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. 44563
• D. 44562

Answer 1

Answer: (C)

44563

Answer 2

We have
A:B $\begin{vmatrix} 1 & 1 & 1 & : & 1 \\\ 1 & 2 &4&:& \alpha \\\ 1&4&10&:&\alpha^2 \\\ \end{vmatrix}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \sim$ $\begin{vmatrix} 1 & 1 & 1 & : & 1 \\\ 0 & 1 &3&:& \alpha-1 \\\ 0&3&9&:&\alpha^2-1 \\\ \end{vmatrix}$
$\begin{bmatrix} applying \ R_2 \ \rightarrow \ R_2 \ - \ R_1 \\\ \& \ R_3 \ \rightarrow \ R_3 \ - \ R_1 \\\ \end{bmatrix}$
$\ \ \ \sim$ $\begin{vmatrix} 1 & 1 & 1 & : & 1 \\\ 0 & 1 &3&:& \alpha-1 \\\ 0&0&0&:&\alpha^2-3\alpha+2 \\\ \end{vmatrix}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [applying \ \ R_3 \ \rightarrow \ R_3 \ - \ 3R_2]$
But the system is consistent
$\therefore \ \ \alpha^2 - \ 3 \alpha + \ 2 \ =0$
$\Rightarrow \ \ (\alpha \ - \ 2)(\alpha \ - \ 1)=0 \ \Rightarrow \ \alpha \ =2 \ or \ \alpha \ = \ 1$