### 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