Use Gauss - jordan elimination to solve for the values of the unknown.
x + y-z = 20
2x + 3y -z = 15
3x -5y + 2z = 10
1 1 -1 20
2 3 -1 15
3 -5 2 10
======================
1 1 -1 20
0 1 1 -25
3 -5 2 10
================
1 1 -1 20
0 1 1 -25
0 -8 5 -50
=============
1 1 -1 20
0 1 1 -25
0 0 13 -250
================
1 1 -1 20
0 1 1 -25
0 0 1 -250/13 yuuk
===================
1 1 -1 20
0 1 0 -75/13
0 0 1 -250/13
==================
1 1 0 10/13
0 1 0 -75/13
0 0 1 -250/13
=================
1 0 0 85/13
010 -75/13
0 0 1 -250/13 Caramba !!!
To solve the system of equations using Gauss-Jordan elimination, we will write the augmented matrix and then perform row operations to transform it into row-echelon form. Let's proceed step by step:
Step 1: Write the augmented matrix
The augmented matrix for the system of equations is:
| 1 1 -1 | 20 |
| 2 3 -1 | 15 |
| 3 -5 2 | 10 |
Step 2: Perform row operations to transform the matrix into row-echelon form
Row 2 = Row 2 - 2 * Row 1
Row 3 = Row 3 - 3 * Row 1
The new matrix becomes:
| 1 1 -1 | 20 |
| 0 1 1 | -25 |
| 0 -8 5 | -50 |
Row 3 = 8 * Row 2 + Row 3
The new matrix becomes:
| 1 1 -1 | 20 |
| 0 1 1 | -25 |
| 0 0 13 | -250 |
Step 3: Perform row operations to transform the matrix into reduced row-echelon form
Row 3 = (1/13) * Row 3
The new matrix becomes:
| 1 1 -1 | 20 |
| 0 1 1 | -25 |
| 0 0 1 | -19.23 |
Row 2 = Row 2 - Row 3
Row 1 = Row 1 - Row 3
The new matrix becomes:
| 1 1 0 | 39.23 |
| 0 1 0 | -6.77 |
| 0 0 1 | -19.23 |
Step 4: Read off the solutions
From the row-echelon form, we can conclude that:
x = 39.23
y = -6.77
z = -19.23
Therefore, the values of the unknown variables are x = 39.23, y = -6.77, and z = -19.23.
To solve the system of equations using Gauss-Jordan elimination, we will use a matrix representation. Let's define the coefficient matrix and the augmented matrix as follows:
Coefficient matrix:
1 1 -1
2 3 -1
3 -5 2
Augmented matrix:
1 1 -1 | 20
2 3 -1 | 15
3 -5 2 | 10
Step 1: Row 1 becomes the pivot row. We need to make the leading coefficient of row 1 equal to 1.
R1 * (1/1) = R1
1 1 -1 | 20
2 3 -1 | 15
3 -5 2 | 10
Step 2: To eliminate the entries below the leading coefficient of row 1, we use row operations.
R2 - 2*R1 = R2
2 1 -3 | 15
2 3 -1 | 15
3 -5 2 | 10
R3 - 3*R1 = R3
3 -4 5 | -10
2 3 -1 | 15
3 -5 2 | 10
Step 3: Make the leading coefficient of row 2 equal to 1.
(1/2)*R2 = R2
1 1 -3 | 15
1 3/2 -1/2 | 7.5
3 -5 2 | 10
Step 4: Eliminate the entries above and below the leading coefficient of row 2.
-R1 + R2 = R2
1 1 -3 | 15
0 1/2 5/2 | -7.5
3 -5 2 | 10
-3*R1 + R3 = R3
1 1 -3 | 15
0 1/2 5/2 | -7.5
0 -8 11 | -35
Step 5: Make the leading coefficient of row 3 equal to 1.
(2/5)*R2 = R2
1 1 -3 | 15
0 1 5 | -15
0 -8 11 | -35
Step 6: Eliminate the entries above and below the leading coefficient of row 3.
-1*R2 + R1 = R1
1 0 -8 | 30
0 1 5 | -15
0 -8 11 | -35
8*R2 + R3 = R3
1 0 -8 | 30
0 1 5 | -15
0 0 51 | -195
Step 7: Solve for the unknowns.
From the resulting matrix, the solutions for x, y, and z are:
x = 30
y = -15
z = -3.82
Therefore, the values of the unknowns are x = 30, y = -15, and z = -3.82.