Consider the following system of linear equations: $x+2y+z=1,$ $2x+y+z=\alpha$,$4x+5y+3z={\alpha }^2$. Then the system has

Given system can be written as

$\begin{array}{l}\left[\begin{array}{lll}1& 2& 1\\ 2& 1& 1\\ 4& 5& 3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ \alpha \\ {\alpha }^2\end{array}\right]\\ \mathrm{\Delta }=\left|\begin{array}{lll}1& 2& 1\\ 2& 1& 1\\ 4& 5& 3\end{array}\right|=0\\ {\mathrm{\Delta }}_1=\left|\begin{array}{lll}1& 2& 1\\ \alpha & 1& 1\\ {\alpha }^2& 5& 3\end{array}\right|={\alpha }^2-\alpha -2\end{array}$

$\begin{array}{l}{\mathrm{\Delta }}_2=\left|\begin{array}{lll}1& 1& 1\\ 2& \alpha & 1\\ 4& {\alpha }^2& 3\end{array}\right|={\alpha }^2-\alpha -2\\ {\mathrm{\Delta }}_3=\left|\begin{array}{lll}1& 2& 1\\ 2& 1& \alpha \\ 4& 5& {\alpha }^2\end{array}\right|=-3\left({\alpha }^2-\alpha -2\right)\end{array}$

The system possesses infinitely many solutions if $\mathrm{\Delta }=$0 and ${\mathrm{\Delta }}_i=0$ for all i = 1, 2, 3 

i.e., when ${\alpha }^2-\alpha -2=0\Rightarrow \text{ when }\alpha =-1,2$

If a assumes any value other than -1 and 2, then, the system has no solution