2x+3y-z=1....(eqn 1)
x-y+3z= -8......(eqn 2)
2x+y+2z=-6..(eqn. 3)
Subtracting eqn.3 from eqn. 1,we get
2y-3z=7....(eqn. 4)
Multiplying eqn 2 by 2 and subtracting it from eqn.1,we get
5y-7z=17....(eqn. 5)
Multiplying eqn. 5 by 2 and subtracting it from eqn 4 multipled by 5
we get, -z=1
=>z= -1
putting this value of z in eqn. 4,we get
2y+3=7
=>2y=7-3=4
=>.y=2
Putting the values of y and z in eqn 2,we get
x=-3
Therefore,x= -3,y=2 and z= -1
2007-01-25 14:56:09
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answer #1
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answered by alpha 7
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I'd do this one by Gaussian reduction or by Cramer's rule. Hang on a sec while I get you some links that explain how to do this. Good links would be better than writing it out here, as they would be easier to read and would explain it better....
For Cramer's rule, looks a bit intimidating at first, all those determinants, but it's the most efficient way of solving systems, used in many computer programs.
http://www.chass.utoronto.ca/~osborne/MathTutorial/MATF.HTM
http://www.analyzemath.com/Tutorial-System-Equations/cramers_rule.html
I personally prefer gaussian reduction or gauss jordan reduction, as it's more intellectualy satisfying, like working out a puzzle.
In gaussian reduction we get the matrix upper triangular, obtain one varialbe, then work backwards to obtain the others. In Gauss jordan reduction we go a bit further, get everything in the matrix to be 0's except the main diagonal. Then you can read the answers directly.
http://www.math.hmc.edu/calculus/tutorials/linearsystems/
http://vc.wscc.cc.tn.us/math1022/week7/Solvematrix/solving.htm
For gauss jordan reduction, this one is a bit technical but I like it because it specifies the order in which you should attack the rows and columns, so that you don't just work around in circles.
http://www.math.okstate.edu/~myersr/3013/quiz/q2s/
So in your example you'd work with the matrix
...2...3...-1...|...1
...1..-1....3...|...-8
...2....1...2...|...-6
If you want to use Cramer's rule, the first thing you'd do is find the determinant of
2...3...-1
1..-1...3
2...1....2 This would be the denominator of all your answers.
Then you'd replace the "x" column in this matrix with the "answers" column and this would be the numerator of your answer for x.
Replace the "y" column in this matrix with the "answers" column and this would be the numerator of our answer for y
Do the same for z.
To do it by gauss jordan elimination you start again with the matrix
2...3...-1...|...1
1..-1....3...|...-8
2....1....2..|...-6
Work left to right, and top to bottom. I'd interchange the first and second rows, to get the 1...--1....3...|...-8 on top, then multiply this row by (-2) and add the result, -2...2...-6...|...16 to the other two rows to get zeroes in the first column. This gives...
1...-1...3...|...-8
0...5....-7...|...17.....
0...3...-4....|...10
When I'm doing this by hand I make small notes as superscripts as the numbers to be addded. I always add, if it is necessary to subtract I multiply by a negative and add.
The next step gets a little sticky, I like to get a little creative so as to aviod using fractions until it absolutely cannot be avoided. So I'd multiply the bottom row by -2 to get 0...-6...8...|...-20 and add it to the middle row to get a -1 in the middle position where the 5 is, then multiply the middle row by -1 to get a 1 there.
You see where it gets kind of hard to explain here, this is why I referred you to the sites above. Hopefully this and the online sites is enough to get you going.
2007-01-25 22:11:25
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answer #2
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answered by Joni DaNerd 6
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