Question

Determina e tal que:

Answer 1

Resposta:

$\left[\begin{array}{ccc}e-1&2&e\\0&1&-1\\3e&e+1&2e\end{array}\right] \\\\\\\\a22=-1*\left[\begin{array}{cc}e-1&e\\3e&2e\end{array}\right] -((e-1)*2e - e* 3e)=2e^2-2e-3e^2 = e^2+2e\\a23=1*\left[\begin{array}{cc}e-1&2\\3e&e+1\end{array}\right] (e-1*e+1)-6e=e^2-1-6e\\e^2+2e+e^2-1-6e\\2e^2-4e-1 = -3e^2-8e\\5e^2+4e-1 = 0$

delta=36

(-4 - 6)/2*5 =-1

-4+6/2*5 = 1/5

substituindo -1 fica

$\left[\begin{array}{ccc}-2&2&-1\\0&1&-1\\-3&0&-2\end{array}\right] -1*((-2*-2)-(-1*-3))+1(0-(2*-3))\\-4+3+6=5\ determinante\ da\ primeira\\\-3+8=5\ determinante\ da\ segunda$