The number of values of k for which the linear equations $4x+ky+2z=0;$ $kx+4y+z=0$; $2x+2y+z=0$ possess a non-zero solution is :

Question

Accepted Answer

To determine the number of values of $ k $ for which the given system of linear equations possesses a non-zero solution, we need to analyze the conditions under which the system has a non-trivial solution. The system of equations is:

1. $ 4x + ky + 2z = 0 $
2. $ kx + 4y + z = 0 $
3. $ 2x + 2y + z = 0 $

For the system to have a non-zero solution, the determinant of the coefficient matrix must be zero. The coefficient matrix $ A $ is:

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

We calculate the determinant of $ A $:

$$
\text{det}(A) = \begin{vmatrix}
4 & k & 2 \
k & 4 & 1 \
2 & 2 & 1
\end{vmatrix}
$$

Expanding along the first row, we have:

$$
\text{det}(A) = 4 \begin{vmatrix} 4 & 1 \ 2 & 1 \end{vmatrix} - k \begin{vmatrix} k & 1 \ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} k & 4 \ 2 & 2 \end{vmatrix}
$$

Calculating each of these 2x2 determinants:

$$
\begin{vmatrix} 4 & 1 \ 2 & 1 \end{vmatrix} = (4)(1) - (1)(2) = 4 - 2 = 2
$$

$$
\begin{vmatrix} k & 1 \ 2 & 1 \end{vmatrix} = (k)(1) - (1)(2) = k - 2
$$

$$
\begin{vmatrix} k & 4 \ 2 & 2 \end{vmatrix} = (k)(2) - (4)(2) = 2k - 8
$$

Substituting back, we get:

$$
\text{det}(A) = 4(2) - k(k - 2) + 2(2k - 8)
$$

Simplifying:

$$
= 8 - (k^2 - 2k) + (4k - 16)
$$

$$
= 8 - k^2 + 2k + 4k - 16
$$

$$
= -k^2 + 6k - 8
$$

Setting the determinant to zero for a non-trivial solution:

$$
-k^2 + 6k - 8 = 0
$$

Multiplying through by -1:

$$
k^2 - 6k + 8 = 0
$$

This is a quadratic equation, which we solve using the quadratic formula:

$$
k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $ a = 1 $, $ b = -6 $, and $ c = 8 $.

$$
k = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}
$$

$$
k = \frac{6 \pm \sqrt{36 - 32}}{2}
$$

$$
k = \frac{6 \pm \sqrt{4}}{2}
$$

$$
k = \frac{6 \pm 2}{2}
$$

This gives two solutions:

$$
k = \frac{8}{2} = 4 \quad \text{and} \quad k = \frac{4}{2} = 2
$$

Thus, there are 2 values of $ k $ (namely $ k = 4 $ and $ k = 2 $) for which the determinant is zero, and the system has a non-zero solution. Therefore, the correct answer is Option 1: 2.

**Explanation for Other Options:**

- **Option 2 (1 value):** This is incorrect because we found two distinct values of $ k $ that satisfy the condition.
- **Option 3 (Zero values):** This is incorrect because there are indeed values of $ k $ that make the determinant zero, allowing for a non-zero solution.
- **Option 4 (3 values):** This is incorrect because the quadratic equation only provides two distinct solutions for $ k $.

Thus, the correct answer is indeed Option 1: 2.