Question

Consider the following matrix as an augmented matrix of a linear system. State in words the next two elementary row operations that should be preformed in the process of solving the system.

Answer 1

\begin{matrix}1 & -4 & -3 & 0 & 7 \\ 0 & 1 & 4 & 0 & 6 \\ 0 & 0 & 1 & 0 & 2 \\0 & 0 & 0 & 1 & -5\end{matrix}

Answer 2

@whpalmer4

Answer 3

look in the second row, you want to turn that 4 into a 0

so you need to subtract 4 times row 3 from row 2, then replace row 2 with that result

this will turn the 4 into 0

Answer 4

consider your pivot entries first, pivot columns and zero them out to obtain the rref (reduced row echelon form) of your matrix.

Answer 5

you keep doing this until you have a diagonal of 1s only (and nothing but zeros) for everything excluding that last column

Answer 6

((Sorry, OS is lagging for me, didn't see that this question was already being answered))

Answer 7

that's fine Spacelimbus, the more answers the better

Answer 8

I thought that I could have any nonzero value for all values above the diagonal 1s, as long as the below values were zero?

Answer 9

well this system in it's original given form is in what it is called row echelon form

from here you can use back substitution to solve for each variable

Answer 10

to get it into reduced row echelon form, you would need to keep row reducing until you have zeroed out each value you can (that's not a pivot)

Answer 11

granted there are some occasions where you cannot row reduce further and you will have nonzero values above the pivots...but not in this case

in this case, you can zero out each value above the pivots

Answer 12

the first one is incorrect, but the other values looked right (you deleted it though lol so I didn't have much a chance to check)

Answer 13

but they're not interested in the final solution really

they just want the next two steps to get there (to rref form)

Answer 14

I was able to solve without further simplification. Should this not be possible?\[(21, -2, 2, -5)\]

Answer 15

Yes, I know they're not interested in final solution, hence they didn't ask for it. :P

Answer 16

21 isn't correct for the first variable

Answer 17

the rest is correct though

Answer 18

First is 5?

Answer 19

yes

Answer 20

are you use back substitution to get this?

Answer 21

Yes.

Answer 22

ok do so without using that

Answer 23

so keep going with the row reductions

Answer 24

Okay.

Answer 25

you basically want something of this form

\[\begin{bmatrix}1 & 0 & 0 & 0 & ? \\ 0 & 1 & 0 & 0 & ? \\ 0 & 0 & 1 & 0 & ? \\0 & 0 & 0 & 1 & ?\end{bmatrix}\]

Answer 26

and the final answer will be the right column after you get it into that form