Yes, Yes, or any method you choose. Have you learned augmented matrix yet?
2,-5,3,8
3,-1,4,7
1,3,2,-3
Solution:Result of solution using Gauss-Jordan elimination
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Your matrix
X1 X2 X3 b
1 2 -5 3 8
2 3 -1 4 7
3 1 3 2 -3
Find the pivot in the 1st column and swap the 3rd and the 1st rows
X1 X2 X3 b
1 1 3 2 -3
2 3 -1 4 7
3 2 -5 3 8
Multiply the 1st row by 3
X1 X2 X3 b
1 3 9 6 -9
2 3 -1 4 7
3 2 -5 3 8
Subtract the 1st row from the 2nd row and restore it
X1 X2 X3 b
1 1 3 2 -3
2 0 -10 -2 16
3 2 -5 3 8
Multiply the 1st row by 2
X1 X2 X3 b
1 2 6 4 -6
2 0 -10 -2 16
3 2 -5 3 8
Subtract the 1st row from the 3rd row and restore it
X1 X2 X3 b
1 1 3 2 -3
2 0 -10 -2 16
3 0 -11 -1 14
Make the pivot in the 2nd column by dividing the 2nd row by -10
X1 X2 X3 b
1 1 3 2 -3
2 0 1 1/5 -8/5
3 0 -11 -1 14
Multiply the 2nd row by 3
X1 X2 X3 b
1 1 3 2 -3
2 0 3 3/5 -24/5
3 0 -11 -1 14
Subtract the 2nd row from the 1st row and restore it
X1 X2 X3 b
1 1 0 7/5 9/5
2 0 1 1/5 -8/5
3 0 -11 -1 14
Multiply the 2nd row by -11
X1 X2 X3 b
1 1 0 7/5 9/5
2 0 -11 -11/5 88/5
3 0 -11 -1 14
Subtract the 2nd row from the 3rd row and restore it
X1 X2 X3 b
1 1 0 7/5 9/5
2 0 1 1/5 -8/5
3 0 0 6/5 -18/5
Make the pivot in the 3rd column by dividing the 3rd row by 6/5
X1 X2 X3 b
1 1 0 7/5 9/5
2 0 1 1/5 -8/5
3 0 0 1 -3
Multiply the 3rd row by 7/5
X1 X2 X3 b
1 1 0 7/5 9/5
2 0 1 1/5 -8/5
3 0 0 7/5 -21/5
Subtract the 3rd row from the 1st row and restore it
X1 X2 X3 b
1 1 0 0 6
2 0 1 1/5 -8/5
3 0 0 1 -3
Multiply the 3rd row by 1/5
X1 X2 X3 b
1 1 0 0 6
2 0 1 1/5 -8/5
3 0 0 1/5 -3/5
Subtract the 3rd row from the 2nd row and restore it
X1 X2 X3 b
1 1 0 0 6
2 0 1 0 -1
3 0 0 1 -3
Solution set:
x1 = 6
x2 = -1
x3 = -3
2x − 5y + 3z = 8
3x − y + 4z = 7
x + 3y + 2z = −3
Can I multiple the 2nd equation by 3 and the 3rd by 3?
Or should I take care of 1&3 and then deal with 2. since both of them can be multiplied by three and five
6 answers
Multiplying Eq2 and Eq3 by 3 won't help, because it does not eliminate a variable. Yes, you can use Eq1 and Eq3 and then Eq2 and Eq3. Eliminate
X in each case.
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X in each case.
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okay.
@Bob pursley its a new method I just learned
@Henry I see.
@Bob pursley its a new method I just learned
@Henry I see.
Eq1: 2x - 5y + 3z = 8.
Eq3: 2x + 6y + 4z = -6.(multiplied by 2).
Diff. = -11y - z = 14.
Eq2: 3x - y + 4z = 7.
Eq3: 3x + 9y + 6z = -9.(been multiplied by 3).
Diff. = -10y - 2z = 16.
-22y - 2z = 28.(been multiplied by 2).
-10y - 2z = 16.
Diff. = -12y = 12,
Y = -1.
-10*(-1) -2z = 16.
Z = -3.
Eq1: 2x - 5y + 3z = 8.
2x -5*(-1) + 3*(-3) = 8,
X = 6.
Eq3: 2x + 6y + 4z = -6.(multiplied by 2).
Diff. = -11y - z = 14.
Eq2: 3x - y + 4z = 7.
Eq3: 3x + 9y + 6z = -9.(been multiplied by 3).
Diff. = -10y - 2z = 16.
-22y - 2z = 28.(been multiplied by 2).
-10y - 2z = 16.
Diff. = -12y = 12,
Y = -1.
-10*(-1) -2z = 16.
Z = -3.
Eq1: 2x - 5y + 3z = 8.
2x -5*(-1) + 3*(-3) = 8,
X = 6.
okay, thanks, Henry. I just got back home from class.
You are welcome!