Solve the system of non homogeneous equation by echelon matrix form 5x_1+20x_3=50 10x_2+20x_3=-30 5x_1-10x_3=80
To solve the given system of equations using the echelon matrix form, we can start by expressing the system in augmented matrix form. The equations are:
1. \( 5x_1 + 20x_3 = 50 \)
2. \( 10x_2 + 20x_3 = -30 \)
3. \( 5x_1 - 10x_3 = 80 \)
The augmented matrix for this system is:
\[
\begin{bmatrix}
5 & 0 & 20 & | & 50 \\
0 & 10 & 20 & | & -30 \\
5 & 0 & -10 & | & 80
\end{bmatrix}
\]
We will perform row operations to convert this matrix into echelon form.
### Step 1: Row Operations
1. Divide the first row by 5 to simplify:
\[
R_1 \leftarrow \frac{1}{5}R_1
\]
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 10 & 20 & | & -30 \\
5 & 0 & -10 & | & 80
\end{bmatrix}
\]
2. Subtract 5 times the first row from the third row:
\[
R_3 \leftarrow R_3 - 5R_1
\]
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 10 & 20 & | & -30 \\
0 & 0 & -30 & | & 25
\end{bmatrix}
\]
3. Divide the second row by 10:
\[
R_2 \leftarrow \frac{1}{10}R_2
\]
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 1 & 2 & | & -3 \\
0 & 0 & -30 & | & 25
\end{bmatrix}
\]
4. Finally, divide the third row by -30:
\[
R_3 \leftarrow -\frac{1}{30}R_3
\]
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 1 & 2 & | & -3 \\
0 & 0 & 1 & | & -\frac{5}{6}
\end{bmatrix}
\]
### Step 2: Back Substitution
Now we can express the system of equations from the echelon form as:
1. \( x_1 + 4x_3 = 10 \)
2. \( x_2 + 2x_3 = -3 \)
3. \( x_3 = -\frac{5}{6} \)
Substituting \( x_3 \) into the second equation:
\[
x_2 + 2\left(-\frac{5}{6}\right) = -3
\]
\[
x_2 - \frac{10}{6} = -3
\]
\[
x_2 - \frac{5}{3} = -3
\]
\[
x_2 = -3 + \frac{5}{3}
\]
\[
x_2 = -\frac{9}{3} + \frac{5}{3} = -\frac{4}{3}
\]
Now substituting \( x_3 \) back into the first equation:
\[
x_1 + 4\left(-\frac{5}{6}\right) = 10
\]
\[
x_1 - \frac{20}{6} = 10
\]
\[
x_1 - \frac{10}{3} = 10
\]
\[
x_1 = 10 + \frac{10}{3}
\]
\[
x_1 = \frac{30}{3} + \frac{10}{3} = \frac{40}{3}
\]
### Final Solution
The solution to the system of equations is:
\[
x_1 = \frac{40}{3}, \quad x_2 = -\frac{4}{3}, \quad x_3 = -\frac{5}{6}
\]
Update (2024-12-03):
To solve the given system of equations using the echelon form, we first write the equations in augmented matrix form:
1. The equations are:
\[
\begin{align*}
5x_1 + 0x_2 + 20x_3 &= 50 \\
0x_1 + 10x_2 + 20x_3 &= -30 \\
5x_1 + 0x_2 - 10x_3 &= 80 \\
\end{align*}
\]
2. This can be represented in matrix form as:
\[
\begin{bmatrix}
5 & 0 & 20 & | & 50 \\
0 & 10 & 20 & | & -30 \\
5 & 0 & -10 & | & 80
\end{bmatrix}
\]
3. Now, we'll perform row operations to transform this augmented matrix to echelon form.
### Step 1: Row Operations
- **R1**: Divide Row 1 by 5 (to simplify)
\[
R_1 \leftarrow \frac{1}{5}R_1 \rightarrow \begin{bmatrix} 1 & 0 & 4 & | & 10 \end{bmatrix}
\]
- Updated Matrix:
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 10 & 20 & | & -30 \\
5 & 0 & -10 & | & 80
\end{bmatrix}
\]
- **R3**: Subtract 5 times Row 1 from Row 3
\[
R_3 \leftarrow R_3 - 5R_1 \rightarrow \begin{bmatrix} 0 & 0 & -30 & | & 30 \end{bmatrix}
\]
- Updated Matrix:
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 10 & 20 & | & -30 \\
0 & 0 & -30 & | & 30
\end{bmatrix}
\]
- **R2**: Divide Row 2 by 10
\[
R_2 \leftarrow \frac{1}{10}R_2 \rightarrow \begin{bmatrix} 0 & 1 & 2 & | & -3 \end{bmatrix}
\]
- Updated Matrix:
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 1 & 2 & | & -3 \\
0 & 0 & -30 & | & 30
\end{bmatrix}
\]
- **R3**: Divide Row 3 by -30
\[
R_3 \leftarrow -\frac{1}{30}R_3 \rightarrow \begin{bmatrix} 0 & 0 & 1 & | & -1 \end{bmatrix}
\]
- Final Echelon Matrix:
\[
\begin{bmatrix}
1 & 0 & 4 & | & 10 \\
0 & 1 & 2 & | & -3 \\
0 & 0 & 1 & | & -1
\end{bmatrix}
\]
### Step 2: Back Substitution
Now we can back substitute to find the values of the variables.
From Row 3:
\[
x_3 = -1
\]
From Row 2:
\[
x_2 + 2(-1) = -3 \implies x_2 - 2 = -3 \implies x_2 = -1
\]
From Row 1:
\[
x_1 + 4(-1) = 10 \implies x_1 - 4 = 10 \implies x_1 = 14
\]
### Final Solution
Thus, the solution to the system is:
\[
x_1 = 14, \quad x_2 = -1, \quad x_3 = -1
\]
This solution can also be written as the ordered triplet:
\[
(x_1, x_2, x_3) = (14, -1, -1)
\]