Ask your own question, for FREE!
Mathematics 86 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

Please write out the matrix.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

#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

jimthompson5910 (jim_thompson5910):

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)

OpenStudy (anonymous):

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

jimthompson5910 (jim_thompson5910):

Look at the third row, what do you see?

OpenStudy (anonymous):

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

jimthompson5910 (jim_thompson5910):

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

jimthompson5910 (jim_thompson5910):

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.

OpenStudy (anonymous):

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

jimthompson5910 (jim_thompson5910):

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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Latest Questions
xcoledd1: what is the song
11 minutes ago 5 Replies 1 Medal
strugglinginSchool27: which graph represents the solution to this system of inequalities? 3x - 5y u2264
15 minutes ago 2 Replies 0 Medals
RTXfrshman: How is everyone's day going
1 second ago 67 Replies 4 Medals
Manny300303199: I got a 10inch 4000 watt sub for my truck. Do you think that it is to loud?
31 minutes ago 0 Replies 0 Medals
strugglinginSchool27: 24x + 12y u2265 180 15x + 15y u2264 150
41 minutes ago 3 Replies 2 Medals
xcoledd1: i need help with a drawing
16 minutes ago 24 Replies 2 Medals
Manny300303199: Does anybody know the window tint laws in Germany?
33 minutes ago 21 Replies 1 Medal
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!