APPLICATION PROBLRMS SOLVE BY GAUSS ELIMINATION OR JRDAN ELIMINATION METHOD

1. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS

2. 1 −1 1
7 4 0
0 4 2
0
15
20
R2+(-7R1)
R2/11,
R3/2

3. 1 −1 1
0 1 −7/11
0 2 1
0
15/11
10 R3-2R2
1 −1 1
0 1 −7/11
0 0 25/11
0
15/11
80/11 R3X(11/25)

4. 1 −1 1
0 1 −7/11
0 0 1
0
15/11
80/25
(15/11) +(7/11)(80/25)=3.4
80/25 =3.2

5. 1 −1 0
0 1 0
0 0 1
−3.2
3.4
3.2
R1+R2
1 0 0
0 1 0
0 0 1
0.2
3.4
3.2 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS

6. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS

7. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS

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10. 6/19/2020BY SANAULLAH MEMON

11. 0 0 0 1 −1 35
1 1 0 0 0 85
1 0 1 0 0 75
0 1 −1 1 0 70
0 0 0 0 1 25
1 0 1 0 0 75
1 1 0 0 0 85
0 0 0 1 −1 35
0 1 −1 1 0 70
0 0 0 0 1 25
R13
R2-R1

12. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 1 −1 35
0 0 0 1 0 60
0 0 0 0 1 25
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 1 −1 35
0 1 −1 1 0 70
0 0 0 0 1 25
R4-R2
ECHELON FORM
R4-R5

13. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 1 0 60
0 0 0 1 0 60
0 0 0 0 1 25
R3-R4
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 0 0 0
0 0 0 1 0 60
0 0 0 0 1 25
R34

14. 6/24/2020BY SANAULLAH MEMON LECTURER MUET SZAB CAMPUS
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 1 0 60
0 0 0 0 0 0
0 0 0 0 1 25
R45
1 0 1 0 0 75
0 1 −1 0 0 10
0 0 0 1 0 60
0 0 0 0 1 25
0 0 0 0 0 0
REDUCED ECHELON
FORM

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24. 6/19/2020BY SANAULLAH MEMON
Find the values of k such that the system of equations
02zy2x0,kz3y4x0,3zkyx =++=++=++
have a non – trivial solution.

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Fornon–trivialsolution2k2
–9k=0➔k(2k–9)=0➔k=0ork=9/2

28. 6/19/2020BY SANAULLAH MEMON
For what value of λ the following system of linear equations
( )
( )
( ) 0λ222x12x
032x2xλ514x
032x24x1xλ5
=−++
=+−+
=++−

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32. 6/19/2020BY SANAULLAH MEMON
For the non-trivial solution, rank of A < 3 (the number of unknowns), so we must have
==
−
100
2
10
This implies that

33. 6/19/2020BY SANAULLAH MEMON
,3x22xor03x22xand3x21xor03x21xor03x
2
10
31x ==−==−=−+ 





where 3x is arbitrary. If 01 =− or 1= , then above system becomes:










000
000
2111
,x
2
1
xx0x
2
1
xx 321321 





+−==++ where 2x and 3x are both arbitrary.