The system of linear equations x + λy - z = 0, λx - y - z = 0, x + y - λz = 0 has a non-trivial solu

Q: Wait, why do we set the determinant to zero? I thought determinants were for finding the inverse.

That is a great observation! You are right that we use determinants to find inverses, but think about it this way: if a matrix has an inverse, it means the transformation is 'bijective'—it maps the origin to the origin uniquely. If the determinant is zero, the matrix collapses space into a lower dimension (like a line or a plane). In our system $AX=0$, a zero determinant means the matrix is not full-rank, allowing for a non-zero vector $X$ to map to the zero vector. That is exactly what a non-trivial solution is!

Q: Can I just use Cramer's Rule here?

You could, but Cramer's Rule is designed for systems where $D \neq 0$. If you try to use it for a homogeneous system where $D=0$, you will end up with $0/0$ forms, which are indeterminate. It is much more elegant to use the property of the determinant directly to find the values of $\lambda$ that make the system singular.

Q: Why do we call it a 'non-trivial' solution?

In a homogeneous system $AX=0$, the solution $x=0, y=0, z=0$ is always present. We call this the 'trivial' solution because it is the most obvious one. A 'non-trivial' solution is any solution where at least one variable is non-zero. Finding these is the key to understanding when the system has more than just the origin as a solution.

Question

The system of linear equations $x + \lambda y - z = 0$, $\lambda x - y - z = 0$, $x + y - \lambda z = 0$ has a non-trivial solution for:

Options:

(A) exactly two values of $\lambda$

(B) exactly three values of $\lambda$

(C) infinitely many values of $\lambda$

(D) exactly one value of $\lambda$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate space with three planes, each defined by a linear equation. In a homogeneous system, every plane passes through the origin $(0,0,0)$.

Because they all share the origin, the origin is always a solution. However, we are searching for a **non-trivial solution**.

This implies that the planes do not merely meet at a single point; rather, they are aligned such that they intersect along an entire line. This is the moment where the geometry becomes truly beautiful.

### The Determinant as a Gatekeeper

To determine when these planes intersect along a line, we examine the coefficient matrix $A$. The determinant of this matrix, denoted as $|A|$, acts as a gatekeeper.

If $|A| 
eq 0$, the matrix is invertible, and the only solution is the trivial one—the origin. If $|A| = 0$, the matrix is singular, the system collapses, and we obtain the non-trivial solutions we seek.

Our mission is to find the values of $\lambda$ that make the determinant zero. We construct the matrix $A$ as follows:

$$A = \begin{pmatrix} 1 & \lambda & -1 \ \lambda & -1 & -1 \ 1 & 1 & -\lambda \end{pmatrix}$$

We must now solve the equation $|A| = 0$.

### The Dance of Symbols

Expanding this determinant along the first row yields the following expression:

$$1 \cdot [(-1)(-\lambda) - (-1)(1)] - \lambda \cdot [(\lambda)(-\lambda) - (-1)(1)] + (-1) \cdot [(\lambda)(1) - (-1)(1)] = 0$$

Let us simplify this step by step. The first term becomes $1 \cdot (\lambda + 1)$. The second term is $-\lambda \cdot (-\lambda^2 + 1)$, which simplifies to $\lambda^3 - \lambda$. The third term is $-1 \cdot (\lambda + 1)$.

Putting it all together, we have:

$$(\lambda + 1) + (\lambda^3 - \lambda) - (\lambda + 1) = 0$$

Notice the elegance here! The $(\lambda + 1)$ and the $-(\lambda + 1)$ cancel out perfectly, leaving us with the simplified equation:

$$\lambda^3 - \lambda = 0$$

### The Final Revelation

We are left with a simple cubic equation: $\lambda^3 - \lambda = 0$. Factoring this, we get:

$$\lambda(\lambda^2 - 1) = 0$$

This further breaks down to:

$$\lambda(\lambda - 1)(\lambda + 1) = 0$$

This gives us three distinct values for $\lambda$: **$0, 1,$ and $-1$**.

We have successfully found the three keys that unlock the non-trivial solutions for this system. It is a perfect example of how complex-looking systems often reduce to simple, elegant truths when we apply the right mathematical tools.

Animated Solution for Mathematics - Matrices and Determinants: The system of linear equations $x + λ y - z = 0$ , $λ x - y - z = 0$ , $x + y - λ z = 0$ has a non-trivial solution for:

Select Answer:

Visualized Solution

Homogeneous System of Equations

Condition for Non-Trivial Solution

Setting up the Determinant

Expanding the Determinant

Simplifying the Terms

Forming the Polynomial

Solving for $λ$

Finding the Roots

Final Conclusion

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

Analyzing the Setup

The Determinant as a Gatekeeper

The Dance of Symbols

The Final Revelation

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Animated Solution for Mathematics - Matrices and Determinants: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The system of linear equations x+λy−z=0, λx−y−z=0, x+y−λz=0 has a non-trivial solution for: