Current Attempt in Progress Solve the following linear system by Gaussian elimination. x_(1)+3x_(2)+4x_(3)=9 -x_(1)-4x_(2)+5x_(3)=9 3x_(1)-7x_(2)+6x_(3)=15

To solve the given system of linear equations using Gaussian elimination, we need to perform a series of row operations on the augmented matrix to reach row-echelon form and then use back-substitution to find the solutions.

The system of equations is:
\[
\begin{array}{c}
x_{1} + 3x_{2} + 4x_{3} = 9 \\
-x_{1} - 4x_{2} + 5x_{3} = 9 \\
3x_{1} - 7x_{2} + 6x_{3} = 15
\end{array}
\]

First, write the augmented matrix for this system:
\[
\left[\begin{array}{ccc|c}
1 & 3 & 4 & 9 \\
-1 & -4 & 5 & 9 \\
3 & -7 & 6 & 15
\end{array}\right]
\]

Step 1:

Make the element below the first pivot (first column, second row) zero.
Add the first row to the second row:
\[
R2 \rightarrow R2 + R1
\]
\[
\left[\begin{array}{ccc|c}
1 & 3 & 4 & 9 \\
0 & -1 & 9 & 18 \\
3 & -7 & 6 & 15
\end{array}\right]
\]

Subtract 3 times the first row from the third row:
\[
R3 \rightarrow R3 - 3R1
\]
\[
\left[\begin{array}{ccc|c}
1 & 3 & 4 & 9 \\
0 & -1 & 9 & 18 \\
0 & -16 & -6 & -12
\end{array}\right]
\]

Step 2:

Make the element below the second pivot (second column, third row) zero.
Add 16 times the second row to the third row:
\[
R3 \rightarrow R3 + 16R2
\]
\[
\left[\begin{array}{ccc|c}
1 & 3 & 4 & 9 \\
0 & -1 & 9 & 18 \\
0 & 0 & 126 & 252
\end{array}\right]
\]

Step 3:

Simplify the third row.
Divide the third row by 126:
\[
R3 \rightarrow \frac{1}{126}R3
\]
\[
\left[\begin{array}{ccc|c}
1 & 3 & 4 & 9 \\
0 & -1 & & 18 \\
0 & 0 & 1 & 2
\end{array}\right]
\]

Step 4:

Back-substitution to find the solution.
From the third row, we have:
\[
x_3 = 2
\]

Substitute \( x_3 = 2 \) into the second row:
\[
- x_2 + 9x_3 = 18 \implies - x_2 + 18 = 18 \implies x_2 = 0
\]

Substitute \( x_2 = 0 \) and \( x_3 = 2 \) into the first row:
\[
x_1 + 3x_2 + 4x_3 = 9 \implies x_1 + 0 + 8 = 9 \implies x_1 = 1
\]

Solution:

\[
x_1 = 1, \quad x_2 = 0, \quad x_3 = 2
\]