Use the echelon method to solve the following system.
x/5+3y =31
2x-y/5=8
Echelon (rung of a ladder in French, meaning steps, rank.
The echelon method is one by which the matrix is reduced to the upper triangular form, resembling steps. The backward pass will enable all unknowns to be evaluated, if the system is soluble.
The row echelon form is the result of the reduction such that the leading element (the first non-zero element from the left) of each row is reduced to 1.0, and that the leading element is to the right of that of the preceding rows.
As an example:
x+y=3 ....(1)
3x-y=1 ....(2)
x+y = 3 ....(1)A
0x-4y = -8 (1)B (2)-3(1)
x+y = 3 ....(1)C
0x+y = 2 ...(2)C
Equations (1)C and (2)C are now in row-echelon form.
from which we can do the backward substitution to get
y=2,
x=1.
From (3)
y=767/5/76
To solve the system using the echelon method, follow these steps:
Step 1: Write the system in matrix form
Start by writing the system of equations in matrix form. Let's call this matrix A, and its augmented matrix will be denoted by [A|B].
\[ A = \begin{bmatrix}
\frac{1}{5} & 3 \\
2 & -\frac{1}{5}
\end{bmatrix}, B = \begin{bmatrix}
31 \\
8
\end{bmatrix} \]
Step 2: Perform row operations to get the matrix in echelon form
The goal is to transform matrix A into echelon form (also known as row echelon form). This is done by using elementary row operations: interchange rows, multiply a row by a non-zero constant, and add or subtract rows.
Start by eliminating the fraction in the first row. Multiply the entire first row by 5 to remove the fraction:
\[ \begin{bmatrix}
1 & 15 \\
2 & -\frac{1}{5}
\end{bmatrix} \begin{bmatrix}
5 & 0 \\
0 & 1
\end{bmatrix} = \begin{bmatrix}
5 & 15 \\
2 & -\frac{1}{5}
\end{bmatrix} \]
Next, we want to eliminate the 2 in the second row. Multiply the first row by -2 and add it to the second row:
\[ \begin{bmatrix}
5 & 15 \\
-2 & -\frac{2}{5}
\end{bmatrix} + \begin{bmatrix}
0 & 0 \\
-10 & 0
\end{bmatrix} = \begin{bmatrix}
5 & 15 \\
-12 & -\frac{2}{5}
\end{bmatrix} \]
Step 3: Continue with row operations to reduce the matrix to row echelon form
To complete the echelon form, multiply the second row by -\frac{1}{12} to make the leading coefficient of the second row 1:
\[ \begin{bmatrix}
5 & 15 \\
-12 & -\frac{2}{5}
\end{bmatrix} \begin{bmatrix}
1 & 0 \\
0 & -\frac{1}{12}
\end{bmatrix} = \begin{bmatrix}
5 & 15 \\
1 & \frac{1}{30}
\end{bmatrix} \]
Step 4: Solve for the variables using back substitution
Now that the matrix is in echelon form, we can solve for the variables by using back substitution.
From the last row, we can determine the value of y:
\[ y = \frac{1}{30} \]
Substituting this value back into the first row, we can solve for x:
\[ 5x + 15(\frac{1}{30}) = 31 \]
\[ 5x + \frac{1}{2} = 31 \]
\[ 5x = \frac{61}{2} \]
\[ x = \frac{61}{10} \]
So the solution to the system is:
\[ x = \frac{61}{10}, y = \frac{1}{30} \]