Only answers with formally and fully justified written proofs will be granted 5

Question

Only answers with formally and fully justified written proofs will be granted 5 stars.

Define f : R times R rightarrow R times R as follows: f(x, y) = (3x: + 5y,x + 2y), for all (x, y) R times R. Is f one-to-one? Prove or give a counterexample. Is f onto? Prove or give a counterexample. Is f a one-to-one correspondence? For A = [3 5 1 2 ] find A - 1 if it exists. Define f : R times R rightarrow R times R as follows: f(x, y) = (x + 2y, 2x + 4y), for all (x, y) R times R. Is f one-to-one? Prove or give a counterexample. Is f onto? Prove or give a counterexample. Is f a one-to-one correspondence? For A = [1 2 2 4] find A - 1, if it exists.

Explanation / Answer

(a) suppose f(x1,y1) = f(x2,y2) then :

x1+2y1 = x2+2y2
2x1+4y1 = 2x2+4y2

We see that second line is the first (factor 2)
For instance for (x1,y1)=(0,1) and (x2,y2)=(1,1/2) f(x1,y1)=f(x2,y2), so f is not one-to-one.

So f is NOT one-to-one


(b)

Let (u,v) in R^2, then

(u,v) = f(x,y) <=>
u = x+2y
v = 2x+4y

=>2u = v

So for any (u,v) in R^2, you can't find (x,y) such that f(x,y)=(u,v)
so f is NOT onto


(c) Since f is not one-to-one f can't be a one-to-one correspondence


(d) As before, A is the matrix representation of f, since f isn't a one-to-one correspondance, A^(-1) doesn't not exist.
We verify furthermore that det(A) = 1*4-2*2 = 0, which validate this hypothesis.

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.