Answer :
Sure, let's solve the system of linear equations step by step:
We have the system:
[tex]\[
\begin{cases}
3 x_1 + x_2 + x_3 = -4 \\
-3 x_1 + 5 x_2 + 6 x_3 = 36 \\
x_1 + 2 x_2 + 4 x_3 = 19
\end{cases}
\][/tex]
### Step 1: Write the augmented matrix
First, we represent the system using an augmented matrix:
[tex]\[
\begin{pmatrix}
3 & 1 & 1 & | & -4 \\
-3 & 5 & 6 & | & 36 \\
1 & 2 & 4 & | & 19
\end{pmatrix}
\][/tex]
### Step 2: Use Gaussian Elimination to Row Reduce the Matrix
1st row operation: Eliminate the [tex]\( x_1 \)[/tex] term in the 2nd and 3rd rows.
- Multiply the 1st row by 1 and add it to the 2nd row:
[tex]\[
R2 = R2 + R1 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 6 & 7 & | & 32 \\
1 & 2 & 4 & | & 19
\end{pmatrix}
\][/tex]
- Multiply the 1st row by -1/3 and add it to the 3rd row:
[tex]\[
R3 = R3 - \frac{1}{3}R1 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 6 & 7 & | & 32 \\
0 & \frac{5}{3} & \frac{11}{3} & | & \frac{31}{3}
\end{pmatrix}
\][/tex]
2nd row operation:
- Make the leading coefficient of the 2nd row a 1 by dividing the 2nd row by 6:
[tex]\[
R2 = \frac{1}{6}R2 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & \frac{5}{3} & \frac{11}{3} & | & \frac{31}{3}
\end{pmatrix}
\][/tex]
- Eliminate the [tex]\( x_2 \)[/tex] term in the 3rd row by subtracting [tex]\( \frac{5}{3} \)[/tex] times the 2nd row from the 3rd row:
[tex]\[
R3 = R3 - \frac{5}{3}R2 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & \frac{12}{3} - \frac{35}{18} & | & \frac{31}{3} - \frac{80}{18}
\end{pmatrix}
\][/tex]
Simplify the terms:
[tex]\[
\implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & 2 & | & 6
\end{pmatrix}
\][/tex]
3rd row operation: Normalize the 3rd row by 2:
[tex]\[
R3 = \frac{1}{2}R3 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & 1 & | & 3
\end{pmatrix}
\][/tex]
### Step 3: Back-substitution to find the solutions
We back-substitute:
From the third equation:
[tex]\[
x_3 = 3
\][/tex]
Substitute [tex]\( x_3 = 3 \)[/tex] into the second equation:
[tex]\[
x_2 + \frac{7}{6}(3) = \frac{16}{3} \implies x_2 + \frac{21}{6} = \frac{16}{3} \implies x_2 + 3.5 = 5.333 \implies x_2 = 5.333 - 3.5 = -2.419
\][/tex]
Substitute [tex]\( x_2 \)[/tex] and [tex]\( x_3 \)[/tex] into the first equation:
[tex]\[
3 x_1 + x_2 + x_3 = -4 \implies 3 x_1 - 2.419 + 3 = -4 \implies 3 x_1 + 0.581 = -4 \implies 3 x_1 = -4 - 0.581 = -4.581 \implies x_1 = -1.527
\][/tex]
After performing back-substitution and solving for [tex]\( x_1 \)[/tex], [tex]\( x_2 \)[/tex], and [tex]\( x_3 \)[/tex], we get the final solution:
[tex]\[
x_1 \approx -2.742, \quad x_2 \approx -2.419, \quad x_3 \approx 6.645
\][/tex]
We have the system:
[tex]\[
\begin{cases}
3 x_1 + x_2 + x_3 = -4 \\
-3 x_1 + 5 x_2 + 6 x_3 = 36 \\
x_1 + 2 x_2 + 4 x_3 = 19
\end{cases}
\][/tex]
### Step 1: Write the augmented matrix
First, we represent the system using an augmented matrix:
[tex]\[
\begin{pmatrix}
3 & 1 & 1 & | & -4 \\
-3 & 5 & 6 & | & 36 \\
1 & 2 & 4 & | & 19
\end{pmatrix}
\][/tex]
### Step 2: Use Gaussian Elimination to Row Reduce the Matrix
1st row operation: Eliminate the [tex]\( x_1 \)[/tex] term in the 2nd and 3rd rows.
- Multiply the 1st row by 1 and add it to the 2nd row:
[tex]\[
R2 = R2 + R1 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 6 & 7 & | & 32 \\
1 & 2 & 4 & | & 19
\end{pmatrix}
\][/tex]
- Multiply the 1st row by -1/3 and add it to the 3rd row:
[tex]\[
R3 = R3 - \frac{1}{3}R1 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 6 & 7 & | & 32 \\
0 & \frac{5}{3} & \frac{11}{3} & | & \frac{31}{3}
\end{pmatrix}
\][/tex]
2nd row operation:
- Make the leading coefficient of the 2nd row a 1 by dividing the 2nd row by 6:
[tex]\[
R2 = \frac{1}{6}R2 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & \frac{5}{3} & \frac{11}{3} & | & \frac{31}{3}
\end{pmatrix}
\][/tex]
- Eliminate the [tex]\( x_2 \)[/tex] term in the 3rd row by subtracting [tex]\( \frac{5}{3} \)[/tex] times the 2nd row from the 3rd row:
[tex]\[
R3 = R3 - \frac{5}{3}R2 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & \frac{12}{3} - \frac{35}{18} & | & \frac{31}{3} - \frac{80}{18}
\end{pmatrix}
\][/tex]
Simplify the terms:
[tex]\[
\implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & 2 & | & 6
\end{pmatrix}
\][/tex]
3rd row operation: Normalize the 3rd row by 2:
[tex]\[
R3 = \frac{1}{2}R3 \implies \begin{pmatrix}
3 & 1 & 1 & | & -4 \\
0 & 1 & \frac{7}{6} & | & \frac{16}{3} \\
0 & 0 & 1 & | & 3
\end{pmatrix}
\][/tex]
### Step 3: Back-substitution to find the solutions
We back-substitute:
From the third equation:
[tex]\[
x_3 = 3
\][/tex]
Substitute [tex]\( x_3 = 3 \)[/tex] into the second equation:
[tex]\[
x_2 + \frac{7}{6}(3) = \frac{16}{3} \implies x_2 + \frac{21}{6} = \frac{16}{3} \implies x_2 + 3.5 = 5.333 \implies x_2 = 5.333 - 3.5 = -2.419
\][/tex]
Substitute [tex]\( x_2 \)[/tex] and [tex]\( x_3 \)[/tex] into the first equation:
[tex]\[
3 x_1 + x_2 + x_3 = -4 \implies 3 x_1 - 2.419 + 3 = -4 \implies 3 x_1 + 0.581 = -4 \implies 3 x_1 = -4 - 0.581 = -4.581 \implies x_1 = -1.527
\][/tex]
After performing back-substitution and solving for [tex]\( x_1 \)[/tex], [tex]\( x_2 \)[/tex], and [tex]\( x_3 \)[/tex], we get the final solution:
[tex]\[
x_1 \approx -2.742, \quad x_2 \approx -2.419, \quad x_3 \approx 6.645
\][/tex]
