Mathematics - Determinant Question with Solution | TestHub

Given: $\left|\begin{array}{ccc}1& \frac{3}{2}& \mathrm{\alpha }+\frac{3}{2}\\ 1& \frac{1}{3}& \mathrm{\alpha }+\frac{1}{3}\\ 2\mathrm{\alpha }+3& 3\mathrm{\alpha }+1& 0\end{array}\right|=0$

Applying, $R_1\to R_1-R_2$

$\Rightarrow \left|\begin{array}{ccc}0& \frac{7}{6}& \frac{7}{6}\\ 1& \frac{1}{3}& \mathrm{\alpha }+\frac{1}{3}\\ 2\mathrm{\alpha }+3& 3\mathrm{\alpha }+1& 0\end{array}\right|=0$

Applying, $C_2\to C_2-C_3$

$\Rightarrow \left|\begin{array}{ccc}0& 0& \frac{7}{6}\\ 1& -\mathrm{\alpha }& \mathrm{\alpha }+\frac{1}{3}\\ 2\mathrm{\alpha }+3& 3\mathrm{\alpha }+1& 0\end{array}\right|=0$

$\Rightarrow 0-0+\frac{7}{6}\left(3\mathrm{\alpha }+1+2{\mathrm{\alpha }}^2+3\mathrm{\alpha }\right)=0$

$\Rightarrow 2{\mathrm{\alpha }}^2+6\mathrm{\alpha }+1=0$

$\Rightarrow \alpha =\frac{-6\pm \sqrt{36-4\left(2\right)\left(1\right)}}{2\times 2}$

$\Rightarrow \alpha =\frac{-6\pm 2\sqrt{7}}{4}$

$\Rightarrow \alpha =\frac{-3\pm \sqrt{7}}{2}$

$\Rightarrow \alpha \approx \frac{-3\pm 2.6}{2}$

$\Rightarrow \alpha \approx \frac{-0.4}{2}, \frac{-5.6}{2}$

$\Rightarrow \alpha \approx -0.2, -2.8$

$\Rightarrow \alpha \in \left(-3,0\right)$