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
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.
.
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
okay.
@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.
okay, thanks, Henry. I just got back home from class.
You are welcome!
To solve the system of equations, you can use various methods such as substitution, elimination, or matrix operations. Let's go through the process step by step.
First, let's choose a method to solve the equations. Both substitution and elimination can be used, but elimination seems more efficient in this case. So, we'll use the elimination method.
Now, let's solve the system of equations.
1. We will start by eliminating the variable x. We can do this by multiplying the first equation by 3 and the second equation by 2, so the coefficients of x in both equations will be the same.
Multiplying the first equation by 3:
3(2x − 5y + 3z) = 3 * 8
6x − 15y + 9z = 24
Multiplying the second equation by 2:
2(3x − y + 4z) = 2 * 7
6x − 2y + 8z = 14
Now we have two equations with the same coefficient for x.
2. Next, we'll eliminate the variable y. To do this, we'll multiply the second equation by 3 and the third equation by 5, so the coefficients of y in both equations will match.
Multiplying the second equation by 3:
3(6x − 2y + 8z) = 3 * 14
18x − 6y + 24z = 42
Multiplying the third equation by 5:
5(x + 3y + 2z) = 5 * -3
5x + 15y + 10z = -15
Now we have two equations with the same coefficient for y.
3. Now, we have two simplified equations:
6x − 15y + 9z = 24 (Equation A)
18x − 6y + 24z = 42 (Equation B)
4. We can eliminate x in these two equations by multiplying Equation A by 3 and Equation B by 1.
Multiplying Equation A by 3:
3(6x − 15y + 9z) = 3 * 24
18x − 45y + 27z = 72
Multiplying Equation B by 1:
1(18x − 6y + 24z) = 1 * 42
18x − 6y + 24z = 42
Now we have two equations with the same coefficient for x.
5. Now, we have two simplified equations:
18x − 45y + 27z = 72 (Equation C)
18x − 6y + 24z = 42 (Equation D)
6. We can further simplify these equations by subtracting Equation D from Equation C.
(18x − 45y + 27z) - (18x − 6y + 24z) = 72 - 42
On simplifying, we get:
-39y + 3z = 30 (Equation E)
7. Finally, we can substitute the value of y from Equation E back into Equation A to solve for z.
6x - 15y + 9z = 24
Now, substitute y = (30 - 3z)/(-39) in Equation A:
6x - 15((30 - 3z)/(-39)) + 9z = 24
Simplify this equation to solve for z.
Once you find the value of z, you can substitute it back into the equations to find the values of x and y.
Remember, this process can involve multiple steps, so ensure that you carefully carry out each step and simplify the equations properly to avoid errors.