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