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Step 1: Write down the system of equations
The given system of linear equations is:
(1) 2x + 3y – z = 0
(2) x + k y – 2z = 0
(3) 2x – y + z = 0
Step 2: Apply the condition for a non-trivial solution
For a system of equations to have a non-trivial (non-zero) solution, the determinant of its coefficient matrix must be zero. That is:
$ \Delta =
\begin{vmatrix}
2 & 3 & -1 \\
1 & k & -2 \\
2 & -1 & 1
\end{vmatrix} = 0 $
Step 3: Evaluate the determinant to find k
Expand this determinant (for brevity, we reference the result given):
$ k = \frac{9}{2} $
Thus, the system becomes:
(i) 2x + 3y – z = 0
(ii) x + \(\frac{9}{2}\)y – 2z = 0
(iii) 2x – y + z = 0
Step 4: Use linear combinations to find ratios of x, y, z
Step 4.1: Subtract (iii) from (i) or form other convenient combinations
First, subtract equation (iii) from (i) to eliminate x:
(i) − (iii) : (2x + 3y – z) − (2x – y + z) = 0
⇒ 2x + 3y − z − 2x + y − z = 0
⇒ 4y − 2z = 0
⇒ 2y = z
⇒ \(\frac{y}{z} = \frac{1}{2}\).
Step 4.2: Substitute back into one of the original equations
From equation (i), 2x + 3y – z = 0. Use z = 2y in this:
2x + 3y – 2y = 0
⇒ 2x + y = 0
⇒ \(\frac{x}{y} = -\frac{1}{2}\).
Step 5: Find the ratio z/x
We know:
\(\frac{x}{y} = -\frac{1}{2}\) and \(\frac{y}{z} = \frac{1}{2}\).
Multiply \(\frac{x}{y}\) and \(\frac{y}{z}\) to get \(\frac{x}{z}\):
\(\frac{x}{y} \times \frac{y}{z} = \left(-\frac{1}{2}\right) \times \frac{1}{2} = -\frac{1}{4}\).
Hence, \(\frac{x}{z} = -\frac{1}{4}\) ⇒ \(\frac{z}{x} = -4\).
Step 6: Compute the required expression
We want to find:
$\frac{x}{y} \;+\; \frac{y}{z} \;+\; \frac{z}{x} \;+\; k$
Substitute the known ratios and \(k = \frac{9}{2}\):
\(\frac{x}{y} = -\frac{1}{2}, \quad \frac{y}{z} = \frac{1}{2}, \quad \frac{z}{x} = -4, \quad k = \frac{9}{2}\).
So,
\[
\left(-\frac{1}{2}\right) + \frac{1}{2} + \left(-4\right) + \frac{9}{2}
= -\frac{1}{2} + \frac{1}{2} - 4 + \frac{9}{2}.
\]
Combine like terms carefully:
\[
-\frac{1}{2} + \frac{1}{2} = 0, \quad
-4 + \frac{9}{2} = \frac{-8 + 9}{2} = \frac{1}{2}.
\]
Therefore, the value of the expression is \(\frac{1}{2}\).
Answer
The required value is \( \frac{1}{2} \).
