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TheFallen018

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**[solved] Basis for set of solutions for linear equation**

Hi,

I have this problem I was working through, but I'm not sure that I've approached it from the right way. The problem consists of 3 parts, which build off of each other. I'm pretty confident about the first two parts, but no so much for the third. Here's the problem

Consider the linear equation $x+y-2z=0$

A). Find all solutions of this equation in vector form:

B). Give two vectors that span the set of solutions:

C). Prove that those two vectors are linearly independent and hence give a basis for the set of solutions

The first part, I basically pointed out that y and z are free variables, and expressed them as such

$y=t$

$z=s$

$x=-t+2s$

Solutions given by $(-t+2s,t,s)$

Then the second part, I just split up that vector into two separate vectors:

$t(-1,1,0)+s(2,0,1)$

For the third part, I used a basic proof that two vectors in ${R}^{n}$ are linearly depended iff one vector is a scalar multiple of the other. Therefore they are linearly independent.

Here's where I'm really not sure about much at all. I basically used the initial system, and turned the two vectors in part 2 into linear equations.

$x+y-2z=0$

$-x+y=0$

$2x+z=0$

I then used these three systems to make a 3x3 matrix which I row reduced. It came out as a matrix with a unique solution, so I thought the basis would be $(1,0,0),(0,1,0),(0,0,1)$, or maybe the initial equations expressed as vectors $(1,1,-2),(-1,1,0),(2,0,1)$.

Though, I'm really not sure here, as these span ${R}^{3}$ and seem a little too broad for what we are after. Any help would be great. Thanks :)

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