Understanding the Key Concept

For a system of homogeneous linear equations, given by $A\mathbf{x} = \mathbf{0}$, where $A$ is the coefficient matrix, $\mathbf{x}$ is the column vector of variables, and $\mathbf{0}$ is the zero vector:

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

This system always has a trivial solution, which is $x=0, y=0, z=0$. However, if the system is to have a non-zero solution (also called a non-trivial solution), then the determinant of the coefficient matrix must be equal to zero. That is, $\det(A) = 0$.

Step-by-Step Solution

1. Formulating the Determinant Condition

The given system of linear equations is:

$$\begin{cases} x + ky + 3z = 0 \quad &(1) \\ 3x + ky - 2z = 0 \quad &(2) \\ 2x + 4y - 3z = 0 \quad &(3) \end{cases}$$

The coefficient matrix $A$ for this system is:

$$A = \begin{pmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{pmatrix}$$

For a non-zero solution to exist, the determinant of this coefficient matrix must be zero:

$$\det(A) = \left| \begin{matrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{matrix} \right| = 0$$

2. Evaluating the Determinant to Find K

We will expand the determinant along the first row for simplicity.

$$1 \cdot \left| \begin{matrix} k & -2 \\ 4 & -3 \end{matrix} \right| - k \cdot \left| \begin{matrix} 3 & -2 \\ 2 & -3 \end{matrix} \right| + 3 \cdot \left| \begin{matrix} 3 & k \\ 2 & 4 \end{matrix} \right| = 0$$

Now, we calculate the $2 \times 2$ determinants:

$$1 \cdot (k \cdot (-3) - (-2) \cdot 4) - k \cdot (3 \cdot (-3) - (-2) \cdot 2) + 3 \cdot (3 \cdot 4 - k \cdot 2) = 0$$ $$1 \cdot (-3k + 8) - k \cdot (-9 + 4) + 3 \cdot (12 - 2k) = 0$$ $$-3k + 8 - k(-5) + 36 - 6k = 0$$ $$-3k + 8 + 5k + 36 - 6k = 0$$

Combine the terms involving $k$ and the constant terms:

$$(-3k + 5k - 6k) + (8 + 36) = 0$$ $$-4k + 44 = 0$$

Solving for $k$:

$$4k = 44$$ $$k = 11$$

Tip: Be extremely careful with signs when evaluating determinants, especially with negative numbers. A common mistake is mismanaging the minus sign in the cofactor expansion or in the $ad-bc$ calculation for $2 \times 2$ determinants.

3. Substituting K and Forming the Reduced System

Now that we have found $k=11$, we substitute this value back into the original system of equations. This ensures that the system now has infinitely many solutions (a non-zero solution is one of them). The system becomes:

$$\begin{cases} x + 11y + 3z = 0 \quad &(1') \\ 3x + 11y - 2z = 0 \quad &(2') \\ 2x + 4y - 3z = 0 \quad &(3') \end{cases}$$

Since this system has a non-zero solution, the equations are linearly dependent. This means we can express some variables in terms of others. Our goal is to find the ratio $x:y:z$.

4. Solving the System for Relationships between x, y, z

We need to find relationships between $x, y,$ and $z$. We can use any two of the three equations since they are linearly dependent. Let's try to eliminate one variable.

Adding equation (1') and (3') seems like a good strategy as the $z$ terms have opposite signs and coefficients that sum to zero ($3z - 3z = 0$):

$$(x + 11y + 3z) + (2x + 4y - 3z) = 0 + 0$$ $$3x + 15y = 0$$

From this, we can express $x$ in terms of $y$:

$$3x = -15y$$ $$x = -5y$$

Now we have a relationship for $x$. Let's use this in one of the original equations to find a relationship for $z$. Substituting $x = -5y$ into equation (1'):

$$(-5y) + 11y + 3z = 0$$ $$6y + 3z = 0$$

Now, express $z$ in terms of $y$:

$$3z = -6y$$ $$z = -2y$$

So, we have found that $x = -5y$ and $z = -2y$. This means for any non-zero value of $y$, we can find corresponding $x$ and $z$ values that satisfy the system. For a non-zero solution, $y$ cannot be zero (if $y=0$, then $x=0$ and $z=0$, which is the trivial solution).

5. Calculating the Required Expression

We need to find the value of $\frac{xz}{y^2}$. Substitute the expressions for $x$ and $z$ that we just found into the given expression:

$$\frac{xz}{y^2} = \frac{(-5y)(-2y)}{y^2}$$ $$= \frac{10y^2}{y^2}$$

Since we are looking for a non-zero solution, $y \neq 0$, so $y^2 \neq 0$. Therefore, we can cancel $y^2$ from the numerator and denominator:

$$= 10$$

The value of $\frac{xz}{y^2}$ is $10$.

Final Answer Check: If $y=1$, then $x=-5$ and $z=-2$. Let's check with original equations with $K=11$:

1. $x+11y+3z = -5 + 11(1) + 3(-2) = -5 + 11 - 6 = 0$. (Correct)
2. $3x+11y-2z = 3(-5) + 11(1) - 2(-2) = -15 + 11 + 4 = 0$. (Correct)
3. $2x+4y-3z = 2(-5) + 4(1) - 3(-2) = -10 + 4 + 6 = 0$. (Correct) The values are consistent. Then $\frac{xz}{y^2} = \frac{(-5)(-2)}{(1)^2} = \frac{10}{1} = 10$.

The final answer is $\boxed{\text{10}}$.

Summary and Key Takeaway

This problem demonstrates a crucial concept in linear algebra: a homogeneous system of linear equations ($A\mathbf{x}=\mathbf{0}$) has a non-trivial (non-zero) solution if and only if the determinant of its coefficient matrix is zero. Once this condition is used to find any unknown parameters (like $k$ in this case), the system becomes dependent. This allows us to express variables in terms of each other (i.e., find their ratios), which is often sufficient to evaluate expressions involving these variables. Always remember to handle determinant calculations carefully and ensure your variable substitutions are accurate.