Solve the following system of linear equations by Gaussian Elimination:
2 x−y+5 z = 10
x+ y−3 z = −2
2 x+4 y+z = 1
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solve the following system of linear equation using Gaussian back substitution {2x+3y=4,x+2y=2
To solve the given system of linear equations using Gaussian elimination, follow these steps:
Step 1: Write the augmented matrix
First, write the augmented matrix by combining the coefficients of the variables and the constant terms from the equations:
| 2 -1 5 | 10 |
| 1 1 -3 | -2 |
| 2 4 1 | 1 |
Step 2: Starting with the first column, create zeros below the diagonal
The goal is to create zeros below the diagonal in the first column. To do this, perform row operations. Start by making the first element in the first column (pivot) equal to 1:
R1 = R1/2 (divide row 1 by 2)
| 1 -0.5 2.5 | 5 |
| 1 1 -3 | -2 |
| 2 4 1 | 1 |
Now, make all the elements below the pivot in the first column equal to zero:
R2 = R2 - R1 (subtract row 1 from row 2)
R3 = R3 - 2R1 (subtract 2 times row 1 from row 3)
| 1 -0.5 2.5 | 5 |
| 0 1.5 -5.5 | -7 |
| 0 5 -4 | -9 |
Step 3: Continue with the second column and create zeros below the pivot
Make the element in the second row and second column (pivot) equal to 1:
R2 = R2/1.5 (divide row 2 by 1.5)
| 1 -0.5 2.5 | 5 |
| 0 1 -3.67 | -4.67 |
| 0 5 -4 | -9 |
Next, make all the elements below the pivot in the second column equal to zero:
R1 = R1 + 0.5R2 (add 0.5 times row 2 to row 1)
R3 = R3 - 5R2 (subtract 5 times row 2 from row 3)
| 1 0 1 | 2.5 |
| 0 1 -3.67 | -4.67 |
| 0 0 10.17 | 5.33 |
Step 4: Solve for the variables
From the last row, we can solve for the variable z:
10.17z = 5.33
z ≈ 0.523
Substitute the value of z back into the second row equation to solve for y:
1y - 3.67(0.523) = -4.67
y ≈ -0.827
Finally, substitute the values of z and y into the first row equation to solve for x:
x - 0.5(-0.827) + 2.5(0.523) = 2.5
x ≈ 1.07
Therefore, the solution to the system of linear equations is:
x ≈ 1.07
y ≈ -0.827
z ≈ 0.523