For the system of linear equations x + y + z = 1 , x + y + z = 1 , x + y + z = , which one of the following statements is NOT correct? [2023]

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Solution

Q.
For the system of linear equations $α x + y + z = 1, x + α y + z = 1, x + y + α z = β,$ which one of the following statements is NOT correct [2023]

1 It has infinitely many solutions if $α = 2$ and $β = - 1$

2 $x + y + z = \frac{3}{4}$ if $α = 2$ and $β = 1$

3 It has infinitely many solutions if $α = 1$ and $β = 1$

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

Ans.
(1)

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

$\Rightarrow α (α^{2} - 1) - 1 (α - 1) + (1 - α) = 0$

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

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

For $α = 2, β = 1;$ $Δ = 4$

$Δ_{1} =$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} |$ $= 3 - 1 - 1 = 1 \Rightarrow x = \frac{1}{4}$

$Δ_{2} =$ $| \begin{matrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} |$ $= 2 - 1 = 1 \Rightarrow y = \frac{1}{4}$

$Δ_{3} =$ $| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{matrix} |$ $= 2 - 1 = 1 \Rightarrow z = \frac{1}{4}$

For $α = 2 \Rightarrow$ unique solution and $x + y + z = \frac{3}{4}$

For $α = - 2, β = 1$

$Δ =$ $| \begin{matrix} - 2 & 1 & 1 \\ 1 & - 2 & 1 \\ 1 & 1 & - 2 \end{matrix} |$ $= - 2 (4 - 1) - 1 (- 2 - 1) + 1 (1 + 2)$

$= - 6 + 3 + 3 = 0$

$Δ_{1} =$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 2 & 1 \\ 1 & 1 & - 2 \end{matrix} |$ $= 3 + 3 + 3 = 9 \Rightarrow x = \frac{9}{0}$

$Δ_{2} =$ $| \begin{matrix} - 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & - 2 \end{matrix} |$ $= 6 + 3 + 0 = 9 \Rightarrow y = \frac{9}{0}$

$Δ_{3} =$ $| \begin{matrix} - 2 & 1 & 1 \\ 1 & - 2 & 1 \\ 1 & 1 & 1 \end{matrix} |$ $= 6 + 0 + 3 = 9 \Rightarrow z = \frac{9}{0}$

For $α = - 2 \Rightarrow$ no solution

For $α = 1$ and $β = 1$

$Δ = Δ_{1} = Δ_{2} = Δ_{3} = 0$

$∴ It has infinitely many solutions if α = 1 and β = 1$

For $α = 2$ and $β = - 1$

$Δ = 0$

$Δ_{1} =$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ - 1 & 1 & 2 \end{matrix} |$ $= 1 (3) - 1 (3) + 1 (3) = 3$

$Δ_{2} =$ $| \begin{matrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & - 1 & 2 \end{matrix} |$ $= 2 (3) - 1 (1) + 1 (- 2) = 3$

$Δ_{3} =$ $| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & - 1 \end{matrix} |$ $= 2 (- 3) - 1 (- 2) + 1 (- 1) = - 5$

Hence, it does not have infinitely many solutions if $α = 2$ and $β = - 1$ .

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