This problem tests your understanding of systems of linear equations, specifically homogeneous systems, and their conditions for having non-trivial (non-zero) solutions.

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1. Key Concept: Condition for Non-Zero Solution in Homogeneous Linear Equations

A system of homogeneous linear equations is one where all constant terms are zero. For a system of $n$ linear equations in $n$ variables, represented in matrix form as $AX = 0$: $A = \begin{pmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{pmatrix}, \quad X = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}, \quad 0 = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$ This system always has a trivial solution, $X=0$ (i.e., $x_1=0, x_2=0, \dots, x_n=0$). However, for the system to possess a non-zero solution (also called a non-trivial solution), a specific condition must be met: The determinant of the coefficient matrix $A$ must be equal to zero. $\det(A) = 0$

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2. Identifying the Given System of Equations

We are given the following system of three homogeneous linear equations in three variables ($x, y, z$):

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

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3. Forming the Coefficient Matrix

From the given equations, we can construct the coefficient matrix $A$ by taking the coefficients of $x, y,$ and $z$ from each equation: $A = \begin{pmatrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{pmatrix}$

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4. Applying the Condition for Non-Zero Solution

For the system to have a non-zero solution, the determinant of the coefficient matrix $A$ must be zero. $\det(A) = 0$ $\left| \begin{matrix} 4 & k & 2 \\ k & 4 & 1 \\ 2 & 2 & 1 \end{matrix} \right| = 0$

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5. Step-by-Step Calculation of the Determinant

We will expand the $3 \times 3$ determinant using the cofactor expansion method along the first row ($R_1$). The formula for expanding a $3 \times 3$ determinant $\left| \begin{smallmatrix} a & b & c \\ d & e & f \\ g & h & i \end{smallmatrix} \right|$ along the first row is $a(ei-fh) - b(di-fg) + c(dh-eg)$.

Applying this to our determinant: $4 \cdot \left| \begin{matrix} 4 & 1 \\ 2 & 1 \end{matrix} \right| - k \cdot \left| \begin{matrix} k & 1 \\ 2 & 1 \end{matrix} \right| + 2 \cdot \left| \begin{matrix} k & 4 \\ 2 & 2 \end{matrix} \right| = 0$

Step 5.1: Calculate the $2 \times 2$ sub-determinants.

• The first $2 \times 2$ determinant is $\left| \begin{smallmatrix} 4 & 1 \\ 2 & 1 \end{smallmatrix} \right| = (4 \times 1) - (1 \times 2) = 4 - 2 = 2$.
• The second $2 \times 2$ determinant is $\left| \begin{smallmatrix} k & 1 \\ 2 & 1 \end{smallmatrix} \right| = (k \times 1) - (1 \times 2) = k - 2$.
• The third $2 \times 2$ determinant is $\left| \begin{smallmatrix} k & 4 \\ 2 & 2 \end{smallmatrix} \right| = (k \times 2) - (4 \times 2) = 2k - 8$.

Step 5.2: Substitute these values back into the expansion. $4(2) - k(k - 2) + 2(2k - 8) = 0$

Step 5.3: Simplify the expression. $8 - (k^2 - 2k) + (4k - 16) = 0$ $8 - k^2 + 2k + 4k - 16 = 0$

Step 5.4: Combine like terms. $-k^2 + (2k + 4k) + (8 - 16) = 0$ $-k^2 + 6k - 8 = 0$

Step 5.5: Multiply by -1 to make the leading coefficient positive (optional, but standard practice). $k^2 - 6k + 8 = 0$

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6. Solving the Resulting Quadratic Equation

We now have a quadratic equation in $k$: $k^2 - 6k + 8 = 0$. We can solve this by factoring. We need two numbers that multiply to 8 and add up to -6. These numbers are -2 and -4. $(k - 2)(k - 4) = 0$

This gives us two possible values for $k$:

• $k - 2 = 0 \Rightarrow k = 2$
• $k - 4 = 0 \Rightarrow k = 4$

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7. Conclusion: Number of Values of $k$

The values of $k$ for which the system possesses a non-zero solution are $k=2$ and $k=4$. These are two distinct values.

Therefore, the number of values of $k$ is 2.

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Tips and Common Mistakes:

• Homogeneous vs. Non-Homogeneous Systems: Remember the condition $\det(A)=0$ is specific to homogeneous systems for non-trivial solutions. For non-homogeneous systems ($AX=B$ where $B \ne 0$), Cramer's Rule applies, and $\det(A) \ne 0$ implies a unique solution, while $\det(A) = 0$ can lead to either no solution or infinitely many solutions.
• Determinant Calculation Errors: Be very careful with signs and arithmetic when expanding determinants, especially for $3 \times 3$ matrices. A common mistake is forgetting the alternating signs in cofactor expansion.
• Counting Distinct Values: After finding the roots of the equation, always double-check if the question asks for the values themselves or the number of distinct values. If, for example, the quadratic had a repeated root, there would only be one distinct value of $k$.

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Summary and Key Takeaway:

For a system of homogeneous linear equations to have a non-zero solution, the determinant of its coefficient matrix must be zero. By forming the coefficient matrix, calculating its determinant, and setting it to zero, we obtain an equation for the unknown parameter $k$. Solving this equation yields the required values of $k$. In this case, we found two distinct values of $k$, namely $2$ and $4$.

The final answer is $\boxed{A}$.