Gauss elimination method 3x+4y+5z=18, 2x-y+8z=13, 5x-2y+7z=20
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Answer
To solve the given system of linear equations using the Gauss elimination method, we'll follow these steps:
Step 1: Write the augmented matrix for the given system of equations.
Step 2: Perform row operations to obtain an upper triangular matrix.
Step 3: Solve for the variables using back-substitution.
Step 4: Verify the solution obtained.
Let's go through these steps one by one.
Step 1: Write the augmented matrix for the given system of equations
The augmented matrix for the given system of equations is:
[ 3 4 5 | 18 ]
[ 2 -1 8 | 13 ]
[ 5 -2 7 | 20 ]
Step 2: Perform row operations to obtain an upper triangular matrix.
To eliminate the x terms in the second and third rows, we'll perform the following row operations:
R2 = R2 - (2/3)*R1
R3 = R3 - (5/3)*R1
The updated augmented matrix becomes:
[ 3 4 5 | 18 ]
[ 0 -11/3 22/3 | 13 ]
[ 0 -22/3 4/3 | 20 - 30/3 ]
Next, we'll eliminate the y term in the third row by performing the following row operation:
R3 = R3 - (2/11)*R2
The updated augmented matrix becomes:
[ 3 4 5 | 18 ]
[ 0 -11/3 22/3 | 13 ]
[ 0 0 20/11 | 20/11 - 6/11 ]
Step 3: Solve for the variables using back-substitution.
From the last row, we have:
(20/11)z = (20/11) - (6/11)
Simplifying, we get:
z = 1
Substituting the value of z back into the second row, we have:
(-11/3)y + (22/3)z = 13
(-11/3)y + (22/3)*1 = 13
(-11/3)y + 22/3 = 13
(-11/3)y = 13 - 22/3
(-11/3)y = (39 - 22)/3
(-11/3)y = 17/3
y = -17/11
Finally, substituting the values of y and z back into the first row, we have:
3x + 4y + 5z = 18
3x + 4*(-17/11) + 5*1 = 18
3x - (68/11) + 5 = 18
3x - 68/11 + 55/11 = 18
3x - 13/11 = 18
3x = 18 + 13/11
3x = (198 + 143)/11
3x = 341/11
x = 341/33
So, the solution to the given system of equations is:
x = 341/33
y = -17/11
z = 1
Step 4: Verify the solution obtained.
Substituting the values of x, y, and z back into the original equations, we can verify if the solution satisfies the equations.
For the first equation: 3x + 4y + 5z = 18
3*(341/33) + 4*(-17/11) + 5*1 = 18
341/11 - 68/11 + 55/11 = 198/11
(341 - 68 + 55)/11 = 198/11
328/11 = 198/11
Both sides are equal, so the solution satisfies the first equation.
Perform the same verification process for the remaining two equations to confirm that the solution obtained is correct.
To solve the system of equations using Gauss elimination method, we need to follow these steps:
Step 1: Write the augmented matrix
First, we need to write the system of equations in the form of an augmented matrix. The augmented matrix is a matrix that includes the coefficients of the variables and the constants on the right side of the equations.
The augmented matrix for the given system of equations is:
```
[ 3 4 5 | 18 ]
[ 2 -1 8 | 13 ]
[ 5 -2 7 | 20 ]
```
Step 2: Perform row operations to create zeros below the leading entries
We want to create zeros below the leading entries (the leftmost nonzero entries) in each column. For this, we perform row operations.
First, we want to get a leading entry of 1 in the first column, so we divide the first row by the coefficient of x, which is 3:
```
[ 1 4/3 5/3 | 6 ]
[ 2 -1 8 | 13 ]
[ 5 -2 7 | 20 ]
```
Next, we create zeros below the leading entry of the first column. To do this, we subtract 2 times the first row from the second row and 5 times the first row from the third row:
```
[ 1 4/3 5/3 | 6 ]
[ 0 -11/3 2/3 | 1 ]
[ 0 -22/3 8/3 | -10 ]
```
Step 3: Continue performing row operations to create a row-echelon form
Now, we continue performing row operations to create a row-echelon form, where each leading entry is to the right and below the leading entry of the row above it.
Next, we want to create a leading entry of 1 in the second row. We multiply the second row by -3/11 to make the leading coefficient -1:
```
[ 1 4/3 5/3 | 6 ]
[ 0 1 -2/11 | -3/11 ]
[ 0 -22/3 8/3 | -10 ]
```
Next, we create zeros below the leading entry of the second row. To do this, we subtract 4/3 times the second row from the first row and add 22/3 times the second row to the third row:
```
[ 1 0 27/11 | 8/11 ]
[ 0 1 -2/11 | -3/11 ]
[ 0 0 38/11 | -56/11 ]
```
Step 4: Back substitution
Now, we have obtained the row-echelon form. We can solve for z by dividing the third row by 38/11:
```
[ 1 0 27/11 | 8/11 ]
[ 0 1 -2/11 | -3/11 ]
[ 0 0 1 | -56/38 ]
```
Replacing the values back into the original equations, we can solve for x and y:
From the third row, z = -56/38 = -28/19.
From the second row, y - (2/11)(-28/19) = -3/11.
Simplifying, we get y = -3/11 + (2/11)(28/19) = 49/19.
From the first row, x - (27/11)(-28/19) + (27/11)(49/19) = 8/11.
Simplifying, we get x = 8/11 + (27/11)(28/19) - (27/11)(49/19) = 3.
Therefore, the solution to the system of equations is x = 3, y = 49/19, z = -28/19.