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3. Use Gauss-Jordan elimination to solve: 2x1 + x2 – X3 = 1 5x + 2x2 + 2x3 = -4 3x1 + x2 + x3 = 5 - Do not employ pivoting. Check your answers by substituting them into the original equations.

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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.

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