For a real number a, if the system $\begin{bmatrix}1&α&α^2\\α &1 &α\\α^2&α&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\-1\\1\end{bmatrix}$ of the linear equations, has infinitely many solutions, then $1+α+α^2=$
Answer & explanation
Correct answer: option 1
The given system of equation is
$AX = B$
where $A=\begin{bmatrix}1&α&α^2\\α &1 &α\\α^2&α&1\end{bmatrix},X=\begin{bmatrix}x\\y\\z\end{bmatrix}$ and $B=\begin{bmatrix}1\\-1\\1\end{bmatrix}$
If it has infinitely many solutions, then
$|A|=0⇒\begin{bmatrix}1&α&α^2\\α &1 &α\\α^2&α&1\end{bmatrix}=0$
$⇒(1-α^2)-a (α-α^3) + α^2 (α^2-α^2)=0$
$⇒1-2α^2+α^=0⇒ (α^2-1)^2=0⇒α=±1$
When $α = 1$, the given system reduces to
$x+y+z=1$
$x+y+z = -1$
and, $x+y+z=1$
Clearly, it is an inconsistent system.
When $α = -1$, the given system reduces
$x-y+z=1$
$-x+y-z=-1$
$x-y+z=1$
This system of equations is equivalent to $x-y+z = 1$ which has infinitely many solutions.
For this value of $α, 1+α+α^2=1-1+(-1)^2 = 1$.