Key Concept: Non-Trivial Solutions for Homogeneous Systems

For a system of homogeneous linear equations of the form:

$$\begin{align*} a_1x + b_1y + c_1z &= 0 \\ a_2x + b_2y + c_2z &= 0 \\ a_3x + b_3y + c_3z &= 0 \end{align*}$$

to have a non-trivial solution (i.e., a solution other than $x=0, y=0, z=0$), the determinant of the coefficient matrix must be equal to zero. Let the coefficient matrix be $A = \begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{pmatrix}$. Then, for a non-trivial solution, we must have $\det(A) = 0$.

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Step 1: Determine the Value of 'k' for a Non-Trivial Solution

The given system of equations is:

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

To find the value of $k$ for which this system has a non-trivial solution, we set the determinant of the coefficient matrix to zero.

The coefficient matrix is:

$$A = \begin{pmatrix} 2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1 \end{pmatrix}$$

Setting $\det(A) = 0$:

$$\left| \begin{matrix} 2 & 3 & -1 \\ 1 & k & -2 \\ 2 & -1 & 1 \end{matrix} \right| = 0$$

Now, we expand the determinant along the first row:

$$2 \left| \begin{matrix} k & -2 \\ -1 & 1 \end{matrix} \right| - 3 \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| + (-1) \left| \begin{matrix} 1 & k \\ 2 & -1 \end{matrix} \right| = 0$$

Calculate the $2 \times 2$ determinants:

• $2(k \times 1 - (-2) \times (-1))$
• $-3(1 \times 1 - (-2) \times 2)$
• $-1(1 \times (-1) - k \times 2)$

Substituting these back:

$$2(k - 2) - 3(1 - (-4)) - 1(-1 - 2k) = 0$$ $$2(k - 2) - 3(1 + 4) - 1(-1 - 2k) = 0$$ $$2k - 4 - 3(5) + 1 + 2k = 0$$ $$2k - 4 - 15 + 1 + 2k = 0$$ $$4k - 18 = 0$$ $$4k = 18$$ $$k = \frac{18}{4}$$ $$k = \frac{9}{2}$$

Explanation: We perform this step first because the problem explicitly states a "non-trivial solution," which is the direct condition to find the unknown parameter $k$. Without $k$, we cannot fully define the system and solve for the ratios of $x, y, z$. Tip: Be very careful with signs when expanding determinants, especially with the negative terms in the cofactor expansion. A common mistake is to miss a sign change.

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Step 2: Solve the System of Equations with the Found 'k'

Now that we have $k = \frac{9}{2}$, we substitute this value back into the original system of equations. The system becomes:

1. $2x + 3y – z = 0$ (Equation (i))
2. $x + \frac{9}{2}y – 2z = 0 \implies 2x + 9y – 4z = 0$ (Equation (ii)')
3. $2x – y + z = 0$ (Equation (iii))

Since the determinant of the coefficient matrix is zero, these equations are linearly dependent. This means we cannot find unique values for $x, y, z$, but we can find the ratios between them (e.g., $x:y:z$). We need to find $x/y$, $y/z$, and $z/x$.

Let's use a combination of these equations to find relationships between the variables:

Subtract Equation (iii) from Equation (i):

$$(2x + 3y – z) - (2x – y + z) = 0 - 0$$ $$2x + 3y – z - 2x + y - z = 0$$ $$4y - 2z = 0$$ $$4y = 2z$$ $$z = 2y \quad (\text{Equation (iv)})$$

From this, we can directly find one of the required ratios:

$$\frac{y}{z} = \frac{1}{2}$$

Now, substitute $z = 2y$ into Equation (i):

$$2x + 3y - (2y) = 0$$ $$2x + y = 0$$ $$y = -2x \quad (\text{Equation (v)})$$

From this, we can find another required ratio:

$$\frac{x}{y} = -\frac{1}{2}$$

Explanation: We are looking for ratios of variables. By combining equations, we aim to eliminate one variable at a time or establish a direct relationship between two variables. The choice of equations (i) and (iii) was strategic because subtracting them directly eliminated $x$ and simplified the expression quickly. We then used this relationship ($z=2y$) in another equation (i) to find a relationship between $x$ and $y$. Tip: When solving for ratios, it's often easiest to express one variable in terms of another. For example, $z=2y$ means $z$ is twice $y$.

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Step 3: Calculate the Remaining Ratio

We have $y/z = 1/2$ and $x/y = -1/2$. We need $z/x$. We can find $z/x$ by multiplying the reciprocals of the ratios we already found: From $z=2y$ and $y=-2x$: Substitute $y=-2x$ into $z=2y$:

$$z = 2(-2x)$$ $$z = -4x$$

Therefore:

$$\frac{z}{x} = -4$$

Explanation: By chaining the relationships we found ($z$ in terms of $y$, and $y$ in terms of $x$), we can establish a direct relationship between $z$ and $x$. This is a more robust way than multiplying fractions which might lead to errors if not careful.

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Step 4: Substitute Values into the Expression and Final Calculation

Now we have all the components needed for the expression:

• $k = \frac{9}{2}$
• $\frac{x}{y} = -\frac{1}{2}$
• $\frac{y}{z} = \frac{1}{2}$
• $\frac{z}{x} = -4$

Substitute these values into the given expression:

$$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} + k$$ $$= \left(-\frac{1}{2}\right) + \left(\frac{1}{2}\right) + (-4) + \left(\frac{9}{2}\right)$$ $$= 0 - 4 + \frac{9}{2}$$ $$= -4 + \frac{9}{2}$$

To add these, find a common denominator:

$$= -\frac{8}{2} + \frac{9}{2}$$ $$= \frac{9 - 8}{2}$$ $$= \frac{1}{2}$$

Explanation: This is the final step where all previously calculated values are brought together. Careful arithmetic is crucial here. Common Mistake: Careless addition/subtraction of fractions can lead to errors. Double-check your calculations.

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Summary and Key Takeaway

This problem effectively tests your understanding of the conditions for non-trivial solutions in homogeneous systems of linear equations and your ability to solve such systems for ratios of variables. The key steps were:

1. Recognizing that a non-trivial solution implies the determinant of the coefficient matrix is zero.
2. Accurately calculating the determinant to find the unknown parameter $k$.
3. Substituting $k$ back into the system and strategically solving for the ratios of variables ($x/y$, $y/z$, $z/x$) by eliminating variables.
4. Finally, substituting all derived values into the given expression and performing the arithmetic.

The final answer is $\frac{1}{2}$.

The correct option is (C).