Solve the system

frctl

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Joined
Jun 29, 2019
Messages
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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
 
Could post your work using the same format as the problem?
 
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
 
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.
 
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
 
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....
 
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).

?
 
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
 
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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