Solve using Gauss Jordan method - Write in terms of (x, y, z) = ( )
3y + 2z = 1
2x − y − 3z = 3
2x + 2y − z = 4
To solve the given system of equations using the Gauss-Jordan method, we'll perform row operations to transform the augmented matrix into reduced row-echelon form.
First, we'll set up the augmented matrix using the coefficients of the variables:
| 0 3 2 | 1 |
| 2 -1 -3 | 3 |
| 2 2 -1 | 4 |
The goal is to transform the left side of the augmented matrix into the identity matrix (I).
Step 1: Swap rows if necessary to put a non-zero entry at the top left corner (pivot position). In this case, we'll swap Row 1 and Row 2 to get a non-zero entry.
| 2 -1 -3 | 3 |
| 0 3 2 | 1 |
| 2 2 -1 | 4 |
Step 2: Use row operations to create zeros below and above the pivot position.
First, we'll multiply Row 2 by 2 and subtract it from Row 1:
| 0 -7 -7 | 1 |
| 0 3 2 | 1 |
| 2 2 -1 | 4 |
Then, we'll multiply Row 2 by 2/3 to normalize it:
| 0 -7 -7 | 1 |
| 0 1 2/3 | 1/3 |
| 2 2 -1 | 4 |
Finally, we'll multiply Row 2 by 2 and subtract it from Row 3:
| 0 -7 -7 | 1 |
| 0 1 2/3 | 1/3 |
| 2 -4 -5/3 | 2 |
Step 3: Use row operations to create a diagonal of 1's as pivots and zeros above and below them.
First, we'll multiply Row 3 by 7 and add 4 times Row 2:
| 0 -7 -7 | 1 |
| 0 1 2/3 | 1/3 |
| 0 0 1 | 10 |
Next, we'll multiply Row 3 by -7 and add 7 times Row 2:
| 0 -7 -7 | 1 |
| 0 1 0 | 21/3 |
| 0 0 1 | 10 |
Finally, we'll multiply Row 3 by 7 and add 7 times Row 1:
| 0 0 0 | 22 |
| 0 1 0 | 21/3 |
| 0 0 1 | 10 |
Step 4: Back-substitution.
From the last row, we have 0 = 22, which is inconsistent. Therefore, the system of equations is inconsistent and has no solution.
Hence, there are no values of x, y, and z that satisfy the given system of equations.