For what value of k do the following system of equations possess a non trivial (i.e., not all zero)

Q: Wait, why can't we just use Cramer's Rule to solve for $x, y,$ and $z$?

That is a brilliant question! Cramer's Rule is a powerful tool, but it is specifically designed for systems with a *unique* solution, which occurs only when the determinant of the coefficient matrix is non-zero. In this problem, we are explicitly looking for a non-trivial solution, which forces the determinant to be zero. When the determinant is zero, Cramer's Rule breaks down because you would be dividing by zero. We must instead rely on the dependency between the equations to find the line of intersection.

Q: Why do we introduce a parameter $b$ at the end? Can't we just find specific values for $x, y,$ and $z$?

Because the system has infinitely many solutions, there isn't just one 'correct' point. The solutions form a line in 3D space. By setting $x = b$, we are parameterizing this line. Any point on this line is a valid solution, so $b$ acts as a placeholder for any rational number. It allows us to describe the entire set of solutions elegantly.

Q: How do I know if I should use Gaussian elimination or the determinant method?

Both are valid! However, when you are asked to find a specific parameter $k$ that allows for a non-trivial solution, the determinant method is usually faster and more direct. Gaussian elimination is excellent for solving the system once you have the value of $k$, but the determinant is the 'master key' that unlocks the value of $k$ itself.

Question

For what value of $k$ do the following system of equations possess a non trivial (i.e., not all zero) solution over the set of rationals $\mathbb{Q}$? $x + ky + 3z = 0, 3x + ky - 2z = 0, 2x + 3y - 4z = 0$. For that value of $k$, find all the solutions for the system.

Sigma Percentile · Accepted Answer

### The Geometry of the Void: Understanding Homogeneous Systems

Welcome, future engineer. Today, we are not just solving a system of equations; we are exploring the architecture of 3D space.

When you look at the system $x + ky + 3z = 0$, $3x + ky - 2z = 0$, and $2x + 3y - 4z = 0$, I want you to stop seeing just variables and coefficients. I want you to see three planes.

Because the constant terms are all zero, these planes are not floating aimlessly in space; they are all anchored, pinned to the origin $(0,0,0)$.

Usually, three planes in space intersect at a single point. But here, we are asked for a 'non-trivial' solution.

This is a fancy way of saying that these planes are 'cooperating.' They are not just meeting at the origin; they are colliding along a common line. This is the heart of the problem.

### The Gatekeeper: The Determinant

How do we force these planes to share a line? We need the determinant of the coefficient matrix, $\Delta$, to be zero.

Think of the determinant as the 'volume' of the parallelepiped formed by the normal vectors of these planes. If that volume is zero, the vectors are coplanar, and the planes must intersect in a way that creates a line or a plane of solutions.

Let us construct our matrix $A$:

$$\Delta = \begin{vmatrix} 1 & k & 3 \ 3 & k & -2 \ 2 & 3 & -4 \end{vmatrix} = 0$$

Now, let us perform the expansion along the first row. I know this looks like a tedious algebraic chore, but stay with me. It is a dance of signs and coefficients:

$$1[k(-4) - 3(-2)] - k[3(-4) - 2(-2)] + 3[3(3) - 2(k)] = 0$$

Let us break this down. The first term is $1(-4k + 6)$. The second term is $-k(-12 + 4)$, which simplifies to $-k(-8)$, or $+8k$. The final term is $3(9 - 2k)$, which is $27 - 6k$.

Putting it all together:

$$-4k + 6 + 8k + 27 - 6k = 0$$

Look at the $k$ terms: $-4k + 8k - 6k$ gives us $-2k$. The constants $6 + 27$ give us $33$.

So, we are left with $-2k + 33 = 0$. Solving this, we find our critical value: **$k = \frac{33}{2}$**. This is the magic value that aligns our planes perfectly.

### Tracing the Line of Intersection

Now that we have $k = \frac{33}{2}$, the system is no longer a mystery. We substitute this value back into our equations. The system becomes:

1) $x + \frac{33}{2}y + 3z = 0$

2) $3x + \frac{33}{2}y - 2z = 0$

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

Notice the beauty of equations (1) and (2). They both contain the term $\frac{33}{2}y$. This is not a coincidence; it is a gift.

If we subtract equation (1) from equation (2), the $y$ terms vanish entirely:

$$(3x - x) + \left(\frac{33}{2}y - \frac{33}{2}yight) + (-2z - 3z) = 0$$

$$2x - 5z = 0 \Rightarrow z = \frac{2x}{5}$$

We have successfully linked $z$ to $x$. Now, let us find $y$. Substitute $z = \frac{2x}{5}$ into equation (3):

$$2x + 3y - 4\left(\frac{2x}{5}ight) = 0$$

$$2x + 3y - \frac{8x}{5} = 0$$

Multiply by $5$ to clear the fraction, or just combine the $x$ terms: $2x - \frac{8x}{5} = \frac{10x - 8x}{5} = \frac{2x}{5}$.

So, $3y + \frac{2x}{5} = 0$, which leads us to $y = -\frac{2x}{15}$.

### The Final Parametric Elegance

We have arrived. Since $x$ can be any rational number, let us define it as a parameter $b$.

Our solution set is not a single point, but a family of points defined by:

$$x = b, \quad y = -\frac{2b}{15}, \quad z = \frac{2b}{5}$$

For any rational $b$, this triplet $\left(b, -\frac{2b}{15}, \frac{2b}{5}ight)$ satisfies all three equations. You have just mapped the line of intersection in 3D space.

You didn't just solve for $k$; you uncovered the geometric relationship between these planes. That is the power of linear algebra—it turns abstract equations into a tangible, visual reality. Keep this intuition, and no system will ever be too complex for you to conquer.

Visualized Solution

System of Homogeneous Equations

Condition for Non-Trivial Solution

Setting up the Determinant $Δ$

Expanding the Determinant

Solving for $k$

Substituting $k$ to find Solutions

Eliminating $y$ to Relate $x$ and $z$

Relating $x$ and $y$

Final Parametric Solution

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

The Geometry of the Void

The Gatekeeper

Tracing the Line of Intersection

The Final Parametric Elegance

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Visualized Solution

System of Homogeneous Equations

Condition for Non-Trivial Solution

Setting up the Determinant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ

Expanding the Determinant

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k to find Solutions

Relating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Final Parametric Solution

The Sigma Insight: Solution of System of Linear Equations (Matrix Method and Cramer's Rule)

The Geometry of the Void

The Gatekeeper

Tracing the Line of Intersection

The Final Parametric Elegance

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