For a real number α, if the system 1 α α² α 1 α α² α 1 x y z = 1 -1 1 of linear equations, has infinitely many solutions, then 1 + α + α² =

For a real number α, if the system 1 α α² α 1 α α² α 1 x y z = 1 -1 1 of linear equations, has infinitely many solutions, then 1 + α + α² = A = 1 α α² α 1 α α² α 1 , b⃗ = 1 -1 1 Ax⃗ = b⃗ has infinitely many solutions Find: 1 + α + α² Infinitely many solutions ⇒ det A = 0 (and the system must be consistent) det A = 1(1-α²) - α(α-α³) + α²(α²-α²) = 1 - 2α² + α⁴ = (α²-1)² (α²-1)² = 0 ⇒ α = ± 1 α = 1: A = 1 1 1 1 1 1 1 1 1 , b⃗ = (1,-1,1)^T x+y+z = 1 and x+y+z = -1 ⇒ inconsistent α = -1: A = 1 -1 1 -1 1 -1 1 -1 1 all rows ⇒ x - y + z = 1 ⇒ infinitely many solutions therefore α = -1 1 + α + α² = 1 - 1 + 1 = 1

For a real number $\alpha$ , if the system $\begin{bmatrix} 1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ of linear equations, has infinitely many solutions, then $1 + \alpha + \alpha^{2} =$

Answer: $1$

Determinants For a real number

\alpha

, if the system

\begin{bmatrix} 1 & \alpha & \alpha^{2} \\ \alpha & 1 & \alpha \\ \alpha^{2} & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

of linear equations, has infinitely many solutions, then

1 + \alpha + \alpha^{2} =

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Written solution

$A = \begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix}, \quad \vec{b} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ $A\vec{x} = \vec{b} \text{ has infinitely many solutions}$ $\text{Find: } 1 + \alpha + \alpha^2$ $\text{Infinitely many solutions} \Rightarrow \det A = 0$ $\text{(and the system must be consistent)}$ $\det A = 1(1-\alpha^2) - \alpha(\alpha-\alpha^3) + \alpha^2(\alpha^2-\alpha^2)$ $= 1 - 2\alpha^2 + \alpha^4 = (\alpha^2-1)^2$ $(\alpha^2-1)^2 = 0 \Rightarrow \alpha = \pm 1$ $\alpha = 1: A = \begin{bmatrix} 1&1&1\\1&1&1\\1&1&1 \end{bmatrix}, \quad \vec{b} = (1,-1,1)^T$ $x+y+z = 1 \text{ and } x+y+z = -1 \Rightarrow \text{inconsistent}$ $\alpha = -1: A = \begin{bmatrix} 1&-1&1\\-1&1&-1\\1&-1&1 \end{bmatrix}$ $\text{all rows} \Rightarrow x - y + z = 1 \Rightarrow \text{infinitely many solutions}$ $\therefore \alpha = -1$ $1 + \alpha + \alpha^2 = 1 - 1 + 1 = 1$

Written solution

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