Answer:
The solution of the system of equations is
.
Step-by-step explanation:
Gauss–Jordan elimination is a procedure for converting a matrix to reduced row echelon form using elementary row operations.
It relies upon three elementary row operations one can use on a matrix:
- Swap the positions of two of the rows
- Multiply one of the rows by a nonzero scalar.
- Add or subtract the scalar multiple of one row to another row.
To find the solution of the system

using Gauss-Jordan elimination you must:
Step 1: Transform the augmented matrix to the reduced row echelon form.
In an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.
This is the augmented matrix that represents the system.
![\left[ \begin{array}{ccc} -3 & 5 & -34 \\\\ 3 & 4 & -11 \\\\ 4 & -8 & 52 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%20-3%20%26%205%20%26%20-34%20%5C%5C%5C%5C%203%20%26%204%20%26%20-11%20%5C%5C%5C%5C%204%20%26%20-8%20%26%2052%20%5Cend%7Barray%7D%20%5Cright%5D)
Using elementary matrix operations, we get that
Row Operation 1: Add row 1 to row 2 
Row Operation 2: Divide row 1 by −3 
Row Operation 3: Subtract row 1 multiplied by 4 from row 3 
Row Operation 4: Divide row 2 by 9 
Row Operation 5: Add row 2 multiplied by 5/3 to row 1 
Row Operation 6: Add row 2 multiplied by 4/3 to row 3 
This is the reduced row echelon form matrix
![\left[ \begin{array}{ccc} 1 & 0 & 3 \\\\ 0 & 1 & -5 \\\\ 0 & 0 & 0 \end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%203%20%5C%5C%5C%5C%200%20%26%201%20%26%20-5%20%5C%5C%5C%5C%200%20%26%200%20%26%200%20%5Cend%7Barray%7D%20%5Cright%5D)
Step 2: Interpret the reduced row echelon form
The reduced row echelon form of the augmented matrix corresponds to the system
