APM2613 Assignment TWO of 2023

APM2613 ASSIGNMENT 2 2023

Question 1

4𝑥1 + 𝑥2 − 𝑥3 = 5

−𝑥1 + 3𝑥2 + 𝑥3 = −4

2𝑥1 + 2𝑥2 + 5𝑥3 = 1

(1.1)

4 1 −1 𝑥1 5
[−1 3 1 ] [𝑥2 ] = [−4]
2 2 5 𝑥3 1

Let
4 1 −1 𝑥1 5
𝐴 = [−1 3 1 ] , 𝒙 = [𝑥2 ] , 𝒃 = [−4]
2 2 5 𝑥3 1

Therefore,
𝐴𝒙 = 𝒃

(1.2) MATLAB direct method




The actual solution is:

1.447761194029851
𝒙 = [−0.835820895522388]
−0.044776119402985

, (1.3)

(a) Gaussian elimination without pivoting.

4 1 −1 𝑥1 5
[−1 3 1 ] [𝑥2 ] = [−4]
2 2 5 𝑥3 1

4 1 −1 5
[−1 3 1 | −4]
2 2 5 1
1 1
𝑅2 → 𝑅2 + 4 𝑅1 − 4 to be used in the 𝐿 matrix in (1.3) (c) as row 2 column 1

1 1
𝑅3 → 𝑅3 − 2 𝑅1 2
to be used in the 𝐿 matrix in (1.3) (c) as row 3 column 1

4 1 −1 5
1 1 1 1
−1 + (4) 3 + (1) 1 + (−1)| −4 + (5)
4 4 4 | 4
1 1 1 1
(4) 2 − (1) 5 − (−1) 1 − (5) ]
[ 2−2 2 2 2

4 1 −1 5
13 3 11
0 | −
4 4| 4
3 11 3
[0 2 2
− ]
2

3/2
𝑅3 → 𝑅3 − 𝑅
13/4 2

6 6
𝑅3 → 𝑅3 − 13 𝑅2 13
to be used in the 𝐿 matrix in (1.3) (c) as row 3 column 2

4 1 −1 5
13 3 11
0 | −
4 4 | 4
6 3 6 13 11 6 3 3 6 11
0 − (0) − ( ) − ( ) − − (− )
[ 13 2 13 4 2 13 4 2 13 4 ]

4 1 −1 5
13 3 11
0 | −
4 4| 4
67 3
[0 0
13
− ]
13

4 1 −1
13 3
Note that the matrix 𝑈 in (1.3) (c) will be [0 4 4 ]
67
0 0 13