Solve the system

frctl

Junior Member
Joined
Jun 29, 2019
Messages
153
Solve the system:
x1 + 2x2 - x3 = 1
2x1 - x2 + x3 = 3
-x + 2x2 + 3x3 = 7

My work:
two times the first row subtracted from the second row
1 2 -2 1
0 5 -5 -1
-1 2 3 3

Where I am stuck:
I'm not how to eliminate the -1 in the last row
Also I want to turn the 5 to a 1 in the second row
 

lev888

Full Member
Joined
Jan 16, 2018
Messages
428
Could post your work using the same format as the problem?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
8,527
Solve the system:
x1 + 2x2 - x3 = 1
2x1 - x2 + x3 = 3
-x + 2x2 + 3x3 = 7

My work:
two times the first row subtracted from the second row
1 2 -2 1
0 5 -5 -1
-1 2 3 3

Where I am stuck:
I'm not how to eliminate the -1 in the last row
Also I want to turn the 5 to a 1 in the second row
Look HERE
 

frctl

Junior Member
Joined
Jun 29, 2019
Messages
153
Why did you multiply the inverse by 1, 3, 7?
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,777
Solve the system:
x1 + 2x2 - x3 = 1
2x1 - x2 + x3 = 3
-x + 2x2 + 3x3 = 7

My work:
two times the first row subtracted from the second row
1 2 -2 1
0 5 -5 -1
-1 2 3 3

Where I am stuck:
I'm not how to eliminate the -1 in the last row
Also I want to turn the 5 to a 1 in the second row
Just add 1st row to the 3rd Row

Divide the second row by 5.
 

frctl

Junior Member
Joined
Jun 29, 2019
Messages
153
Add row1 to row3 and replace row3
1 2-2 1
0 5 -5-1
0 4 1 4
Divide row2 by 5
1 2 -2 1
0 1 -1 -1/5
0 4 1 4

Does this appear correct?
I am left with a 4 in row3
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,777
Add row1 to row3 and replace row3
1 2-2 1
0 5 -5-1
0 4 1 4
Divide row2 by 5
1 2 -2 1
0 1 -1 -1/5
0 4 1 4

Does this appear correct?
I am left with a 4 in row3
I don't see any problem .... continue....
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,861
Solve the system:
x1 + 2x2 - x3 = 1
2x1 - x2 + x3 = 3
-x + 2x2 + 3x3 = 7
two times the first row subtracted from the second row
1 2 -2 1
0 5 -5 -1
-1 2 3 3
Hello frctl. The elements highlighted in red above are not correct.

That first row operation (R2 -> -2*R1 + R2) changes elements in row2 only. Why did you change elements in row1 and row3?

Can you post a picture of your paperwork? I would like to see how you organize your work, when doing matrix arithmetic (row operations).

Here's one way to organize the arithmetic for adding -2 times row1 to row2. First, write row1. Next, proofread what you wrote (i.e., confirm that you copied each number correctly). Next, multiply each number by -2, and then double-check that arithmetic. After you've confirmed that -2*R1 is correct, write row2 underneath it. Now proofread, to confirm that you copied each number correctly. Lastly, add those two rows, and then double-check your arithmetic. Written out, the steps would look similar to this:

Code:
R2 -> -2*R1 + R2

      R1    1       2      -1       1
   -----------------------------------
   -2*R1   -2      -4       2      -2
+     R2    2      -1       1       3
   -----------------------------------
   R2 ->    0      -5       3       1
Organizing work like that (or something similar) helps to reduce errors because it makes mistakes easier to see (both as you go and when revisiting work later).

😎
 

frctl

Junior Member
Joined
Jun 29, 2019
Messages
153
Thank you Otis
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,765
Solve the system:
x1 + 2x2 - x3 = 1
2x1 - x2 + x3 = 3
-x + 2x2 + 3x3 = 7

My work:
two times the first row subtracted from the second row
1 2 -2 1
0 5 -5 -1
-1 2 3 3
You have an error in the 1st and last row (actually you have all entries wrong because you did NOT do two times the first row subtracted from the second row)

Where I am stuck:
I'm not how to eliminate the -1 in the last row
Also I want to turn the 5 to a 1 in the second row
So divide row 2 by 5
 

Otis

Senior Member
Joined
Apr 22, 2015
Messages
1,861
… all entries wrong …
Actually, frctl got the leading zero correctly, but the rest of row2 is wrong.

So divide row 2 by 5
After frctl corrects the 5 in R2C2, then row2 can be divided by -5.

😎
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
3,765
Actually, frctl got the leading zero correctly, but the rest of row2 is wrong.


After frctl corrects the 5 in R2C2, then row2 can be divided by -5.

😎
Oh, I thought -0 and 0 were different.
The op asked how to get rid of the 5 (which the op did not obtain yet) and the answer is to divide by 5

Only joking. have a great night!
 
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