Determine if vector is in span of other vectors

TheFallen018

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Hey guys,

I'm working on this problem here, and I'm a little fuzzy on how to go about this. I feel like I should put the vectors into a matrix as columns, and row reduce. Here is the question

Consider the vectors u = (3;-7; 2), v = (-1; 1; 3) and w = (-10; 18; 8) in R^3

Prove that w is in the span of u and v, by solving this linear system. When you apply elementary row operations, say explicitly which ones you use.


My problem here is that I'm not sure if matrix row reduction is the right way to go about it, if we are trying to find if a third vector is in the span of only another two in R^3.

The matrix reduces just fine, but I'm not sure if there's more to it. Thanks
 
"w is in the span of u and v" if and only if there exist numbers, a and b, such that w= au+ bv. In this case, u = (3;-7; 2), v = (-1; 1; 3) and w = (-10; 18; 8) so the equation is (-10; 18; 8)= a(3; -7; 2)+ b(-1; 1; 3)= (3a- b;-7a+ b; 2a+ 3b)= (-10; 18; 8).

So we need to determine whether there exist two numbers, a and b, that satisfy the three equations 3a- b= -10, -7a+ b= 18, and 2a+ 3b= 8.

You say "I'm not sure if matrix row reduction is the right way to go about it". Well, the problem says "When you apply elementary row operations" so clearly you are expected to do that. Of course, that is not the only way to answer this. The most basic way would be to solve two of those three equations for a and b then see if those values satisfy the third equation. For example, I see immediately that if we add the first two equations, b is eliminated leaving -4a= 8 so a= -2. Then -7a+ b= 14+ b= 18. a= -2, b= 4 satisfy the first two equations. Do they satisfy the third? 2a+ 3b= 2(-2)+ 3(4)= -4+ 12= 8. Yes, a= -2, b= 4 satisfy all three equations. Yes, w is in the span of u and v.

But you can do this as a "row reduction". The matrix set up would be \begin{bmatrix}3 & -1 & -10 \\ -7 & 1 & 18 \\ 2 & 3 & 8 \end{bmatrix}.

Following the standard, rather mechanical, row reduction, we would first divide the first row by 3 then add 7 times the (new) first row to the second row and subtract 2 times the (new) first row from the third row to get \begin{bmatrix}1 & -\frac{1}{3} & -\frac{10}{3} \\ 0 & -\frac{4}{3} & -\frac{16}{3} \\ 0 & \frac{11}{3} & \frac{44}{3}\end{bmatrix}

The rest is pretty obvious. Divide the second row by \(\displaystyle -\frac{4}{3}\), add \(\displaystyle -\frac{1}{3}\) times the new second row to the first row, and subtract \(\displaystyle \frac{11}{3}\) times the new second row from the third row to get \begin{bmatrix}1 & 0 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \end{bmatrix}
 
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