Q43 / 120 · VITEEE 2006 · Mathematics
The values of $\alpha$ for which the system of equation
$x+y+z=1, x+2 y+4 z=\alpha, x+4 y+10 z=\alpha^{2}$ is consistent are given by
- A$1,-2$
- B$-1,2$
- Correct answerC1,2
- D1,1
Solution
$\mathrm{A}: \mathrm{B}=\left|\begin{array}{ccccc}1 & 1 & 1 & : & 1 \\ 1 & 2 & 4 & : & \alpha \\ 1 & 4 & 10 & : & \alpha^{2}\end{array}\right|$
$\sim\left|\begin{array}{cccc}1 & 1 & 1 & 1 \\ 0 & 1 & 3: & \alpha-1 \\ 0 & 3 & 9 & \alpha^{2}-1\end{array}\right|$
$\left[\begin{array}{l}\text { applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \\ \& \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\end{array}\right]$
$\begin{array}{r}
\sim\left|\begin{array}{ccccc}
1 & 1 & 1 & : & 1 \\
0 & 1 & 3 & : & \alpha-1 \\
0 & 0 & 0 & : & \alpha^{2}-3 \alpha+2
\end{array}\right| \\
& {\left[\text { applying } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2}\right]}
\end{array}$
But the system is consistent
$\begin{array}{l}
\therefore \alpha^{2}-3 \alpha+2=0 \\
\Rightarrow(\alpha-2)(\alpha-1)=0 \Rightarrow \alpha=2 \text { or } \alpha=1
\end{array}$
Hence, the correct answer is option: 'C'.