First slide

System of linear equations

Question

The system of equations $α x + y + z = α - 1, x + α y + z = α - 1, x + y + α z = α - 1$ has no solution, if $α =$

Difficult

A

-2
B

1
C

2
D

either -2 or -1

Solution

Let $Δ = |\begin{matrix} α & 1 & 1 \\ 1 & α & 1 \\ 1 & 1 & α \end{matrix}|$
   $C_{1} \to C_{1} + C_{2} + C_{3}$
   $= |\begin{matrix} α + 2 & 1 & 1 \\ α + 2 & α & 1 \\ α + 2 & 1 & α \end{matrix}|$
= $(α + 2) |\begin{matrix} 1 & 1 & 1 \\ 1 & α & 1 \\ 1 & 1 & α \end{matrix}|$
$R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3}$
   $= (α + 2) |\begin{matrix} 0 & 1 - α & 0 \\ 0 & α - 1 & 1 - α \\ 1 & 1 & α \end{matrix}| = (α + 2) [o (α (α - 1) - 1 (1 - α)) - (1 - α) [0 - (1 - α)]]$
$= (α + 2) {(1 - α)}^{2}$
If $α = 1$ the system of equations has infinitely many solutions If $α = - 2$ , on adding 3 equations we obtain $0 = - 9$
$∴$ The system has no solution for $α = - 2$

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