- Home
- Standard 12
- Mathematics
Consider the following system of questions $\alpha x+2 y+z=1$ ; $2 \alpha x+3 y+z=1$ ; $3 x+\alpha y+2 z=\beta$ . For some $\alpha, \beta \in R$. Then which of the following is NOT correct.
It has no solution if $\alpha=-1$ and $\beta \neq 2$
It has no solution for $\alpha=-1$ and for all $\beta \in R$
It has no solution for $\alpha=3$ and for all $\beta \neq 2$
It has a solution for all $\alpha \neq-1$ and $\beta=2$
Solution
$D=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{array}\right|=0 \Rightarrow \alpha=-1,3$
$D_x=\left|\begin{array}{lll}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{array}\right|=0 \Rightarrow \beta=2$
$D_y=\left|\begin{array}{ccc}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{array}\right|=0$
$D_z=\left|\begin{array}{ccc}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{array}\right|=0$
$\beta=2, \alpha=-1$
$\alpha=-1, \beta=2$ Infinite solution
