Write the system as a matrix and solve it by Gauss-Jordan elimination. (If the system is inconsistent, enter INCONSISTENT. If the system has infinitely many solutions, show a general solution in terms of x, y, or z.)
x + 2y − z = 5
2x − y + z = 2
3x − 4y + 3z = −1
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To solve the system of equations using Gauss-Jordan elimination, let's write the system as an augmented matrix:
| 1 2 -1 | 5 |
| 2 -1 1 | 2 |
| 3 -4 3 | -1 |
Now, let's perform the row operations to get the matrix in reduced row-echelon form:
Step 1: Replace R2 with R2 - 2R1.
| 1 2 -1 | 5 |
| 0 -5 3 | -8 |
| 3 -4 3 | -1 |
Step 2: Replace R3 with R3 - 3R1.
| 1 2 -1 | 5 |
| 0 -5 3 | -8 |
| 0 -10 6 | -16 |
Step 3: Replace R3 with R3 + 2R2.
| 1 2 -1 | 5 |
| 0 -5 3 | -8 |
| 0 0 12 | -32 |
Step 4: Replace R1 with R1 + R2.
| 1 -3 2 | -3 |
| 0 -5 3 | -8 |
| 0 0 12 | -32 |
Step 5: Replace R2 with R2/(-5).
| 1 -3 2 | -3 |
| 0 1 -(3/5)| 8/5 |
| 0 0 12 | -32 |
Step 6: Replace R3 with R3/12.
| 1 -3 2 | -3 |
| 0 1 -(3/5)| 8/5 |
| 0 0 1 | -8/3 |
Step 7: Replace R2 with R2 + (3/5)R3.
| 1 -3 2 | -3 |
| 0 1 0 | 0 |
| 0 0 1 | -8/3 |
Step 8: Replace R1 with R1 - 2R3.
| 1 -3 0 | 13/3 |
| 0 1 0 | 0 |
| 0 0 1 | -8/3 |
Step 9: Replace R1 with R1 + 3R2.
| 1 0 0 | 13/3 |
| 0 1 0 | 0 |
| 0 0 1 | -8/3 |
Now we have the matrix in reduced row-echelon form. The solution to the system of equations is:
x = 13/3
y = 0
z = -8/3
Therefore, the system has a unique solution.
To solve the system of equations using Gauss-Jordan elimination, we can represent the system as an augmented matrix.
First, let's write down the augmented matrix:
| 1 2 -1 | 5 |
| 2 -1 1 | 2 |
| 3 -4 3 | -1 |
The goal of Gauss-Jordan elimination is to transform the augmented matrix into reduced row-echelon form, where each leading coefficient (the first non-zero value in each row) is 1 and is located to the right of leading coefficients in previous rows. This will help us easily solve for the variables.
Let's start the elimination process:
Step 1: Swap rows if necessary to make the leading coefficient of the first row non-zero.
We don't need to swap rows in this case since the first row's leading coefficient is already 1.
| 1 2 -1 | 5 |
| 2 -1 1 | 2 |
| 3 -4 3 | -1 |
Step 2: Use row operations to eliminate the leading coefficients below the first row.
Multiply the first row by -2 and add it to the second row:
| 1 2 -1 | 5 |
| 0 -5 3 | -8 |
| 3 -4 3 | -1 |
Multiply the first row by -3 and add it to the third row:
| 1 2 -1 | 5 |
| 0 -5 3 | -8 |
| 0 -10 6 | -16 |
Step 3: Perform further row operations to eliminate the leading coefficients above the first row.
Multiply the second row by -2 and add it to the first row:
| 1 0 1 | -1 |
| 0 -5 3 | -8 |
| 0 -10 6 | -16 |
Multiply the second row by -2 and add it to the third row:
| 1 0 1 | -1 |
| 0 -5 3 | -8 |
| 0 0 0 | 0 |
Now, the augmented matrix is in reduced row-echelon form. Let's interpret it back into equations:
x + z = -1
-5y + 3z = -8
0 = 0
From the bottom row, we can see that 0 = 0. This means that one of the variables (z, in this case) is a free variable. We can choose any value for z.
Let's express the remaining variables in terms of z:
From the second row, we have -5y + 3z = -8. Solving for y:
-5y = -8 - 3z
y = (8 + 3z) / 5
From the first row, we have x + z = -1. Solving for x:
x = -1 - z
Therefore, the system has infinitely many solutions, and the general solution in terms of x, y, and z is:
x = -1 - z
y = (8 + 3z) / 5
z is any real number