Past exam of the algebra course of the University of Oxford / 2007 / A / 4 / iii

Table of contents
- Solution iii

Solution

For

d (X) = X^{n} + a_{n - 1} X^{n - 1} + \dots + a_{0}

, the determinant of its companion matrix is

(- 1)^{n} a_{0}

. Therefore

α invertible ⟺ a_{0} \neq = 0.

Thus invertibility is readable from the constant coefficient, with no factorization or elimination.

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