# Math Help - Row/Column space and reduced row echelon form.

1. ## Row/Column space and reduced row echelon form.

Hello, i have a problem that asks me to put in reduced row echelon form and I have no clue how to even begin.

Let A be a 4X5 matrix. If a1,a2,a4 are linearly independent and

a3=a1-3a2

a5=a1-2a2-3a4

determine the reduced row echelon form of A.

Any help is greatly appreeciated.

Thank You,
Diggidy

2. Originally Posted by Diggidy
Hello, i have a problem that asks me to put in reduced row echelon form and I have no clue how to even begin.

Let A be a 4X5 matrix. If a1,a2,a4 are linearly independent and

a3=a1-3a2

a5=a1-2a2-3a4

determine the reduced row echelon form of A.

Any help is greatly appreeciated.

Thank You,
Diggidy
Create a matrix that follows your criteria and see what happens. It should be easy enough.

3. I am working on this same problem and am having trouble understanding linear independence. Can anyone please explain how a1, a2, a4 being linearly independent will effect the reduced row echelon form of this matrix?

Thank you

4. try this: do you know of any "ready-to-eat" basis sets for R^4? let's say you did: B = {b1,b2,b3,b4}.

try setting the 1st column of A to b1, the 2nd column of A to b2, and the 4th column of A to b3, and the other two columns as per instructed.

you know that the basis elements are linearly independent. plug in the numbers and see what the matrix looks like. is it row-reduced?

5. Originally Posted by mybrohshi5
I am working on this same problem and am having trouble understanding linear independence. Can anyone please explain how a1, a2, a4 being linearly independent will effect the reduced row echelon form of this matrix?

Thank you
You won't have a 1 in every pivot position since there are 2 dependent vectors and the determinant is 0.

You can just make a matrix following the stipulations and reduce it to see what happens.

6. ok i think i got it. can anyone let me know if this is correct? Thanks in advanced!

A = $\begin{bmatrix}1&0&0&0&0\\0&1&3&0&2\\0&0&0&1&2\\0& 0&0&0&0 \end{bmatrix}= 0$

7. i don't see any negative numbers...that can't be right. are you sure you set a3 = a1-3a2, and a5=a1-2a2-3a4?

the a's are column vectors, not scalars.

8. Ok here we go, now i think i got it and i do have a negative number this time

$\begin{bmatrix}1&0&1&0&2\\0&1&2&0&-1\\0&0&0&1&3\\0&0&0&0&0 \end{bmatrix}= 0$

9. Originally Posted by mybrohshi5
Ok here we go, now i think i got it and i do have a negative number this time

$\begin{bmatrix}1&0&1&0&2\\0&1&2&0&-1\\0&0&0&1&3\\0&0&0&0&0 \end{bmatrix}= 0$
Try

$\displaystyle\begin{bmatrix}1&0&1&5&-14\\1&-3&10&\frac{1}{2}&\frac{11}{2}\\1&1&-2&1&-4\\1&2&-5&0&-3\end{bmatrix}$

10. if a1 = (1,0,0,0) and a2 = (0,1,0,0) then a1 - 2a2 = (1,0,0,0) - 2(0,1,0,0) = (1,0,0,0) + (0,-2,0,0) = (1,-2,0,0)

that is NOT what i see as your third column. column vectors add and subtract just as other vectors do: coordinate-by-coordinate.

if a4 = (0,0,1,0), then a1-2a2-3a4 = (1,0,0,0) - 2(0,1,0,0) - 3(0,0,1,0) = (1,0,0,0) + (0,-2,0,0) + (0,0,-3,0) = (1,-2,-3,0).

again, this is not what i see as your 4th column.

and why the =0 at the end? the matrix is NOT equal to 0.