Solve the equation by Gauss elimination method
6x+3y-z=7
5x-9y+4z=4
8x+5y-z=11
https://www.youtube.com/watch?v=2j5Ic2V7wq4
To solve the system of equations using Gauss elimination method, we need to perform row operations to make a triangular form of the augmented matrix.
Step 1: Write down the augmented matrix
The augmented matrix for the given system of equations is:
[6 3 -1 | 7]
[5 -9 4 | 4]
[8 5 -1 | 11]
Step 2: Perform row operations to eliminate variables
Since the first column's coefficient of the first row is not zero, we use this value as the pivot.
Row 2: Multiply Row 1 by -5 and add it to Row 2
Row 3: Multiply Row 1 by -8 and add it to Row 3
Updated matrix:
[6 3 -1 | 7]
[0 -42 9 | -29]
[0 -19 7 | -45]
Step 3: Continue eliminating variables
The second column's coefficient of the second row is now zero, so we'll use the coefficient of the second column's third row (which is -19) as the pivot.
Row 3: Multiply Row 2 by -42 and add it to Row 3
Updated matrix:
[6 3 -1 | 7]
[0 -42 9 | -29]
[0 0 -295 | -901]
Step 4: Solve for variables by back substitution
We can now solve for the variables in reverse order.
From the last row, we have
-295z = -901
Solving for z, we get z = 901/295
From the second row, we have
-42y + 9z = -29
Substituting the value of z, we get
-42y + 9(901/295) = -29
Simplifying, -42y = -29 - 9(901/295)
Solving for y, we get y = (-29 - 9(901/295))/(-42)
From the first row, we have
6x + 3y - z = 7
Substituting the values of y and z, we get
6x + 3((-29 - 9(901/295))/(-42)) - (901/295) = 7
Simplifying, solve for x.
After substituting the values obtained for x, y, and z, you will have the solution to the system of equations.
To solve the given system of equations using the Gauss elimination method, we will follow these steps:
Step 1: Set up the augmented matrix
Write down the coefficients of the variables and the constant terms from the equations to form an augmented matrix. The augmented matrix is a matrix that represents the system of equations.
The given system of equations can be represented as follows:
| 6 3 -1 | 7 |
| 5 -9 4 | 4 |
| 8 5 -1 | 11 |
Step 2: Perform row operations to simplify the matrix
The goal is to get the matrix into row-echelon form or reduced row-echelon form. We can achieve this by performing row operations while preserving the equality of the system.
We can start by making the first coefficient in the first row equal to 1:
| 1 1/2 -1/6 | 7/6 |
| 5 -9 4 | 4 |
| 8 5 -1 | 11 |
Next, we can use row 1 to eliminate the first coefficient in rows 2 and 3:
| 1 1/2 -1/6 | 7/6 |
| 0 -11/2 19/6 | 1/6 |
| 0 1/2 -2/3 | 1/3 |
Now, we want to make the second coefficient in the second row equal to 1:
| 1 1/2 -1/6 | 7/6 |
| 0 1 -38/11 | -1/11 |
| 0 1/2 -2/3 | 1/3 |
Next, we use row 2 to eliminate the second coefficient in row 3:
| 1 1/2 -1/6 | 7/6 |
| 0 1 -38/11 | -1/11 |
| 0 0 -1/33 | 2/33 |
Step 3: Back substitution
Now, we can solve for the variables starting from the last equation:
From the third equation:
-1/33 * z = 2/33
z = -2
Substituting z into the second equation:
1 * y - (38/11) * (-2) = -1/11
y + (76/11) = -1/11
y = -1/11 - (76/11)
y = -77/11
Finally, substituting z and y into the first equation:
1 * x + (1/2) * (-77/11) - (1/6) * (-2) = 7/6
x - (77/22) + (1/3) = 7/6
x = 7/6 + (77/22) - (1/3)
x = 429/132
Therefore, the solution to the given system of equations is:
x = 429/132, y = -77/11, z = -2.