Consider the following system of equations x + 2 y + z = 1 2 x + 3 y + z = 1 3 x + y + 2 z = for some. Then which of the following is NOT correct? [2023]

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Solution

Q.
Consider the following system of equations

$α x + 2 y + z = 1$

$2 α x + 3 y + z = 1$

$3 x + α y + 2 z = β$

for some $α, β \in ℝ$ . Then which of the following is NOT correct [2023]

1 It has no solution if $α = - 1$ and $β \neq 2$

2 It has no solution for $α = - 1$ and for all $β \in ℝ$

3 It has a solution for all $α \neq - 1$ and $β = 2$

4 It has no solution for $α = 3$ and for all $β \neq 2$

Ans.
(2)

Given system of equations has no solution if

$| \begin{matrix} α & 2 & 1 \\ 2 α & 3 & 1 \\ 3 & α & 2 \end{matrix} |$ $= 0$

$\Rightarrow α (6 - α) - 2 (4 α - 3) + (2 α^{2} - 9) = 0 \Rightarrow α = 3, - 1$

For no solution, any one of $Δ_{x}, Δ_{y}, Δ_{z}$ should be non-zero.

Now, $Δ_{x} =$ $| \begin{matrix} 2 & 1 & 1 \\ 3 & 1 & 1 \\ α & 2 & β \end{matrix} |$ $\neq 0 \Rightarrow β \neq 2$

Now, $Δ_{y} =$ $| \begin{matrix} α & 1 & 1 \\ 2 α & 1 & 1 \\ 3 & 2 & β \end{matrix} |$ $\neq 0 \Rightarrow - α β + 2 α \neq 0$

For $α = - 1 \Rightarrow β \neq 2$

So, option (2) is incorrect as it has no solution for $α = - 1$ and $β \in R - {2}$ .

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