Can someone explain this solution to me please? (linear algebra)

How did they get to their example solution? I row reduced to a matrix with the first row being (1,2,3) and the second row being (0,0,0) so I got x+y+z=1.

Now am I right in assuming I need to set y and z to be free variables?

Where do I need to go from here?

I think you mean x+2y+3z=1.....................(1)



Yes - that's one way.



So, let y=a and z=b, then x= 1-2a-3b, by (1)

Our set of solutions is therefore {(1-2a-3b,a,b)| a,b in R}

Splitting that up, almost gets us to the example solution. If we then let a= c+1, and simplify, we are there.

Thanks for replying.

Could you please explain why we take a=c+1? I don't quite understand the process going on here, or the motivation for it. Also, what exactly do you mean by 'splitting that up', please?

(1-2a-3b,a,b), separating out the variables, can be written as:

(1,0,0) + a(-2,1,0) + b(-3,0,1)

That's what I meant by splitting that up.

Which is fine as a solution, and I would have stopped there. Note that there are an infinite number of possibilities for that formula - it's the vector formula for a plane.

So see how we can get the example solutions we want to the constant term to be (-1,1,0).

We can get that by adding one lot of (-2,1,0), which is why I set let a=c+1 (c becomes our parameter instead of a). We can do this as the parameters a,b,c all range over the whole of R.

Thank you.

So in the same question the first part is the equations:

5x-y=4
x+y=1
3x-3y=2

I row reduced this to x=5/6 and y=1/6. The third row is 0+0=0 so I discard this. Now in terms of the question we're meant to write this in the form x+nullspace(A) - the equation has a unique solution though? Am I right to just leave it as x=5/6 and y=1/6?

It's not the form they're asking for, though it is the correct solution.

I'd put (5/6, 1/6)