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 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 (row echelon form).
3. Solve for the variables using back-substitution.

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

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):

\( (1, 3, 4, |, 9) \)

Row 2 (R2):

\( (-14, 5, |, 9) \)

Row 3 (R3):

\( (3, -7, 6, |, 15) \)

Operation 1: Add R1 to R2
\[
R2 \rightarrow R2 + R1
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
3 & -7 & 6 & | & 15
\end{pmatrix}
\]

Operation 2: Subtract 3 times R1 from R3
\[
R3 \rightarrow R3 - 3R1
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & -16 & -6 & | & -12
\end{pmatrix}
\]

Operation 3: Add 16 times R2 to R3
\[
R3 \rightarrow R3 + 16R2
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & 0 & 138 & | & 252
\end{pmatrix}
\]

Step 3:

Solve for the variables using back-substitution

From the third row:
\[
138x_3 = 252 \implies x_3 = \frac{252}{138} = \frac{126}{69} = \frac{18}{9} = 2
\]

Substitute \( x_3 = 2 \) into the second row:
\[
-1x_2 + 9(2) = 18 \implies -x_2 + 18 = 18 \implies x_2 = 0
\]

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

Solution:

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