Use the​ Gauss-Jordan method to solve the system.
2xplus
2yminus
2 z equals
22
3xminus
yplus
zequals
negative 19
xminus
4yplus
2zequals
negative 74

To solve the system of equations using the Gauss-Jordan method, we first write the augmented matrix for the system of equations:

\[
\begin{align*}
2x + 2y - 2z &= 22 \\
3x - y + z &= -19 \\
x - 4y + 2z &= -74
\end{align*}
\]

The augmented matrix is:

\[
\begin{pmatrix}
2 & 2 & -2 & | & 22 \\
3 & -1 & 1 & | & -19 \\
1 & -4 & 2 & | & -74
\end{pmatrix}
\]

**Step 1: Make the element in the first row, first column a 1.**

Divide the first row by 2:

\[
\begin{pmatrix}
1 & 1 & -1 & | & 11 \\
3 & -1 & 1 & | & -19 \\
1 & -4 & 2 & | & -74
\end{pmatrix}
\]

**Step 2: Make the elements below the leading 1 in the first column zero.**

Subtract 3 times the first row from the second row:

\[
\text{Row 2} = \text{Row 2} - 3 \cdot \text{Row 1}
\]

\[
\begin{pmatrix}
1 & 1 & -1 & | & 11 \\
0 & -4 & 4 & | & -52 \\
1 & -4 & 2 & | & -74
\end{pmatrix}
\]

Subtract the first row from the third row:

\[
\text{Row 3} = \text{Row 3} - \text{Row 1}
\]

\[
\begin{pmatrix}
1 & 1 & -1 & | & 11 \\
0 & -4 & 4 & | & -52 \\
0 & -5 & 3 & | & -85
\end{pmatrix}
\]

**Step 3: Make the element in the second row, second column a 1.**

Divide the second row by -4:

\[
\begin{pmatrix}
1 & 1 & -1 & | & 11 \\
0 & 1 & -1 & | & 13 \\
0 & -5 & 3 & | & -85
\end{pmatrix}
\]

**Step 4: Make the elements above and below the leading 1 in the second column zero.**

Subtract the second row from the first row:

\[
\text{Row 1} = \text{Row 1} - \text{Row 2}
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -2 \\
0 & 1 & -1 & | & 13 \\
0 & -5 & 3 & | & -85
\end{pmatrix}
\]

Add 5 times the second row to the third row:

\[
\text{Row 3} = \text{Row 3} + 5 \cdot \text{Row 2}
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -2 \\
0 & 1 & -1 & | & 13 \\
0 & 0 & -2 & | & -20
\end{pmatrix}
\]

**Step 5: Make the element in the third row, third column a 1.**

Divide the third row by -2:

\[
\begin{pmatrix}
1 & 0 & 0 & | & -2 \\
0 & 1 & -1 & | & 13 \\
0 & 0 & 1 & | & 10
\end{pmatrix}
\]

**Step 6: Make the elements above the leading 1 in the third column zero.**

Add the third row to the second row:

\[
\text{Row 2} = \text{Row 2} + \text{Row 3}
\]

\[
\begin{pmatrix}
1 & 0 & 0 & | & -2 \\
0 & 1 & 0 & | & 23 \\
0 & 0 & 1 & | & 10
\end{pmatrix}
\]

The final augmented matrix represents the solution to the system:

\[
\begin{cases}
x = -2 \\
y = 23 \\
z = 10
\end{cases}
\]

So, the solution to the system is \( x = -2 \), \( y = 23 \), and \( z = 10 \).

If you have any questions or need further details, feel free to ask!

Here are 5 questions you might find interesting:
1. How do you solve a system of equations using the elimination method?
2. What are the advantages of the Gauss-Jordan method over other methods?
3. Can you explain how to interpret the reduced row echelon form of a matrix?
4. How do you determine the rank of a matrix?
5. What are some applications of solving systems of linear equations in real life?

**Tip:** When solving systems of equations, always check your final answers by substituting them back into the original equations to ensure they satisfy all equations.

Learn how to solve a system of equations using the Gauss-Jordan method step-by-step. This problem covers advanced topics in linear algebra, including the application of matrices and row operations. Suitable for advanced high school students or those studying linear algebra.

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