Let

D = {(a,b,2a): a,b all real numbers}

The set of vectors in R^3 with last entry twice the first.

E = span{(1,0,2),(2,-1,4),(0,-3,0)}

Prove that D = E

I first made an augmented matrix to find out that there is a free variable of x3. Showing that there is a solution where D is a subset of E.

How do I how that E is a subset of D?

To show that E is a subset of D, we need to prove that every vector in E can be written in the form (a, b, 2a) for some real numbers a and b.

Given that E is the span of {(1, 0, 2), (2, -1, 4), (0, -3, 0)}, we can write an arbitrary vector in E as a linear combination of these vectors:

c1(1, 0, 2) + c2(2, -1, 4) + c3(0, -3, 0).

Expanding this equation gives:
(c1 + 2c2, -c2 - 3c3, 2c1 + 4c2).

So, for an arbitrary vector (a, b, 2a) to be in E, we need to find values for a, b, c1, c2, and c3 that satisfy the equation:

a = c1 + 2c2
b = -c2 - 3c3
2a = 2c1 + 4c2

To determine if there is a solution, we can set up an augmented matrix:

[1 2 | a]
[0 -1 | b]
[2 4 | 2a]

Using row operations, we can row reduce this matrix:

R2 -> R2 + R1
R3 -> R3 - 2R1

The resulting matrix is:

[1 2 | a]
[0 -1 | b-a]
[0 0 | 0]

From this row-reduced form, we can see that the variable c3 is free, which means we can choose any real number for it. Let's denote c3 as t.

From R2, we have -t = b-a, which implies a = b+t.

Substituting this value of a into the first equation, we get:
(b + t) = c1 + 2c2,
which implies c1 = b + t - 2c2.

So, any vector (a, b, 2a) can be written as (c1 + 2c2, -c2 - 3t, 2c1 + 4c2) for some values of c1, c2, and c3 (in this case, we replaced c3 with t).

Hence, we have shown that E is a subset of D.