For a real number $\alpha$, if the system \[ \left[\begin{array}{ccc} 1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 1 \\ -1 \\ 1 \end{array}\right] \] of linear equations, has infinitely many solutions, then $1+\alpha+\alpha^{2}=$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Matrices

For a real number

α

, if the system

1 α α^{2} α 1 α α^{2} α 1 x y z = 1 - 1 1

of linear equations, has infinitely many solutions, then

1 + α + α^{2} =

Answer: 1

For infinitely many solutions, the determinant must be zero. Computing the determinant:

D = 1 (1 - α^{2}) - α (α - α^{3}) + α^{2} (α^{2} - 1) = 1 - α^{2} - α^{2} + α^{4} + α^{4} - α^{2} = 2 α^{4} - 3 α^{2} + 1 = (2 α^{2} - 1) (α^{2} - 1) = 0

. This gives

α = \pm 1

α = \pm \frac{1}{2}

. For

α = 1

, all three equations become

x + y + z = 1

x + y + z = - 1

x + y + z = 1

, which are inconsistent. For

α = - 1

, all equations become

x - y + z = 1

, giving infinitely many solutions. Thus

α = - 1

and

1 + α + α^{2} = 1 - 1 + 1 = 1

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer