Matrix/x^2+x -x -x^3+2x^2+5x-1 x^2-x/x invertible/Exercise/Solution

A matrix is invertible if and only if its determinant is $\neq 0$ . The determinant of the matrix is

\begin{matrix} det (\begin{matrix} x^{2} + x & - x \\ - x^{3} + 2 x^{2} + 5 x - 1 & x^{2} - x \end{matrix}) & = (x^{2} + x) (x^{2} - x) + x (- x^{3} + 2 x^{2} + 5 x - 1) \\ = x^{4} - x^{2} - x^{4} + 2 x^{3} + 5 x^{2} - x \\ = 2 x^{3} + 4 x^{2} - x \\ = x (2 x^{2} + 4 x - 1) . \end{matrix}

This equals $0$ for $x = x_{1} = 0$ or for

2 x^{2} + 4 x - 1 = 0 .

This quadratic equation is equivalent to

x^{2} + 2 x - \frac{1}{2} = 0

and so

(x + 1)^{2} - 1 - \frac{1}{2} = 0 .

Hence

x + 1 = \pm \sqrt{\frac{3}{2}}

and therefore

x_{2} = \sqrt{\frac{3}{2}} - 1 and x_{3} = - \sqrt{\frac{3}{2}} - 1.

The only complex numbers for which the matrix is not invertible are thus

0, \sqrt{\frac{3}{2}} - 1, - \sqrt{\frac{3}{2}} - 1.