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Got it?

next is -3 R3 + R2 to get a new row 2

oops, I meant
1 0 0 -11 -25
0 1 0 3 6
0 0 1 0 2
0 0 0 1 -2

oh, gosh, my bad,

ok, I redo it. let clear them all.

I cleared mine!

\(\left[\begin{matrix}1&3&0&-2&7\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\)

1 0 0 -11 -25 (after -3R2 + R1)
0 1 -3 3 0 (after -3R3 + R2)
0 0 1 0 2
0 0 0 1 -2
Is that correct?

ok?

No... Oh my... how did you end up with -11 as your row one column five answer? It used to be -25...

ok??

aaaaaaaaaaaaah, your last one is -7, not 7, ok hence it is -25. I am so so sorry

\[\left[\begin{matrix}1&0&0&-11&-25\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

Step2: 11R4 + R1 to get new row 1

So, multiply row 4 by 11 and add to row one to get the new row one? Why row 4?

\[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&3&6\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

So, what is the next step?

Ah, I see.

So the new row2 would be [0 1 0 0 0]?

yes, and you are done.

\[\left[\begin{matrix}1&0&0&0&-45\\0&1&0&0&0\\0&0&1&0&2\\0&0&0&1&-2\end{matrix}\right]\]

yes, it is.

unless the last column, all of the entries are either 1 or 0.

with 1's are in diagonal line.

ok, don't forget: I am crazy one.

Crazy One?