- #1

twinkerules

- 4

- 0

## Homework Statement

Problem number one: Find a basis for the row space of A consisting of vectors that (a) are not necessarily row vectors of A; and (b) are row vectors of A.

A = [tex]

\begin{bmatrix}

1 & 2 & -1\\

1 & 9 & -1\\

-3 & 8 & 3\\

-2 & 3 & 2

\end{bmatrix}

[/tex]

I was able to solve this one successfully by just doing reduced row echelon form (rref) and using any nonzero row as the basis.

Problem number two: Find a basis for the column space of A consisting of vectors that (a) are not necessarily column vectors of A; and (b) are column vectors of A.

A = [tex]

\begin{bmatrix}

1 & -2 & 7 & 0\\

1 & -1 & 4 & 0\\

3 & 2 & -3 & 5\\

2 & 1 & -1 & 3

\end{bmatrix}

[/tex]

**2. Attempt at a solution.**

For this one I tried doing rref but had the wrong answer. Instead, I transposed the matrix given above, did rref, and then got the correct answer if I transposed the columns again as shown below.

[tex]

\begin{bmatrix}

1 & 1 & 3 & 2\\

-2 & -1 & 2 & 1\\

7 & 4 & -3 & -1\\

0 & 0 & 5 & 3

\end{bmatrix} -->rref \begin{bmatrix}

1 & 0 & 0 & 0\\

0 & 1 & 0 & \frac{1}{5}\\

0 & 0 & 1 & \frac{3}{5}\\

0 & 0 & 0 & 0

\end{bmatrix} Answer: \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\1\\0\\ \frac{1}{5} \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\ \frac{3}{5} \end{bmatrix}

[/tex]

Can you please explain why the column space problem needs to be transposed before doing rref and why the answer is actually the rows of the matrix which need to be transposed back to columns. I can't find a pattern or answer as to why there is a difference in the procedures being performed. I listed problem one above, even though I got it right, because I don't feel confident that I know why that one was solved the way it was. Can you please provide some contrast between the different procedures being performed and how to know which one to do? I apologize in advance for any formatting errors, I have never used tex prior to this day. Thank you kindly.