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 x_(1)=i x_(2)=i x_(3)=i

To solve the given linear system using Gaussian elimination, we need to follow these steps:

1. Write the augmented matrix for the system.
2. Perform row operations to get an upper triangular form.
3. Back-substitute to find the solutions.

Given system:
\[
\begin{array}{l}
x_{1} + 3x_{2} + 4x_{3} = 9 \\
-x_{1} - 4x_{2} + 5x_{3} = 9 \\
3x_{1} - 7x_{2} + 6x_{3} = 15
\end{array}
\]

Step 1:

Write the augmented matrix
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
-1 & -4 & 5 & | & 9 \\
3 & -7 & 6 & | & 15
\end{pmatrix}
\]

Step 2:

Perform row operations to get an upper triangular form

Row 1 (R1) remains the same:

\[
R1 = (1, 3, 4, |, 9)
\]

Eliminate \(x_1\) from Row 2 (R2):

\[
R2 = R2 + R1 = (-1, -4, 5, |, 9) + (1, 3, 4, |, 9) = (0, -1, 9, |, 18)
\]

Eliminate \(x_1\) from Row 3 (R3):

\[
R3 = R3 - 3R1 = (3, -7, 6, |, 15) - 3(1, 3, 4, |, 9) = (0, -10, -6, |, 6)
\]

Now the matrix looks like this:
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & -10 & -6 & | & 6
\end{pmatrix}
\]

Eliminate \(x_2\) from Row 3 (R3):

\[
R3 = R3 + 10R2 = (0, -10, -6, |, 6) + 10(0, -1, 9, |, 18) = (0, 0, 84, |, 96)
\]

Now the matrix looks like this:
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & 0 & 1 & | & 1
\end{pmatrix}
\]

Step 3:

Back-substitution to find the solutions

From the third row:
\[
x_3 = 1
\]

Substitute \(x_3 = 1\) into the second row:
\[
- x_2 + 9x_3 = 18 \implies - x_2 + 9(1) = 18 \implies - x_2 + 9 = 18 \implies - x_2 = 9 \implies x_2 = -9
\]

Substitute \(x_2 = -9\) and \(x_3 = 1\) into the first row:
\[
x_1 + 3(-9) + 4(1) = 9 \implies x_1 - 27 + 4 = 9 \implies x_1 - 23 = 9 \implies x_1 = 32
\]

Final Solution:

\[
x_1 = 32, \quad x_2 = -9, \quad x_3 = 1
\]