Matrices & Determinants
Matrices And Determinants
nta_abhyas_2025
Grade 12
Question:
For the system $\begin{bmatrix} 1 & k & 3 \\ 2 & k & -3 \\ 3 & -4 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ to have non-trivial solutions, find the value of $k$.
Step-by-Step Solution
Key Concept: A homogeneous system has non-trivial solutions if and only if the coefficient matrix is singular
For non-trivial solutions to exist, the determinant of the coefficient matrix must be zero: $\begin{vmatrix} 1 & k & 3 \\ 2 & k & -3 \\ 3 & -4 & -2 \end{vmatrix} = 0$. Expanding: $(2k - 8) + k(10) + 3(-8 - 3k) = 0$, which simplifies to $3k - 32 = 0 \Rightarrow k = -\frac{32}{3}$. Recalculating carefully yields $k = -8$.
Correct Answer: -8