Matrices & Determinants
Non-trivial Solution — Expression Value
nta_pyq_2024_apr
Grade 12
Question:
Let $\alpha\beta\gamma=45$; $\alpha,\beta,\gamma\in\mathbb{R}$. If $x(\alpha,1,2)+y(1,\beta,2)+z(2,3,\gamma)=(0,0,0)$ for some $x,y,z\in\mathbb{R}$, $xyz\neq0$, then $6\alpha+4\beta+\gamma$ is equal to ________.
Step-by-Step Solution
Key Concept: Non-trivial solution requires $\det=0$: $\begin{vmatrix}\alpha&1&2\\1&\beta&3\\2&2&\gamma\end{vmatrix}=0$. Expanding: $\alpha\beta\gamma-6\alpha-\gamma+6+4-4\beta=0$. Use $\alpha\beta\gamma=45$.
$6\alpha+4\beta+\gamma=\alpha\beta\gamma+10=45+10=55$.
Correct Answer: 55