SOLUTION: Solve each system using Gaussian elimination. State whether each system is independent, inconsistent, or dependent.
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Question 44742This question is from textbook College Algebra
: Solve each system using Gaussian elimination. State whether each system is independent, inconsistent, or dependent.
This question is from textbook College Algebra
Found 2 solutions by Nazareth, venugopalramana:
Answer by Nazareth(2) (Show Source): You can put this solution on YOUR website!
Elimination Method
3x - 6y = 9.....multiply this line by 2
2x + y = -4.....mutiply by 3
We have:
6x - 12y = 18...(1)
6x + 3y = -12...(2)
we work by eliminating x therefore (1) - (2)
We have:
6x - 6x = 0
-12y -3y = -15y
and 18 - (-12)= 30
This give -15y = 30
Divide both side by -15, we have:
y = -2
Replcing the value of y in any of the above equation:
for example
we use
3x -6y=9
3x - 6(-2) = 9
3x +12 = 9, since - x - = +
then take away 12 from both sides, leaving
3x = - 3
giving x = -1
Answer:
x = -1
y = -2
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY.COME BACK IF STILL IN DIFFICULTY
--------------------------------------------------------------------------------
x - y + z = 2
-x + y + z = 4
-x + z = 2
augmented matrix is
1.....................-1.......................1.............................2
-1.....................1.......................1.............................4
-1.....................0.......................1.............................2
nr2=r2+r1 and nr3=r3+r1
1..................-1.................1.............2
1-1=0............1-1=0............1+1=2..........2+4=6
1-1=0...........-1+0=-1...........1+1=2..........2+2=4
nr2=r2/2....nr3=r3-r2
1....................-1...............1..............2
0/2=0.............0/2=0..............2/2=1........6/2=3
0-0=0............-1-0=-1..............2-2=0.......4-6=-2
nr1=r1-r3-r2.......nr2=2r2........nr3=-r3
1-0-0=1............-1+1-0=0.........1-0-1=0.......2+2-3=1
0.....................0...................1............3
0.....................1...................0.............2
exchange r2 and r3
1....0........0.......1
0....1........0.......2
0....0........1.......3
hence x=1......y=2.....z=3
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