Determine if the Given Vectors Span R4

[nicknaq]

I have the answer, but it makes no sense to me. Can someone explain it?

QUESTION:
Determine if the vector { [1 3 -1 0]^T, [-2 1 0 0]^T, [0 2 1 -1]^T, [3 6 -3 -2]^T }
spans R^4

ANSWER:

The vectors span R^4. One way to see this is to observe that the matrix A with these vectors as columns is invertible (the derminant is -42), so each Ei=AX for some column in R^n. Henece Ei is span of these vectors by Example 12.

 

[fzero]

Do you know what it means for vectors to span a vector space?

 

[Pen_to_Paper]

Because if you've worked with the adjoint of a matrix (or if you haven't yet)

an equation for A-1 (A inverse) = 1/det(A) * Adjoint (A)

here, don't worry about what the adjoint is, you may or may not work it out in your class; but you can see how from this formula, if det(A) = 0, then A inverse does not existhere are some simple examples that you can work out yourself:

this is a matrix:

0 1 0
2 2 2
4 0 4

you can easily see that if you take subtract (2)*Row1 from row 2, that rows 2 and 3 are linearly dependent

when you compute the determinant you get:
(-1)*(1)*(2*4 - 4*2)

because in this equation, you ignore the middle term and multiply only the outside terms, you can see that two vectors will be dependent because 2*4 - 4* 2 = 0

that was a simple example, look at this one (slightly different)

0 1 0
x 133 2x
10x pi 20x

some row operations:

(row 2) - (133*row1)
(row 3) - (pi*row1)

now we get:

0 1 0
x 0 2x
10x 0 20x

now find the determinant again - because the vectors are dependent, you will get to the step of 20x^2 - 20x^2because the vectors are dependent, the terms are just some scalar multiple of another vector's terms; so in computing the determinant you will end up with zero.

I hope that as you get further in your class this point will become more obvious.

Best of luck!