Question

Reduced echelon form, by elementary row operations.... (Matrices given in comments)

Answer 1

Is the REF of \[A = \left[\begin{matrix}2 & 2 & 0 & 2 \\ 2 & 3 & 0 & 3 \\ 1 & 2 & 0 & 3 \\ 1 & 0 & 0 & 0\end{matrix}\right]\] equal to \[\left[\begin{matrix}0 & 2 & 2 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & -6 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 2

I don't quite think this is correct.... Can anyone help? Thanks :)

Answer 3

\[\left[\begin{matrix}1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 4

Thats what my calculator said, too hahaha how do you get that? What are the row operations you perform? :/

Answer 5

\[r_1\leftrightarrow r_4\]

Answer 6

then use the new row 1 to get rid of the numbers below the leading 1 in the first column

Answer 7

\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 3 \\ 0 & 2 & 0 & 3 \\ 0 & 2 & 0 & 3\end{matrix}\right]\]

Answer 8

doing it this way will not give the same solution as above...but ref is not unique (rref is)

Answer 9

you can then do \[-1r_3+r_4\rightarrow r_4\]

Answer 10

I am using the operations to find the row and column spaces for the matrix.... All the exaples I am given just use ref, rather than rref... How do you know that you need to have a unique solution?? (Thanks, as always, for the help by the way!)

Answer 11

you don't need to have a unique solution...you can have many (infinite) number of equivalent matrices in ref....only rref is unique.

I was only letting you know that our reduction would not be the same as yours and my calculators

Answer 12

\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 3 & 0 & 3 \\ 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 13

\[-1r_3+r_2\rightarrow r_2\]

Answer 14

\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 15

\[-2r_2+r_3\rightarrow r_3\]

Answer 16

\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 17

\[1/3r_3\rightarrow r_3\]

Answer 18

\[\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 19

Oh I see :)

Answer 20

so we actually ended up with rref...;)

Answer 21

So the row space of this matrix is \[\left\{ \left[\begin{matrix}1 \\ 0 \\ 0 \\ 0\end{matrix}\right], \left[\begin{matrix}0 \\ 1 \\0 \\ 0\end{matrix}\right],\left[\begin{matrix}0\\ 0\\0\\1\end{matrix}\right]\right\}\]

And the column space is \[\left\{ \left[\begin{matrix}2 \\ 2 \\ 1 \\ 1\end{matrix}\right], \left[\begin{matrix}2 \\ 3 \\2 \\ 0\end{matrix}\right],\left[\begin{matrix}2\\ 3\\3\\0\end{matrix}\right]\right\}\] ??

(If you can help with that, too hahaha)

Answer 22

yes

Answer 23

Thank youuuu!!!

Answer 24

np

Answer 25

Haaang on a second... I've just gone away and come back to this... In your first step, after making the first column zero (using the leading 1), how does the last row become [0 2 0 3], rather than [0 2 0 2]??

Answer 26

typo

Answer 27

\[A = \left[\begin{matrix}2 & 2 & 0 & 2 \\ 2 & 3 & 0 & 3 \\ 1 & 2 & 0 & 3 \\ 1 & 0 & 0 & 0\end{matrix}\right]\]

\[\left[\begin{matrix} 1 & 0 & 0 & 0\\ 2 & 3 & 0 & 3 \\ 1 & 2 & 0 & 3 \\2 & 2 & 0 & 2 \end{matrix}\right]\]

\[\left[\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 3 & 0 & 3 \\ 0 & 2 & 0 & 3 \\0 & 2 & 0 & 2 \end{matrix}\right]\]

\[-1r_4+4_2\rightarrow r_2\]
\[\left[\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 \\ 0 & 2 & 0 & 3 \\0 & 2 & 0 & 2 \end{matrix}\right]\]

\[\left[\begin{matrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 \\0 & 0 & 0 & 0 \end{matrix}\right]\]

Answer 28

\[-1r_4+r_2\rightarrow r_2\]

Answer 29

then \[-2r_2+r_3\rightarrow r_3\]

\[-2r_2+r_4\rightarrow r_4\]

Answer 30

Sorry, thanks heaps :)