SOLUTION: solve using gauss-jordan elimination
x+y+z=7
x-y+3z=21
5x+y+z=23
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Question 1206899: solve using gauss-jordan elimination
x+y+z=7
x-y+3z=21
5x+y+z=23
Found 3 solutions by MathLover1, ikleyn, math_tutor2020:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
solve using gauss-jordan elimination
Your matrix
Find the pivot in the 1st column in the 1st row
Eliminate the 1st column
Make the pivot in the 2nd column by dividing the 2nd row by
Eliminate the 2nd column
Make the pivot in the 3rd column by dividing the 3rd row by
Eliminate the 3rd column
Solution set:
Answer by ikleyn(52832) (Show Source): You can put this solution on YOUR website!
.
It can be done using special standard solvers in your regular calculator TI-83/84.
On how to do it, read and learn from these sources
http://www.mscs.mu.edu/~marian/1390Lectures/Unit3/4.3Lecture-11N.pdf
(textual), or
https://www.youtube.com/watch?v=_4aKp_ZbTEI
(Youtube-video)
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
This system of equations
converts to this augmented matrix
I have written the matrix into table form to separate each item. Normally the grid lines aren't present in a matrix. Imagine that a vertical line separates the 4th column from the left 3x3 block of values.
The 3x3 block of values are the coefficients of the left hand sides.
The 4th column represents the right hand sides of the system of equations.
The pivot of the first column is found in the first row.
Typically the pivot value is set to 1 or the goal is to get it to 1.
Let's zero out everything below this pivot.
| 1 | 1 | 1 | 7 | |
| 0 | -2 | 2 | 14 | R2 - R1 --> R2 |
| 5 | 1 | 1 | 23 | |
| 1 | 1 | 1 | 7 | |
| 0 | -2 | 2 | 14 | |
| 0 | -4 | -4 | -12 | R3 - 5*R1 --> R3 |
We'll then transform the value in the 2nd row 2nd column into 1.
Multiply everything in the second row by -1/2 = -0.5
| 1 | 1 | 1 | 7 | |
| 0 | 1 | -1 | -7 | -0.5*R2 --> R2 |
| 0 | -4 | -4 | -12 | |
Then we zero out everything in the 2nd column except for the pivot.
| 1 | 0 | 2 | 14 | R1 - R2 --> R1 |
| 0 | 1 | -1 | -7 | |
| 0 | -4 | -4 | -12 | |
| 1 | 0 | 2 | 14 | |
| 0 | 1 | -1 | -7 | |
| 0 | 0 | -8 | -40 | R3 + 4*R2 --> R3 |
Follow similar steps for the 3rd column. The goal is to get a diagonal of 1's if possible and zero everything else out in that 3x3 block of numbers.
| 1 | 0 | 2 | 14 | |
| 0 | 1 | -1 | -7 | |
| 0 | 0 | 1 | 5 | (-1/8)*R3 --> R3 |
| 1 | 0 | 0 | 4 | R1 - 2*R3 --> R3 |
| 0 | 1 | -1 | -7 | |
| 0 | 0 | 1 | 5 | |
| 1 | 0 | 0 | 4 | |
| 0 | 1 | 0 | -2 | R2 + R3 --> R2 |
| 0 | 0 | 1 | 5 | |
--------------------------------------------------------------------------
After doing all that row reduction, we've gone from this matrix
to this matrix
Therefore x = 4, y = -2, z = 5 is the solution to the system.
In other words (x,y,z) = (4,-2,5) which represents the single point of intersection of the planes x+y+z=7, x-y+3z=21, and 5x+y+z=23
I'll leave the verification for the student to do.
Here is a step-by-step calculator that is very useful to row reduce matrices
http://www.math.odu.edu/~bogacki/lat/
It is called "linear algebra toolkit".
Click the "Enter" link and then go to "Row operation calculator". Let me know if you have any questions about this calculator.
More practice with gauss-jordan elimination
https://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.1203611.html
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