Determinants Question 3
Question 3 - 29 January - Shift 1
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 $\alpha, \beta \in \mathbb{R}$. Then which of the following is NOT correct.
(1) It has no solution if $\alpha=-1$ and $\beta \neq 2$
(2) It has no solution for $\alpha=-1$ and for all $\beta \in \mathbb{R}$
(3) It has no solution for $\alpha=3$ and for all $\beta \neq 2$
(4) It has a solution for all $\alpha \neq-1$ and $\beta=2$
Show Answer
Answer: (2)
Solution:
Formula: System of equations with 3 variables, consistency of solutions: inconsistent
$D= \begin{vmatrix} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{vmatrix} =0 \Rightarrow \alpha=-1,3$
$D_x= \begin{vmatrix} 2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{vmatrix} =0 \Rightarrow \beta=2$
$D_y= \begin{vmatrix} \alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{vmatrix} =0$
$D_z= \begin{vmatrix} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{vmatrix} =0$
$\beta=2, \alpha=-1$
$\alpha=-1, \beta=2$ Infinite solution
