Question

X+ay+2z=1 ,x+(2a-1)y+3z=1 ,x+ay+(a-3)z=1,sa se rezolve ecuatia det(A)=0,unde A este matricea asociata sistemului.Sa se rezolve sistemul in multimea numerelor reale in functie de parametrul a apartine nr reale

Answer 1

$\left \{ {{x+ay+2z=1} \atop {x+(2a-1)y+3z=1}}\atop {x+ay+(a-3)z=1}} \right.$

det(A)=0 <=>$\left[\begin{array}{ccc}1&a&2\\1&2a-1&3\\1&a&a-3\end{array}\right]$=(2a-1)(a-3)+2a+3a-2(2a-1)-3a-a(a-3)=2$a^{2}$-6a-a+3+2a+3a-4a+2-3a-$a^{2}$+3a=$a^{2}$-6a+5=0

Δ=36-20=16
 

a1=5 si a2=1