9231 P13 - Jun 2022 - Q02
4240
The cubic equation \(x^3 + 5x^2 + 10x - 2 = 0\) has roots \(\alpha, \beta, \gamma\).
(a) Find the value of \(\alpha^2 + \beta^2 + \gamma^2\).
(b) Show that the matrix \(\begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}\) is singular.
Solution
Checked by expert
(a) By Vieta's formulas, we have:
\(\alpha + \beta + \gamma = -5\)
\(\alpha\beta + \beta\gamma + \gamma\alpha = 10\)
\(\alpha\beta\gamma = 2\)
We use the identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
\(\alpha^2 + \beta^2 + \gamma^2 = (-5)^2 - 2(10) = 25 - 20 = 5\)
(b) To show the matrix is singular, we find its determinant:
\(\begin{vmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{vmatrix} = 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)
Substituting the known values:
\(1 - 5 + 2(2) = 0\)
Thus, the determinant is zero, and the matrix is singular.
