Matrices & Determinants
System of Linear Equations / Non-trivial solutions
Grade 12
Question:
<p>Given the system of linear equations<br>\(x - cy - cz = 0\)<br>\(cx - y + cz = 0\)<br>\(cx + cy - z = 0\)<br>has a non-trivial solution, find the value of \(c\).</p>
<p>\(c = 0\)</p>
<p>\(c = 1\)</p>
<p>\(c = -1\)</p>
<p>\(c = \sqrt{1}\)</p>
Step-by-Step Solution
Key Concept: A homogeneous system has non-trivial solutions if and only if the determinant of the coefficient matrix equals zero. Set up the determinant and solve for c.
<p><strong>Step 1:</strong> Write the coefficient matrix for the system:</p>
<p>$$A = \begin{bmatrix} 1 & -c & -c \\ c & -1 & c \\ c & c & -1 \end{bmatrix}$$</p>
<p><strong>Step 2:</strong> For non-trivial solutions, set $\det(A) = 0$.</p>
<p>Expanding along Row 1:</p>
<p>$$\det(A) = 1\begin{vmatrix} -1 & c \\ c & -1 \end{vmatrix} - (-c)\begin{vmatrix} c & c \\ c & -1 \end{vmatrix} + (-c)\begin{vmatrix} c & -1 \\ c & c \end{vmatrix}$$</p>
<p>$$= 1(1 - c^2) + c(-c - c^2) - c(c^2 + c)$$</p>
<p>$$= 1 - c^2 - c^2 - c^3 - c^3 - c^2$$</p>
<p>$$= 1 - 3c^2 - 2c^3$$</p>
<p><strong>Step 3:</strong> Set $\det(A) = 0$:</p>
<p>$$1 - 3c^2 - 2c^3 = 0$$</p>
<p>$$2c^3 + 3c^2 - 1 = 0$$</p>
<p><strong>Step 4:</strong> Factor: $(c + 1)(2c^2 + c - 1) = 0$</p>
<p>This gives $(c + 1)(2c - 1)(c + 1) = 0$ or $(c + 1)^2(2c - 1) = 0$</p>
<p>Thus $c = -1$ or $c = \frac{1}{2}$</p>
<p><strong>Step 5:</strong> Checking the given options, $c = -1$ is the answer.</p>
<p>$\therefore$ Answer: $c = -1$</p>
Correct Answer: C