Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>If the system of linear equations<br>\(x + ky + 3z = 0\)<br>\(3x + ky - 2z = 0\)<br>\(2x + 4y - 3z = 0\)<br>has a non-zero solution \((x, y, z)\), then \(\dfrac{xz}{y^2}\) is equal to</p>
<p>10</p>
<p>\(-30\)</p>
<p>30</p>
<p>\(-10\)</p>
Step-by-Step Solution
Key Concept: For a homogeneous system to have a non-trivial solution, the coefficient matrix determinant must equal zero. This constraint on k determines the relationship between x, y, and z, allowing us to compute the specific ratio xz/y².
<p><strong>Step 1:</strong> For non-trivial solution, set determinant = 0:</p><p>$$\begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} = 0$$</p><p><strong>Step 2:</strong> Expand along Row 1:</p><p>$$1(k(-3) - (-2)(4)) - k(3(-3) - (-2)(2)) + 3(3(4) - k(2)) = 0$$</p><p>$$1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0$$</p><p>$$-3k + 8 + 5k + 36 - 6k = 0$$</p><p>$$-4k + 44 = 0 \implies k = 11$$</p><p><strong>Step 3:</strong> Substitute k = 11 into the system and solve for ratios:</p><p>$$x + 11y + 3z = 0 \quad (1)$$</p><p>$$3x + 11y - 2z = 0 \quad (2)$$</p><p><strong>Step 4:</strong> From (2) - (1): $2x - 5z = 0 \implies x = \frac{5z}{2}$</p><p><strong>Step 5:</strong> Substitute into (1): $\frac{5z}{2} + 11y + 3z = 0 \implies 11y = -\frac{11z}{2} \implies y = -\frac{z}{2}$</p><p><strong>Step 6:</strong> Calculate the ratio:</p><p>$$\frac{xz}{y^2} = \frac{\frac{5z}{2} \cdot z}{(-\frac{z}{2})^2} = \frac{\frac{5z^2}{2}}{\frac{z^2}{4}} = \frac{5z^2}{2} \cdot \frac{4}{z^2} = 10$$</p><p>∴ Answer: <strong>10</strong></p>
Correct Answer: D