I don't quite understand how to solve matrices with three variables.
If I have the equations:
7x-4y-2z=0
2x-3y+z=-0.25
-2x-2y+5z=0
How can I solve this? I saw the explanation in a site but I didn't understand it. They appeared to make random operations until the matrix was:
1 0 0
0 1 0
0 0 1
I tried with the equations above but I got stuck no matter which order I had.
Please Help.
THAT IS CALLED GAUSS JORDAN REDUCTION
http://www.youtube.com/watch?v=JwT1FrLzEOQ
Here is a calculator
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
Step 2: Enter the values for your matrix, and press the "Solve" button.
7 -4 -2 0
2 -3 1 -1/4
-2 -2 5 0
Divide row1 by 7
1 -4/7 -2/7 0
2 -3 1 -1/4
-2 -2 5 0
Add (-2 * row1) to row2
1 -4/7 -2/7 0
0 -13/7 11/7 -1/4
-2 -2 5 0
Add (2 * row1) to row3
1 -4/7 -2/7 0
0 -13/7 11/7 -1/4
0 -22/7 31/7 0
Divide row2 by -13/7
1 -4/7 -2/7 0
0 1 -11/13 7/52
0 -22/7 31/7 0
Add (22/7 * row2) to row3
1 -4/7 -2/7 0
0 1 -11/13 7/52
0 0 23/13 11/26
Divide row3 by 23/13
1 -4/7 -2/7 0
0 1 -11/13 7/52
0 0 1 11/46
Add (11/13 * row3) to row2
1 -4/7 -2/7 0
0 1 0 31/92
0 0 1 11/46
Add (2/7 * row3) to row1
1 -4/7 0 11/161
0 1 0 31/92
0 0 1 11/46
Add (4/7 * row2) to row1
1 0 0 6/23
0 1 0 31/92
0 0 1 11/46
To solve a system of equations with three variables using matrices, you would typically use the method of Gaussian elimination or matrix row operations.
Here's an explanation of how to solve the given system of equations using Gaussian elimination:
1. Write the system of equations in matrix form: You can represent the coefficients of the variables and the constants on the right-hand side of each equation in a matrix. Let's call this matrix [A|B].
[7 -4 -2 | 0]
[2 -3 1 | -0.25]
[-2 -2 5 | 0]
2. Perform row operations to transform the matrix [A|B] into its row-echelon form:
- Start with the first column (also known as the pivot column) and make the first element (a11) equal to 1 by dividing the entire row by a11.
- Use row operations to eliminate the coefficients below the pivot column (in this case, the first column).
- Repeat these steps for the remaining columns until you have a row-echelon form.
Here are the steps in detail:
Step 1: Make the first element in the first column (a11) equal to 1:
[1 -4/7 -2/7 | 0]
[2 -3 1 | -0.25]
[-2 -2 5 | 0]
Step 2: Use row operations to eliminate the coefficients below the first column:
- Multiply the first row by -2 and add it to the third row:
[1 -4/7 -2/7 | 0]
[2 -3 1 | -0.25]
[0 6 1/7 | 0]
Step 3: Make the second element in the second column (a22) equal to 1:
- Multiply the second row by -1/4:
[1 -4/7 -2/7 | 0]
[-1/2 3/4 -3/4 | 0.0625]
[0 6 1/7 | 0]
Step 4: Use row operations to eliminate the coefficients below the second column:
- Multiply the first row by 2/7 and add it to the second row:
[1 -4/7 -2/7 | 0]
[0 -1/7 -3/7 | 0.0625]
[0 6 1/7 | 0]
- Multiply the second row by -42/7 and add it to the third row:
[1 -4/7 -2/7 | 0]
[0 -1/7 -3/7 | 0.0625]
[0 0 22/7 | -0.875]
Step 5: Make the third element in the third column (a33) equal to 1:
- Multiply the third row by 7/22:
[1 -4/7 -2/7 | 0]
[0 -1/7 -3/7 | 0.0625]
[0 0 1 | -0.393]
Step 6: Use row operations to eliminate the coefficients above the third column:
- Multiply the third row by 2/7 and add it to the first row:
[1 -4/7 0 | -0.112]
[0 -1/7 -3/7 | 0.0625]
[0 0 1 | -0.393]
- Multiply the third row by 3/7 and add it to the second row:
[1 -4/7 0 | -0.112]
[0 -1/7 0 | -0.14]
[0 0 1 | -0.393]
Step 7: Use row operations to eliminate the coefficients in the fourth column:
- Multiply the second row by -4/7 and add it to the first row:
[1 0 0 | -0.2]
[0 -1/7 0 | -0.14]
[0 0 1 | -0.393]
- Multiply the second row by 1/7 and add it to the third row:
[1 0 0 | -0.2]
[0 -1/7 0 | -0.14]
[0 0 1 | -0.393]
The resulting matrix is in row-echelon form. From this, you can see that x = -0.2, y = 0.14, and z = -0.393.
Note that the row-echelon form of the matrix represents the solution in a simplified form. The 1s along the diagonal represent the coefficients of the variables, and the other elements in the right-most column represent the values of the variables.