The number of values of k for which the linear equations `4x""+""k y""+""2z""=""0` `k x""+""4y""+""z""=""0` `2x""+""2y""+""z""=""0` posses a non-zero solution is : (1) 3 (2) 2 (3) 1 (4) zero

Options

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

Explanation

To determine the number of values of \( k \) for which the given linear equations possess a non-zero solution, we will analyze the system of equations using determinants.

The given equations are:

1. \( 4x + ky + 2z = 0 \)
2. \( kx + 4y + z = 0 \)
3. \( 2x + 2y + z = 0 \)

We can express this system in matrix form as follows:

\[ \begin{bmatrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

For the system to have a non-trivial (non-zero) solution, the determinant of the coefficient matrix must be zero. Let's denote the matrix as \( A \):

\[ A = \begin{bmatrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{bmatrix} \]

Now, we will calculate the determinant \( |A| \):

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

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

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

Calculating each of the 2x2 determinants:

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

Substituting these values back into the determinant expression:

\[ |A| = 4(2) - k(k - 2) + 2(2k - 8) \]

Simplifying this:

\[ |A| = 8 - (k^2 - 2k) + (4k - 16) \]
\[ = 8 - k^2 + 2k + 4k - 16 \]
\[ = -k^2 + 6k - 8 \]

Setting the determinant equal to zero for non-trivial solutions:

\[ -k^2 + 6k - 8 = 0 \]

Multiplying through by -1 gives:

\[ k^2 - 6k + 8 = 0 \]

Now, we can factor this quadratic equation:

\[ k^2 - 4k - 2k + 8 = 0 \]
\[ (k - 4)(k - 2) = 0 \]

Thus, the solutions for \( k \) are:

\[ k = 2 \quad \text{and} \quad k = 4 \]

Therefore, there are **2 values** of \( k \) for which the system has a non-zero solution.

**Final Answer:** The number of values of \( k \) is **2**.

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