A matrix is invertible if and only if its determinant is
. The determinant of the matrix is
![{\displaystyle {}{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11fffbd2ae44b65f1d756c56b40c77ce9c483ab3)
This equals
for
or for
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![{\displaystyle {}2x^{2}+4x-1=0\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bcd6b8b6aa6bf7f1aa3bc8b3d932ad6a8e14730)
This quadratic equation is equivalent with
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![{\displaystyle {}x^{2}+2x-{\frac {1}{2}}=0\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac7f923b0ea099b3c1691fa77fc482d041aff6f7)
and so
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![{\displaystyle {}(x+1)^{2}-1-{\frac {1}{2}}=0\,.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f13c11a302dc7b9c65237f03d5d022f3bc880f86)
Hence
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![{\displaystyle {}x+1=\pm {\sqrt {\frac {3}{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5026b5dd284b8e270b785ed5c6d23301c4e2a73e)
and therefore
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The only complex numbers for which the matrix is not invertible are thus
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