If $(a,b,c)\in \mathbb R^3$, under what condition do we have $(a,b,c)\in f(\mathbb R^3)$

Question

Let's say that $f(x,y) = (x+y, 3x-y,2x+y)$.

If $(a,b,c)\in \mathbb R^3$, under what condition do we have $(a,b,c)\in f(\mathbb R^3)$

How would I solve this without using matrices, and only linear systems? I don't really understand how to go about it. What exactly do we want to show?

if $(a,b,c)\in f(\mathbb R^3)$ then $(a,b,c)\in (x+y, 3x-y,2x+y)$. I don't really know where to go from there? Should I be finding the solutions to $(a,b,c) = (x+y, 3x-y,2x+y)$? Wouldn't this be the equivalent of finding the inverse function? Any help is appreciated, thank you.

Answer 1

You have a system of equations $$ \begin{array}{llll} x &+ & y &= a \\ 3x &- &y &= b \\ 2x &+ & y &= c \end{array}$$

Now row reduce:

$$ \begin{array}{llll} x &+ & y &= a \\ &- &4y &= b - 3a \\ &- & y &= c - 2a \end{array}$$

$$ \begin{array}{llll} x &+ & y &= a \\ & &y &= -c + 2a \\ & & 4y &= -b +3a \end{array}$$

$$ \begin{array}{llll} x &+ & y &= a \\ & &y &= -c + 2a \\ & & 0 &= -5a -b + 4c \end{array}$$

This system is consistent if and only if $-5a - b + 4c = 0$.

Answer 2

$a+b=(x+y)+(3x-y)=4x$ $\implies$ $\displaystyle x=\frac{a}{4}+\frac{b}{4}$.

$a=x+y$ and $\displaystyle x=\frac{a}{4}+\frac{b}{4}$ $\implies$ $\displaystyle y=a-\frac{a}{4}-\frac{b}{4}=\frac{3a}{4}-\frac{b}{4}$.

But we need $2x+y=c$. So,

$$2\left(\frac{a}{4}+\frac{b}{4}\right)+\frac{3a}{4}-\frac{b}{4}=c$$

i.e., $5a+b-4c=0$.