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

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

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

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

This quadratic equation is equivalent with

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

and so

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

Hence

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

and therefore

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

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

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

To the exercise