Matrices & Determinants
Non-trivial Solution Condition
nta_pyq_2024_apr
Grade 12
Question:
If the system of equations $x+(\sqrt{2}\sin\alpha)y+(\sqrt{2}\cos\alpha)z=0$, $x+(\cos\alpha)y+(\sin\alpha)z=0$, $x+(\sin\alpha)y-(\cos\alpha)z=0$ has a non-trivial solution, then $\alpha\in\left(0,\dfrac{\pi}{2}\right)$ is equal to:
$\dfrac{11\pi}{24}$
$\dfrac{5\pi}{24}$
$\dfrac{7\pi}{24}$
$\dfrac{3\pi}{4}$
Step-by-Step Solution
Key Concept: Set determinant = 0. Expanding: $1+\sqrt{2}\cos2\alpha-\sqrt{2}\sin2\alpha=0\Rightarrow\cos2\alpha-\sin2\alpha=-\frac{1}{\sqrt{2}}\Rightarrow\cos(2\alpha+\pi/4)=-\frac{1}{2}$.
$\cos(2\alpha+\pi/4)=-1/2\Rightarrow2\alpha+\pi/4=2\pi/3\Rightarrow\alpha=5\pi/24$.
Correct Answer: 2