The greatest value of c ∈ R for which the system of linear equations x - cy - cz = 0, cx - y + cz =

Q: Wait, why do we reject the trivial solution $(0,0,0)$?

We don't reject it; it's always there! But the problem asks for a *non-trivial* solution. If the determinant is non-zero, the only solution is the trivial one. We need the determinant to be zero to 'collapse' the system into a state where infinite solutions exist.

Q: Is there a shortcut to calculating this determinant?

You could use row operations like $R_1 \to R_1 + R_2 + R_3$, but for a $3 \times 3$ with variables, direct expansion is often safer and less prone to sign errors. Don't fear the algebra!

Q: Why did we get a cubic equation? Does that mean there are three values for $c$?

A cubic equation has three roots. However, some might be repeated. In our case, we found a repeated root. Always check for multiplicity! The roots were $c = -1$ and $c = \frac{1}{2}$.

Question

The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0, cx - y + cz = 0, cx + cy - z = 0$ has a non-trivial solution, is :

Options:

(A) $\frac{1}{2}$

(B) $-1$

(C) $0$

(D) $2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When you look at the system $x - cy - cz = 0$, $cx - y + cz = 0$, and $cx + cy - z = 0$, you are observing a homogeneous system of linear equations. In linear algebra, such a system always possesses the trivial solution, where $x=y=z=0$.

However, the problem requires a **non-trivial solution**. This implies that the three planes defined by these equations do not intersect at a single point; instead, they are linearly dependent, collapsing into a line or a plane of solutions.

### The Gatekeeper: The Determinant

To determine when this collapse occurs, we must utilize the determinant of the coefficient matrix. For a system $AX = 0$ to have a non-trivial solution, the determinant of the coefficient matrix $A$ must be equal to zero.

We construct the matrix $A$ by extracting the coefficients:

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

Setting $\det(A) = 0$ is our mathematical declaration that the system is singular and possesses non-trivial solutions.

### The Algebraic Dance

We expand the determinant along the first row to solve for $c$:

$$\det(A) = 1((-1)(-1) - (c)(c)) - (-c)((c)(-1) - (c)(c)) + (-c)((c)(c) - (-1)(c)) = 0$$

Simplifying the expression step-by-step, we obtain:

$$1(1 - c^2) + c(-c - c^2) - c(c^2 + c) = 0$$

Expanding these terms leads to:

$$1 - c^2 - c^2 - c^3 - c^3 - c^2 = 0$$

Combining like terms results in the cubic equation:

$$-2c^3 - 3c^2 + 1 = 0$$

Multiplying by $-1$, we arrive at the standard form:

$$2c^3 + 3c^2 - 1 = 0$$

### The Final Resolution

To solve the cubic equation $2c^3 + 3c^2 - 1 = 0$, we test for simple roots. Observing that $c = -1$ satisfies the equation, we identify $(c + 1)$ as a factor.

Performing polynomial division, we factor the expression as:

$$(c + 1)(2c^2 + c - 1) = 0$$

Factoring the quadratic term further, we obtain:

$$(c + 1)(2c - 1)(c + 1) = 0 \implies (c + 1)^2(2c - 1) = 0$$

The roots of the equation are $c = -1$ and $c = \frac{1}{2}$. Comparing these values, the **greatest value** is:

$$\mathbf{c = \frac{1}{2}}$$

Animated Solution for Mathematics - Matrices and Determinants: The greatest value of $c \in R$ for which the system of linear equations $x - cy - cz = 0, c x - y + cz = 0, c x + cy - z = 0$ has a non-trivial solution, is :

Select Answer:

Visualized Solution

System of Linear Equations

Condition for Non-Trivial Solution

Setting up the Determinant

Expanding the Determinant

Simplifying the Terms

Forming the Cubic Equation

Testing for a Rational Root

Factoring the Polynomial

Factoring the Quadratic Part

Finding All Possible Values of $c$

Selecting the Greatest Value

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

Analyzing the Setup

The Gatekeeper

The Algebraic Dance

The Final Resolution

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