Question 2 (2 marks) Attempt 1 Use Gaussian elimination with partial pivoting to
ID: 3114880 • Letter: Q
Question
Question 2 (2 marks) Attempt 1 Use Gaussian elimination with partial pivoting to solve the system of linear equations. Do not use Gauss-Jordan elimination. Use exact arithmetio. 22 +342 Enter each component of the right hand side vector after completion of the upper triangularization of the linear system. (g1, 92, 93, 94), and, enter each component of the final solution after completion of the backward substitution stage (x1, X2, X3, X4) Enter each component as an exact expressiorn For example: -2/3 OR -45 OR 2 Do not enter floating point values like -0.666667 OR 2.0 91Skipped 92 Skipped g3= skipped 94 Skipped X1 Skipped X2- skipped X3= skipped X4= SkippedExplanation / Answer
The augmented matrix of the given system of linear equations is A =
1
2
2
2
0
2
4
3
0
-5
0
2
1
3
-2
0
1
3
-2
-2
To solve this system of linear equations, we will reduce A to upper triangular matrix as under:
Add -2 times the 1st row to the 2nd row
Interchange the 2nd row and the 3rd row
Multiply the 2nd row by ½
Add -1 times the 2nd row to the 4th row
Multiply the 3rd row by -1
Add -5/2 times the 3rd row to the 4th row
Multiply the 4th row by -2/27
Add -4 times the 4th row to the 3rd row
Add -3/2 times the 4th row to the 2nd row
Add -2 times the 4th row to the 1st row
Add -1/2 times the 3rd row to the 2nd row
Add -2 times the 3rd row to the 1st row
Add -2 times the 2nd row to the 1st row
Then the maytrix A changes t0
1
0
0
0
2
0
1
0
0
-3
0
0
0
0
1
0
0
0
1
1
Then, g1 =1,g2=1,g3= 1= and g4 =1.
Also, x1 = 2, x2 =-3, x3 = 1 and x4=1
1
2
2
2
0
2
4
3
0
-5
0
2
1
3
-2
0
1
3
-2
-2
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