2 - k - 2 + 6k + 10 + (k-1)(6 - 2k) \]

Now expanding \( (k-1)(6 - 2k) \):

\[ = 6k - 2k^2 - 6 + 2k = -2k^2 + 8k - 6 \]

Combining all terms:

\[ |A| = -k^2 - k - 2 + 6k + 10 - 2k^2 + 8k - 6 \]
\[ = -3k^2 + 13k + 2 \]

### Step 3: Set the determinant equal to zero

Now we set the determinant equal to zero to find the values of \( k \):

\[ -3k^2 + 13k + 2 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = -3, b = 13, c = 2 \):

\[ k = \frac{-13 \pm \sqrt{13^2 - 4(-3)(2)}}{2(-3)} \]
\[ = \frac{-13 \pm \sqrt{169 + 24}}{-6} \]
\[ = \frac{-13 \pm \sqrt{193}}{-6} \]

### Step 5: Determine integral values of \( k \)

The roots are not guaranteed to be integers, so we need to check for integer solutions. We can analyze the quadratic equation \( -3k^2 + 13k + 2 = 0 \) to find integral values.

### Step 6: Count the integral solutions

By checking the discriminant \( 193 \) which is not a perfect square, we conclude that there are no integral solutions.

Thus, the number of integral values of \( k \) such that the system has a non-trivial solution is **0**.