Question 1 of 1 View Policies 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

To solve the given linear system by 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. Use back-substitution to find the solutions.

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

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):

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

Row 2 (R2):

\( (-1, -4, 5, |, 9) \)

Row 3 (R3):

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

Operation 1: Make the element below the first pivot (first column, second row) zero by adding \( R1 \) to \( R2 \).
\[
R2 \rightarrow R2 + R1
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
3 & -7 & 6 & | & 15
\end{pmatrix}
\]

Operation 2: Make the element below the first pivot (first column, third row) zero by subtracting \( 3R1 \) from \( R3 \).
\[
R3 \rightarrow R3 - 3R1
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & -16 & -6 & | & -12
\end{pmatrix}
\]

Operation 3: Make the element below the second pivot (second column, third row) zero by adding \( 16R2 \) to \( R3 \).
\[
R3 \rightarrow R3 + 16R2
\]
\[
\begin{pmatrix}
1 & 3 & 4 & | & 9 \\
0 & -1 & 9 & | & 18 \\
0 & 0 & 126 & | & 252
\end{pmatrix}
\]

Step 3:

Use back-substitution to find the solutions

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

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

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

Final Solution:

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