Matrices & Determinants
Determinant Equation — Finding Parameter Range
nta_pyq_2024_jan
Grade 12
Question:
The values of $\alpha$, for which $\begin{vmatrix}1&\frac{3}{2}&\alpha+\frac{3}{2}\\1&\frac{1}{3}&\alpha+\frac{1}{3}\\2\alpha+3&3\alpha+1&0\end{vmatrix}=0$, lie in the interval
$(-2,1)$
$(-3,0)$
$\left(-\frac{3}{2},\frac{3}{2}\right)$
$(0,3)$
Step-by-Step Solution
Key Concept: Expand the determinant along suitable row/column (or use row operations) to obtain a quadratic in $\alpha$. Solve and check which interval contains both roots.
Expanding the determinant: $(2\alpha+3)\cdot\frac{7\alpha}{6}+(3\alpha+1)\cdot\frac{7}{6}=0\Rightarrow 2\alpha^2+6\alpha+1=0\Rightarrow\alpha=\frac{-3+\sqrt{7}}{2},\frac{-3-\sqrt{7}}{2}$. Since $\sqrt{7}\approx2.65$: $\alpha\approx-0.18$ and $\alpha\approx-2.82$, both in $(-3,0)$.
Correct Answer: 2