What is the solution to finding the subspace of R^5?

[maria69]

Please help?

I have done all the parts of question 1 but i really can't solve (iv) part. i know that the answer should be [-8;8;-1;0;1].
if someone can pleaseeeeeeeeeeeee help me?
thank you very much Maria!
http://img3.imageshack.us/img3/5706/72162896rd8.jpg

 

[Mark44]

For part iv, you want to find the subspace of R^5 determined by the three equations. Set up a 3 x 5 matrix with the coefficients of the three equations (where each equation has all its terms on one side and zero on the other).
[1 1 1 1 1]
[1 2 3 -4 -5]
[0 0 1 1 1]

Use row operations to reduce this matrix to row-echelon form (the first nonzero entry in a row has zeroes in the rows above and below it). After the matrix is row-reduced you can almost read off the vectors in a basis for the subspace.

For example, if you ended up with this matrix (I'm just making up numbers here as an example)
[1 0 0 0 1]
[0 1 0 0 -1]
[0 0 0 1 1]
this represents the system of equations
x1 = - x5
x2 = + x5
x3 = x3
x4 = -x5
x5 = x5
Any solution x = (x1, x2, x3, x4, x5) can be written as x3 * (0, 0, 1, 0, 0) + x5 * (-1, 1, 0, -1, 1), so a basis for this subspace is these two vectors.

To verify that the vector they give belongs to S2, just show that its coordinates satisfy all three equations.

To find another vector in a basis, use the Gram-Schmidt process.

 

[maria69]

can you please show me how to do it because i really can't understand what to do? i am really confused and it is really important to have the solution today. i am trying so hard to solve it but i don't think that i know how to.
thanks very much!
maria

 

[Mark44]

OK, let's take it a step at a time. Do you know how to row-reduce this matrix?
[1 1 1 1 1]
[1 2 3 -4 -5]
[0 0 1 1 1]

The first step is to use row 1 to eliminate the leading 1 in the 2nd row.

 

[maria69]

i don't really remember actually

 

[maria69]

if i can just see the whole solution i am really sure that i can understand for the iv part

 

[Mark44]

i don't really remember actually

Then you better get started jogging your memory if you want to do this problem today...