For a real number α, if the system of linear equations, has infinitely many solutions, then 1 + α + a2 = ?

Engineering

Mathematics

Introduction to Determinants

Properties of Determinant

Solving System of Linear Equation Cramers Rule

Question

For a real number α, if the system $[\begin{array}{ccc} 1 & α & α^{2} \\ α & 1 & α \\ α^{2} & α & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 1 \\ - 1 \\ 1 \end{array}]$ of linear equations, has infinitely many solutions, then 1 + α + a² = ?

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Consider the matrix

[\begin{array}{ccc} 1 & α & α^{2} \\ α & 1 & α \\ α^{2} & α & 1 \end{array}]

determinant of

∣ \begin{array}{ccc} 1 & α & α^{2} \\ α & 1 & α \\ α^{2} & α & 1 \end{array} ∣

= 1(1 – α²) – α(α – α³) + α²(α² – α²) = 0

= 1 – α² – α² + α⁴ = 0

= α⁴– 2α² + 1 = 0

= (α² – 1)² = 0

⇒ α = ± 1

For

α = 1 [\begin{array}{ccc} 1 & α & α^{2} \\ α & 1 & α \\ α^{2} & α & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{cc} 1 \\ - 1 & α \\ 1 \end{array}]

becomes

[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 1 \\ - 1 \\ 1 \end{array}]

x + y + z = 1, x + y + z = – 1, x + y + z = 1 thus this system has no solution.

For

α = - 1 [\begin{array}{ccc} 1 & α & α^{2} \\ α & 1 & α \\ α^{2} & α & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{cc} 1 \\ - 1 & α \\ 1 \end{array}]

becomes

[\begin{array}{ccc} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{array}] [\begin{array}{c} x \\ y \\ z \end{array}] = [\begin{array}{c} 1 \\ - 1 \\ 1 \end{array}]

x – y + z = 1, – x + y – z = – 1, x – y + z = 1 thus this system has solution.

∴ α = – 1

Substituting α = – 1 in 1 + α + α² = 1 – 1 + 1 = 1

∴ 1 + α + α² = 1