Question

Let the system of linear equations $\left[\begin{array}{ccc}1& \lambda & {\lambda }^2+5\\ 1& \gamma & {\gamma }^2+5\\ 1& 5& 30\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}2\\ 4\\ -1\end{array}\right]$ has a unique solution. Then which of the following is TRUE?

• A. $\lambda =5,\gamma =6$
• B. $\lambda =7,\gamma =5$
• C. $\lambda =\gamma$
• D. $\lambda =7,\gamma =6$

Answer 1

The correct option is D $\lambda =7,\gamma =6$

$\Delta =\left|\begin{array}{ccc}1& \lambda & {\lambda }^2+5\\ 1& \gamma & {\gamma }^2+5\\ 1& 5& 30\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1,$ we get
$\Delta =\left|\begin{array}{ccc}1& \lambda & {\lambda }^2+5\\ 0& \gamma -\lambda & {\gamma }^2-{\lambda }^2\\ 0& 5-\lambda & 25-{\lambda }^2\end{array}\right|$

$\Rightarrow \Delta =(\gamma -\lambda )(5-\lambda )\left|\begin{array}{ccc}1& \lambda & {\lambda }^2+5\\ 0& 1& \gamma +\lambda \\ 0& 1& 5+\lambda \end{array}\right|$

Expanding along $C_1,$ we get
$\Delta =(\gamma -\lambda )(5-\lambda )(5-\gamma )$
For unique solution, $\Delta \ne 0$
$\Rightarrow \lambda \ne \gamma$ and $\lambda ,\gamma \ne 5$