Question

The values of $\alpha$ for which the system of equation x + y + z = 1, x + 2y + 4z = $\alpha$, x + 4y + 10z = ${\alpha }^2$ is consistent are given by

• A. 1, -2
• B. -1, 2
• C. 1, 2
• D. 1, 1

Answer 1

Answer: (C)

We have
A:B $\left|\begin{array}{ccccc}1& 1& 1& :& 1\\ 1& 2& 4& :& \alpha \\ 1& 4& 10& :& {\alpha }^2\end{array}\right|$
$\sim$ $\left|\begin{array}{ccccc}1& 1& 1& :& 1\\ 0& 1& 3& :& \alpha -1\\ 0& 3& 9& :& {\alpha }^2-1\end{array}\right|$
$\left[\begin{array}{c}applying{R}_2\to R_2-R_1\\ \&R_3\to R_3-R_1\end{array}\right]$
$\sim$ $\left|\begin{array}{ccccc}1& 1& 1& :& 1\\ 0& 1& 3& :& \alpha -1\\ 0& 0& 0& :& {\alpha }^2-3\alpha +2\end{array}\right|$
$[applying{R}_3\to R_3-3R_2]$
But the system is consistent
$\therefore {\alpha }^2-3\alpha +2=0$
$\Rightarrow (\alpha -2)(\alpha -1)=0\Rightarrow \alpha =2or\alpha =1$