I need someone to explain to me step by step the process of this problem its in my book as an example but i am not understanding it. Ive researched the explanation of the ECHELON METHOD...BUt i am not seeingit...this is the problem that they have:
Use the echeclon method to solve the system of three equations in three unknowns
2x+4y+z = 22
4x - 4y -z = 2
4x +y =4z = 12
Matrix would be the prefered method, echelon is harder.
First, take eq1)
2x+4y+z = 22 add this to equation 2 to eliminate z
2x+4y+z = 22
4x - 4y -z = 2
6x=24 solve for x.
Now multipy equation 2) by 4
16x-16y-4z=8
add to equation three
4x +y +4z = 12 or
20x-15y=20 you know x, so solve for y.
Put x,y in any of the equations, solve for z.
This was NOT a good problem for echelon.
To solve the given system of equations using the echelon method, we follow these steps:
Step 1: Write down the augmented matrix for the system of equations.
The augmented matrix is formed by writing the coefficients of the variables on the left-hand side of the equal sign and the constants on the right-hand side. The matrix for the given system of equations is:
[ 2 4 1 | 22 ]
[ 4 -4 -1 | 2 ]
[ 4 1 -4 | 12 ]
Step 2: Perform row operations to simplify the matrix.
Using row operations, we aim to eliminate variables systematically. The goal is to create an upper triangular matrix, also known as row echelon form. Let's perform these operations:
2R2 - R1 → R2 (This eliminates x from the second row)
4R1 - R3 → R3 (This eliminates x from the third row)
After performing these operations, the matrix becomes:
[ 2 4 1 | 22 ]
[ 0 -12 -3 | -42 ]
[ 0 5 -5 | -56 ]
Step 3: Continue performing row operations to simplify the matrix further.
In this step, we aim to create a matrix in reduced row echelon form. Let's proceed with the following operations:
(-1/12)R2 → R2 (To simplify the coefficient -12 into 1)
(5/12)R2 +R3 → R3 (To eliminate the value in the second row)
After performing these operations, the matrix becomes:
[ 2 4 1 | 22 ]
[ 0 1 1/4 | 14/4 ]
[ 0 0 0 | 0 ]
Step 4: Interpret the matrix to find the solution.
The matrix is now in reduced row echelon form. From the matrix, we can write the equations:
2x + 4y + z = 22 (Equation A)
y + 1/4z = 14/4 (Equation B)
0 = 0 (Equation C)
Equation C does not provide any additional information; it indicates that the equation simplifies to 0 = 0, which is always true. This shows that there are infinitely many solutions for the given system of equations.
Now, let's consider Equation B:
y + 1/4z = 14/4
Rearranging the equation, we get:
y = 14/4 - 1/4z
Finally, we can substitute the value of y in Equation A to find x:
2x + 4(14/4 - 1/4z) + z = 22
Now you can solve this equation to find the values of x and z by substituting the appropriate coefficients and constants.