Resposta:
[tex]\boxed{\boxed{x=-\frac{9}{2} \lor x=0}}[/tex]
Explicação passo-a-passo:
[tex]A\cdotp B=\begin{bmatrix}x & 5\\1 & -2\end{bmatrix} \cdotp \begin{bmatrix}\frac{1}{2} & 1\\-1 & x\end{bmatrix} \\ \\ =\begin{bmatrix}\left(\frac{x}{2} -5\right) & ( x+5x)\\\left(\frac{1}{2} +2\right) & ( 1-2x)\end{bmatrix} \\ \\ =\begin{bmatrix}\left(\frac{x}{2} -5\right) & ( 6x)\\\left(\frac{5}{2}\right) & ( 1-2x)\end{bmatrix}[/tex]
O enunciado afirma que o determinante dessa matriz é -5. Resolvendo:
[tex]det( A\cdotp B) =\begin{vmatrix}\left(\frac{x}{2} -5\right) & ( 6x)\\\left(\frac{5}{2}\right) & ( 1-2x)\end{vmatrix} \Leftrightarrow \begin{vmatrix}\left(\frac{x}{2} -5\right) & ( 6x)\\\left(\frac{5}{2}\right) & ( 1-2x)\end{vmatrix} =-5\Leftrightarrow \\ \\ \Leftrightarrow \left[\left(\frac{x}{2} -5\right) \cdotp ( 1-2x)\right] -\left( 6x\cdotp \frac{5}{2}\right) =-5\Leftrightarrow \\ \\ \Leftrightarrow \left(\frac{x}{2} -x^{2} -5+10x\right) -15x=-5\Leftrightarrow \\ \\ \Leftrightarrow -x^{2} -\frac{9x}{2} -5=-5\Leftrightarrow x^{2} +\frac{9x}{2} =0\Leftrightarrow x\left( x+\frac{9}{2}\right) =0 \\ \therefore \boxed{\boxed{x=-\frac{9}{2} \lor x=0}}[/tex]
