Question

Determine o valor de x para que o determinante da matriz A seja igual a 8.

Answer 1

$A=\left[\begin{array}{cc}x&-3\\x+2&x-2\end{array}\right]$

$\det{A}=x(x-2)-(-3)(x+2)\ \therefore$

$\det{A}=x^2-2x+3x+6\ \therefore$

$\boxed{\det{A}=x^2+x+6}$

$\det{A}=8\ \therefore\ x^2+x+6=8\ \therefore\ x^2+x-2=0$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-1\pm\sqrt{1^2-4(1)(-2)}}{2(1)}=\dfrac{-1\pm3}{2}$

$\boxed{x=-2}\ \text{ou}\ \boxed{x=+1}$