Question

the augmented matrix of a linear system has been reduced by row operations to the form shown. in each case, continue the appropriate row operations and describe the solution set of the original system

Answer 1

Please write out the matrix.

Answer 2

1 7 3 -4
0 1 -1 3
0 0 0 1
0 0 1 -2

Answer 3

#2
1 -1 0 0 -5
0 1 -2 0 -7
0 0 1 -3 2
0 0 0 1 4

I also have the answer to both questions if that helps

Answer 4

1 -1 0 0 -5
0 1 -2 0 -7
0 0 1 -3 2
0 0 0 1 4



1 -1 0 0 -5
0 1 -2 0 -7
0 0 1 0 14 R3+3R4
0 0 0 1 4



1 -1 0 0 -5
0 1 0 0 21 R2+2R3
0 0 1 0 14
0 0 0 1 4



1 0 0 0 16 R1+R2
0 1 0 0 21
0 0 1 0 14
0 0 0 1 4


So the solution is (16,21,14,4)

Answer 5

Yes! That's what the book has. The first one says no solution. Any idea as to why?

Answer 6

Look at the third row, what do you see?

Answer 7

i thought it had something to do with that. sadly I still don't quite get the intuition behind it. Is it because row 3 is saying 0*x1 +0*x2 +0*x3= 1?
I know it doesnt look right, but I cant understand why. I've been in the class for just 2 days

Answer 8

You are correct, the row 0 0 0 1 means exactly what you have

You can simplify the entire left side to get 0+0+0, which adds to 0


So the entire third row translates into 0 = 1, but this is false. So there are no solutions

Answer 9

Basically, no matter what I plug in for x1, x2, x3, there is NO way to make 0*x1 +0*x2 +0*x3 equal to 1. It's never going to happen.

Answer 10

Thanks! Before you go, just a quick question. do you have any trick that you use to see how you are going to operate the matrix? I know the form I want it to be in, I just don't see how to get there

Answer 11

you basically want the final matrix to be in the form

1 0 0 0 x
0 1 0 0 y
0 0 1 0 z
0 0 0 1 w

where your solution is (x,y,z,w)


So the goal is to get that diagonal of 1s and zeros everywhere else (ignore the right most column)


To do that, you use the pivot for that column.


When I went from

1 -1 0 0 -5
0 1 -2 0 -7
0 0 1 -3 2
0 0 0 1 4

to

1 -1 0 0 -5
0 1 -2 0 -7
0 0 1 0 14 R3+3R4
0 0 0 1 4


I turned that -3 into a zero by adding Row 3 (R3) to three times Row 4 (R4). Notice that this gives us -3+3*1 = 0 in the fourth column and 2+3*4 = 14 in the fifth column. I then kept going and turned that -2 into 0 and turned that -1 into 0 as well.