Question

Last one, I promise!
Question in next post.

Answer 1

Let A =\(\left[\begin{matrix}0 & 1 & 7 & 8 \\ 1 & 3 & 3 & 8 \\ -2 & -5 & 1 & -8 \end{matrix}\right]\) Write A in termos of A=EFGR, where R is in the row echelon form and E, F, G are elementary matrices.

I've reduced to row echelon form and found that R = \(AE^{-1}F^{-1}G^{-1}\) but that's it. I cannot seem to remember elemetary matrices.

Answer 2

In this case, R = \(\left[\begin{matrix}1 & 3 & 3 & 8 \\ 0 & 1 & 7 & 8\\ 0 & 0 & 0 & 0\end{matrix}\right]\)

Answer 3

Sorry for mistypes. I'm kinda drowsy at the moment.

Answer 4

simply multupl GFE both sides ?

Answer 5

That's where I started the problem: \(A=REFG \leftrightarrow R =AE^{-1}F^{-1}G^{-1}\). But I need to find the elementary matrices and that's what I do not recall.

Answer 6

whats the first row operation that you have done ?

Answer 7

I've swapped L2 and L1

Answer 8

whats the corresponding elementary matrix ?

Answer 9

In other words, what matrix needs to be multiplied by A in order to swap L2 and L1 ?

Answer 10

\[\left[\begin{matrix}1 & 3 & 3 & 8 \\ 0 & 1 & 7 & 8 \\-2 & -5 & 1 & -8\end{matrix}\right]\]

Answer 11

That's the first operation I've done.

Answer 12

Careful, these two statements are really completely different:

\( A=EFGR \)

\(R = AE^{-1}F^{-1}G^{-1}\)

Left and right multiplication are specific, so if you right multiply that second statement by \(GFE\) you get:

\[A = R GFE \ne EFGR \]

Answer 13

Oh, right. I see what we are getting into...

Answer 14

Yeah, that was sloppy of me. My bad.

Answer 15

But I do get it now. It's the matrix that I have to multiply which gives the product after each operation

Answer 16

Below red matrix swaps first and second rows
\[\left[\begin{matrix}1 & 3 & 3 & 8 \\ 0 & 1 & 7 & 8\\ 0 & 0 & 0 & 0\end{matrix}\right] = \color{red}{\left[\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]} \left[\begin{matrix}0 & 1 & 7 & 8 \\ 1 & 3 & 3 & 8 \\ -2 & -5 & 1 & -8 \end{matrix}\right]\]

Answer 17

All right, thats E. , then. Now for G this should be \(\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}\right]\)

Answer 18

I meant F.

Answer 19

are you doing 2*row1 + row3 ?

Answer 20

2R1+R3

Answer 21

\[\left[\begin{matrix}1 & 3 & 3 & 8 \\ 0 & 1 & 7 & 8\\ 0 & 1 & 7 & 8\end{matrix}\right] = \color{red}{\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}\right]} \color{red}{\left[\begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}\right]} \left[\begin{matrix}0 & 1 & 7 & 8 \\ 1 & 3 & 3 & 8 \\ -2 & -5 & 1 & -8 \end{matrix}\right]\]

Answer 22

Okay, seems right. For G I've managed to grab \(\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 1 \end{matrix}\right]\)

Answer 23

Which is -R2 + R3.

Answer 24

looks good, so far you have :

\[R=GFEA \]

Answer 25

They are all nonsingular, so now onto calculating the inverse matrices...

Answer 26

Is it? I thought I'd multiply the elments outside the main row by -1.

Answer 27

No, it doesn't work. Getting sloppy again.

Answer 28

Let me do it with the usual way

Answer 29

yeah it works only for the row swapping matrices
1) inverse of a row swapping matrix is itself

Answer 30

but finding other inverses should be trivial

Answer 31

\[\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix}\right]^{-1} = \left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ -2 & 0 & 1 \end{matrix}\right]\]

Answer 32

Yeah, I'm slow writing in LaTeX but that's what I've found and the last one you just do a32=-a32

Answer 33

yeah

Answer 34

for \(a_{32} \in G\) that is.

Answer 35

I guess it finishes the question, then. Just plug in the inverse matrices and R and I'm done!

Answer 36

Thanks for the help!

Answer 37

Np :)

Answer 38

Might as well do a last problem before going to bed! Haha!