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Question 1182942: Solve the given equation by Guass elimination method;
4x₂-3x₃+1x₄=-8
-2x₂+1x₃-3x₄=-4
1x₂-1x₃+2x₄=3
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
4x1 - 3x2 + x3 = -8
-2x1 + x2 - 3x3 = -4
x1 - x2 + 2x3 = 3
Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)
4 -3 1 -8
-2 1 -3 -4
1 -1 2 3
R1 / 4 → R1 (divide the 1 row by 4)
1 -0.75 0.25 -2
-2 1 -3 -4
1 -1 2 3
R2 + 2 R1 → R2 (multiply 1 row by 2 and add it to 2 row); R3 - 1 R1 → R3 (multiply 1 row by 1 and subtract it from 3 row)
1 -0.75 0.25 -2
0 -0.5 -2.5 -8
0 -0.25 1.75 5
R2 / -0.5 → R2 (divide the 2 row by -0.5)
1 -0.75 0.25 -2
0 1 5 16
0 -0.25 1.75 5
R1 + 0.75 R2 → R1 (multiply 2 row by 0.75 and add it to 1 row); R3 + 0.25 R2 → R3 (multiply 2 row by 0.25 and add it to 3 row)
1 0 4 10
0 1 5 16
0 0 3 9
R3 / 3 → R3 (divide the 3 row by 3)
1 0 4 10
0 1 5 16
0 0 1 3
R1 - 4 R3 → R1 (multiply 3 row by 4 and subtract it from 1 row); R2 - 5 R3 → R2 (multiply 3 row by 5 and subtract it from 2 row)
1 0 0 -2
0 1 0 1
0 0 1 3
x1 = -2
x2 = 1
x3 = 3
Make a check:
4·(-2) - 3·1 + 3 = -8 - 3 + 3 = -8
-2·(-2) + 1 - 3·3 = 4 + 1 - 9 = -4
(-2) - 1 + 2·3 = -2 - 1 + 6 = 3
Check completed successfully.
ANSWER
x1 = -2
x2 = 1
x3 = 3
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Note that if you were to receive 10 solutions from 10 different tutors, they would probably be all different.
The solution from the other tutor is fine.
I personally would work as much of the problem as I could without introducing fractions or decimals. The method is straightforward; but it is aggravatingly easy to make simple arithmetic errors along the way, leading of course to wrong answers. Avoiding fractions and decimals makes those sloppy errors less likely.
Here is how I would solve the system.
Original system, directly from the given equations:
4 -3 1 -8
-2 1 -3 -4
1 -1 2 3
Our first objective is to get a 1 in row 1 column 1. Since we can move rows, and since one of the rows has a 1 in column 1, switch rows:
1 -1 2 3
-2 1 -3 -4
4 -3 1 -8
The next objective is to get 0's in column1 in the other two rows. The method for doing that is standard -- add appropriate multiples of row 1 to rows 2 and 3.
1 -1 2 3
0 -1 1 2 row 2, plus 2 times row 1
0 1 -7 -20 row 3, minus 4 times row 1
Our next objective is to get a 1 in row 2 column 2. We could do that by switching rows 2 and 3, or by multiplying row 2 by -1. I arbitrarily chose the latter.
1 -1 2 3
0 1 -1 -2 row 2, multiplied by -1
0 1 -7 -20
Next we use the 1 in row 3 column 3 to get 0's in column 2 in rows 1 and 3
1 0 1 1 row 1, plus row 2
0 1 -1 -2
0 0 -6 -18 row 3, minus row 2
Now we want a 1 in row 3 column 3, so divide row 3 by -6
1 0 1 1
0 1 -1 -2
0 0 1 3 row 3, divided by -6
And last we use the 1 in row 3 column 3 to get 0's in column 3 in rows 1 and 2
1 0 0 -2 row 1, minus row 3
0 1 0 1 row 3, plus row 3
0 0 1 3
Solution: (-2,1,3)
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