Matrices & Determinants
Determinant Identity — Quadratic with Roots
nta_pyq_2023_apr
Grade 12
Question:
If $\begin{vmatrix}x+1&x&x\\x&x+\lambda&x\\x&x&x+\lambda^2\end{vmatrix}=\dfrac{9}{8}(103x+81)$, then $\dfrac{\lambda}{8}$ and $\dfrac{\lambda}{3}$ are the roots of the equation
$4x^2+24x-27=0$
$4x^2-24x-27=0$
$4x^2+24x+27=0$
$4x^2-24x+27=0$
Step-by-Step Solution
Key Concept: Substitute $x=0$ to find $\lambda^3=\frac{9}{8}\times81=\frac{9^3}{2^3}\Rightarrow\lambda=\frac{9}{2}$. Then $\frac{\lambda}{8}=\frac{9}{16}$ and $\frac{\lambda}{3}=\frac{3}{2}$.
$\lambda=\frac{9}{2}$. Roots $\frac{\lambda}{8}=\frac{9}{16}$ and $\frac{\lambda}{3}=\frac{3}{2}$. Quadratic: $4x^2-24x+27=0$.
Correct Answer: 4