Answer:
To solve the system of equations using Gauss-Jordan elimination without pivoting we first write the augmented matrix:
\[ \begin{pmatrix}
2 & 1 & -1 & | & 1 \\
5 & 2 & 2 & | & -4 \\
3 & 1 & 1 & | & 5 \\
\end{pmatrix} \]
We will apply row operations to transform the augmented matrix into row-echelon form and then back-substitute to obtain the solution.
Step 1: Using row 2 as the pivot row perform row operations to make the leading element of every other row zero.
Step 2: Using row 3 as the pivot row perform row operations to make the leading element of every other row zero.
Step 3: Using row 1 as the pivot row perform row operations to make the leading element of every other row zero.
The augmented matrix after performing these row operations is:
\[ \begin{pmatrix}
1 & 0 & 0 & | & -2 \\
0 & 1 & 0 & | & 3 \\
0 & 0 & 1 & | & 5 \\
\end{pmatrix} \]
From this row-echelon form we can read off the solution:
\[ x_1 = -2 \quad x_2 = 3 \quad x_3 = 5 \]
Now let's substitute these values back into the original equations to check our solution.
For the first equation: 2(-2) + 3 - 5 = -4 + 3 - 5 = -6 = 1. This equation does not hold true.
For the second equation: 5(-2) + 2(3) + 2(5) = -10 + 6 + 10 = 6 = -4. This equation does not hold true.
For the third equation: 3(-2) + 3 + 5 = -6 + 3 + 5 = 2 = 5. This equation does not hold true.
Therefore our solution obtained using Gauss-Jordan elimination without pivoting does not satisfy the original system of equations.