Question

Determinati matricea inversei aplicatiei f: R^3->R^3, f(x,y,z):=(x+y,y,z+x) in baza {(0,1,1),(1,1,0),(1,0,1)}

Answer 1

fie (u,v,t) ∈ R³
cautam (x,y,z)∈ R³ a.i. f(x,y,z)= (u,v,t)
u= x+y
v= y
t= z+x

y=v
x=u-v
z=t-u+v

deci matricea functiei inverse este g(u,v,t)= (u-v,v,t-u+v)

O seara buna!