Solve the following system by using the Gauss-Jordan elimination method. x+y+z=90;2x+3y+ 6z= 370; 3x-8y -42 = -340

To solve the given system of equations using the Gauss-Jordan elimination method, we'll start by writing the system as an augmented matrix and then perform row operations to transform the matrix into reduced row-echelon form (RREF).

The given system of equations is:

\[
\begin{aligned}
1. & \quad x + y + z = 90 \\
2. & \quad 2x + 3y + 6z = 370 \\
3. & \quad 3x - 8y - 42 = -340
\end{aligned}
\]

### Step 1: Write the augmented matrix
\[
\begin{pmatrix}
1 & 1 & 1 & | & 90 \\
2 & 3 & 6 & | & 370 \\
3 & -8 & -42 & | & -340
\end{pmatrix}
\]

### Step 2: Perform row operations to get the leading 1 in the first column of the first row
The matrix already has a leading 1 in the first row and first column, so no changes are needed.

### Step 3: Eliminate the elements below the leading 1 in the first column
To eliminate the 2 in the second row and the 3 in the third row, we'll perform the following row operations:

- \( R_2 \rightarrow R_2 - 2R_1 \)
- \( R_3 \rightarrow R_3 - 3R_1 \)

\[
\begin{pmatrix}
1 & 1 & 1 & | & 90 \\
0 & 1 & 4 & | & 190 \\
0 & -11 & -45 & | & -610
\end{pmatrix}
\]

### Step 4: Get a leading 1 in the second column of the second row
The second row already has a leading 1 in the second column, so no changes are needed.

### Step 5: Eliminate the elements above and below the leading 1 in the second column
To eliminate the 1 in the first row and the -11 in the third row, we'll perform the following row operations:

- \( R_1 \rightarrow R_1 - R_2 \)
- \( R_3 \rightarrow R_3 + 11R_2 \)

\[
\begin{pmatrix}
1 & 0 & -3 & | & -100 \\
0 & 1 & 4 & | & 190 \\
0 & 0 & -1 & | & 480
\end{pmatrix}
\]

### Step 6: Get a leading 1 in the third column of the third row
To get a leading 1 in the third row, we multiply the third row by \(-1\):

\[
\begin{pmatrix}
1 & 0 & -3 & | & -100 \\
0 & 1 & 4 & | & 190 \\
0 & 0 & 1 & | & -480
\end{pmatrix}
\]

### Step 7: Eliminate the elements above the leading 1 in the third column
To eliminate the -3 in the first row and the 4 in the second row, we'll perform the following row operations:

- \( R_1 \rightarrow R_1 + 3R_3 \)
- \( R_2 \rightarrow R_2 - 4R_3 \)

\[
\begin{pmatrix}
1 & 0 & 0 & | & -1540 \\
0 & 1 & 0 & | & 2110 \\
0 & 0 & 1 & | & -480
\end{pmatrix}
\]

### Final Solution:
The matrix is now in reduced row-echelon form, and the solution to the system is:

\[
\begin{aligned}
x & = -1540 \\
y & = 2110 \\
z & = -480
\end{aligned}
\]

Would you like further details on any of these steps?

### Related Questions
1. How would the solution change if the third equation were different?
2. Can you explain the difference between Gaussian and Gauss-Jordan elimination?
3. How do you verify the solution to a system of linear equations?
4. What are the potential pitfalls in applying Gauss-Jordan elimination?
5. Can a system of equations be inconsistent or dependent using this method?

### Tip
Always check the final solution by substituting the values back into the original equations to ensure accuracy.

Learn how to solve a system of linear equations using the Gauss-Jordan elimination method with detailed step-by-step instructions. This problem involves three linear equations with three variables (x, y, z) and covers matrix operations and elimination techniques. Suitable for advanced high school and college-level students.

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