Question

Consider the following system of equations
$ \alpha x+2 y+z=1 $
$ 2 \alpha x+3 y+z=1 $
$ 3 x+\alpha y+2 z=\beta$
for some $\alpha, \beta \in R$. Then which of the following is NOT correct.

• A. It has no solution if $\alpha=-1$ and $\beta \neq 2$
• B. It has no solution for $\alpha=-1$ and for all $\beta \in R$
• C. It has a solution for all $\alpha \neq-1$ and $\beta=2$
• D. It has no solution for $\alpha=3$ and for all $\beta \neq 2$

Answer 1

Answer: (B)

$ D=\begin{vmatrix}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{vmatrix}=0 \Rightarrow \alpha=-1,3 $
$ D_x=\begin{vmatrix}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{vmatrix}=0 \Rightarrow \beta=2 $
$ D_y= \begin{vmatrix}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{vmatrix}=0$
$ D_z= \begin{vmatrix} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{vmatrix}=0 $
$ \beta=2, \alpha=-1 $
$\alpha=-1, \beta=2 $ Infinite solution