Resposta:
{x ∈ R / x = ± 3}
Explicação passo a passo:
[tex]\left[\begin{array}{ccc}1&-x&x^2\\-1&0&x\\3&-2&2\\\end{array}\right] \left\begin{array}{ccc}1&-x\\-1&0\\3&-2\end{array}\right] \\ \\ (1\times 0\times 2)+(-x\times x\times 3)+[x^2\times(-1)\times(-2)]-[-x\times(-1)\times\ 2]-[1\timesx\times (-2)]-(x^2\times0\times 3)=-9\\ 0-3x^2+2x^2-2x+2x-0=9\\ -x^2=-9\\ x^2=9\\ x=\pm\sqrt{9} \\ x=\pm3[/tex]
Espero ter ajudado.
