The document describes putting a system of linear equations into triangular form. It contains a 3x3 system of linear equations. The summary explains that through Gaussian elimination by eliminating variables from equations, the system can be transformed into an upper triangular matrix with equations arranged from top to bottom to easily solve for the variables through back substitution.

1. 9.3 Systems of Linear Equations
in Several Variables
1 Peter 3:9 "Do not repay evil for evil or reviling for reviling,
but on the contrary, bless, for to this you were called, that
you may obtain a blessing."

2. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩

3. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩

4. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩
a) it is linear ... degree of each equation is 1

5. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩
a) it is linear ... degree of each equation is 1
b) graphically ... 3 lines in 3-space; x-y-z axes

6. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩
a) it is linear ... degree of each equation is 1
b) graphically ... 3 lines in 3-space; x-y-z axes
c) our calculators can’t graph these, so a graphic
solution is not available to us

7. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩
a) it is linear ... degree of each equation is 1
b) graphically ... 3 lines in 3-space; x-y-z axes
c) our calculators can’t graph these, so a graphic
solution is not available to us
d) since it is linear, a matrix solution works

8. Consider the system: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
Comments: ⎪2x − y − 2z = −7
⎩
a) it is linear ... degree of each equation is 1
b) graphically ... 3 lines in 3-space; x-y-z axes
c) our calculators can’t graph these, so a graphic
solution is not available to us
d) since it is linear, a matrix solution works
e) we are going to learn to do this by hand ...
using Gaussian Elimination

9. Our System: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩

10. Our System: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
looks like this in Triangular Form:

11. Our System: ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
looks like this in Triangular Form:
⎧ x − y + 2z = 2
⎪
⎪ z 1
⎨ y− =
⎪ 4 2
⎪
⎩ z=2
(continued on next slide)

12. ⎧ x − y + 2z = 2
⎪ It is now easy to solve using
⎪ z 1
⎨ y− = back substitution
⎪ 4 2
⎪
⎩ z=2

13. ⎧ x − y + 2z = 2
⎪ It is now easy to solve using
⎪ z 1
⎨ y− = back substitution
⎪ 4 2
⎪
⎩ z=2
z=2

14. ⎧ x − y + 2z = 2
⎪ It is now easy to solve using
⎪ z 1
⎨ y− = back substitution
⎪ 4 2
⎪
⎩ z=2
2 1
z=2 y− =
4 2
y =1

15. ⎧ x − y + 2z = 2
⎪ It is now easy to solve using
⎪ z 1
⎨ y− = back substitution
⎪ 4 2
⎪
⎩ z=2
2 1
z=2 y− = x − 1+ 2 ( 2 ) = 2
4 2
y =1 x = −1

16. ⎧ x − y + 2z = 2
⎪ It is now easy to solve using
⎪ z 1
⎨ y− = back substitution
⎪ 4 2
⎪
⎩ z=2
2 1
z=2 y− = x − 1+ 2 ( 2 ) = 2
4 2
y =1 x = −1
( −1, 1, 2 )

17. So ... how do we put the original system
into Triangular Form?

18. So ... how do we put the original system
into Triangular Form?
1. Use equation 1 and eliminate x from
equation 2 and equation 3

19. So ... how do we put the original system
into Triangular Form?
1. Use equation 1 and eliminate x from
equation 2 and equation 3
2. Use the new equation 2 and equation 3 ...
and eliminate y from equation 3

20. So ... how do we put the original system
into Triangular Form?
1. Use equation 1 and eliminate x from
equation 2 and equation 3
2. Use the new equation 2 and equation 3 ...
and eliminate y from equation 3
3. Get the leading coefficients on all three
equations to be 1

21. ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
⎧ x − y + 2z = 2
⎪
⎨
⎪
⎩

22. ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
⎧ x − y + 2z = 2
⎪
⎨
⎪
⎩

23. ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪
⎩

24. ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪
⎩

25. ⎧ x − y + 2z = 2
⎪
⎨ 3x + y + 5z = 8
⎪2x − y − 2z = −7
⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪ y − 6z = −11
⎩

26. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪ y − 6z = −11
⎩

27. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪ y − 6z = −11
⎩
−4 ⋅ Eq 3 + Eq 2 → Eq 3

28. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪ 23z = 46
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪ y − 6z = −11
⎩
−4 ⋅ Eq 3 + Eq 2 → Eq 3

29. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪ 23z = 46
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎪
⎨ 4y − z = 2
⎪ y − 6z = −11
⎩
−4 ⋅ Eq 3 + Eq 2 → Eq 3

30. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪ 23z = 46
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎧ x − y + 2z = 2 ⎪
⎪ z 1
⎪ ⎨ y − =
⎨ 4y − z = 2 4 2
⎪ y − 6z = −11 ⎪
⎩ ⎪
⎩ z=2
−4 ⋅ Eq 3 + Eq 2 → Eq 3

31. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪ 23z = 46
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎧ x − y + 2z = 2 ⎪
⎪ z 1
⎪ ⎨ y − =
⎨ 4y − z = 2 4 2
⎪ y − 6z = −11 ⎪
⎩ ⎪
⎩ z=2
−4 ⋅ Eq 3 + Eq 2 → Eq 3 finish with back substitution

32. ⎧ x − y + 2z = 2 ⎧ x − y + 2z = 2
⎪ ⎪
⎨ 3x + y + 5z = 8 ⎨ 4y − z = 2
⎪2x − y − 2z = −7 ⎪ 23z = 46
⎩ ⎩
−3⋅ Eq 1+ Eq 2 → Eq 2
leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − y + 2z = 2
⎧ x − y + 2z = 2 ⎪
⎪ z 1
⎪ ⎨ y − =
⎨ 4y − z = 2 4 2
⎪ y − 6z = −11 ⎪
⎩ ⎪
⎩ z=2
−4 ⋅ Eq 3 + Eq 2 → Eq 3 finish with back substitution
( −1, 1, 2 ) and verify

33. Verify by hand:

34. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )

35. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )

36. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )
Verify by matrix:

37. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )
Verify by matrix:
⎡ 1 −1 2 ⎤ ⎡ 2 ⎤
⎢ ⎥ ⎢ ⎥
A = ⎢ 3 1 5 ⎥ B = ⎢ 8 ⎥
⎢ 2 −1 −2 ⎥
⎣ ⎦ ⎢ −7 ⎥
⎣ ⎦

38. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )
Verify by matrix:
⎡ 1 −1 2 ⎤ ⎡ 2 ⎤
⎢ ⎥ ⎢ ⎥
A = ⎢ 3 1 5 ⎥ B = ⎢ 8 ⎥
⎢ 2 −1 −2 ⎥
⎣ ⎦ ⎢ −7 ⎥
⎣ ⎦
⎡ −1 ⎤
−1 ⎢ ⎥
A B = ⎢ 1 ⎥
⎢ 2 ⎥
⎣ ⎦

39. Verify by hand:
⎧−1− 1+ 2 ( 2 ) = 2
⎪
⎨ 3( −1) + 1+ 5 ( 2 ) = 8
⎪2 −1 − 1− 2 2 = −7
⎩ ( ) ( )
Verify by matrix:
⎡ 1 −1 2 ⎤ ⎡ 2 ⎤
⎢ ⎥ ⎢ ⎥
A = ⎢ 3 1 5 ⎥ B = ⎢ 8 ⎥
⎢ 2 −1 −2 ⎥
⎣ ⎦ ⎢ −7 ⎥
⎣ ⎦
⎡ −1 ⎤
−1 ⎢ ⎥
A B = ⎢ 1 ⎥
⎢ 2 ⎥
⎣ ⎦

40. Solve this linear system using Gaussian Elimination.
Verify with matrices.

41. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1
⎪
⎨ x + 2y − z = 13
⎪2x + 4y − 7z = 11
⎩

42. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1
⎪
⎨ x + 2y − z = 13
⎪2x + 4y − 7z = 11
⎩
−1⋅ Eq 1+ Eq 2 → Eq 2

43. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1
⎪
⎨ x + 2y − z = 13
⎪2x + 4y − 7z = 11
⎩
−1⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3

44. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1
⎪
⎨ x + 2y − z = 13
⎪2x + 4y − 7z = 11
⎩
−1⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎪
⎨ 4y − 4z = 12
⎪ 8y − 13z = 9
⎩

45. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1
⎪
⎨ x + 2y − z = 13
⎪2x + 4y − 7z = 11
⎩
−1⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎪
⎨ 4y − 4z = 12
⎪ 8y − 13z = 9
⎩
−2 ⋅ Eq 2 + Eq 3 → Eq 3

46. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1 ⎧ x − 2y + 3z = 1
⎪ ⎪
⎨ x + 2y − z = 13 ⎨ 4y − 4z = 12
⎪2x + 4y − 7z = 11 ⎪ − 5z = −15
⎩ ⎩
−1⋅ Eq 1+ Eq 2 → Eq 2
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎪
⎨ 4y − 4z = 12
⎪ 8y − 13z = 9
⎩
−2 ⋅ Eq 2 + Eq 3 → Eq 3

47. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1 ⎧ x − 2y + 3z = 1
⎪ ⎪
⎨ x + 2y − z = 13 ⎨ 4y − 4z = 12
⎪2x + 4y − 7z = 11 ⎪ − 5z = −15
⎩ ⎩
−1⋅ Eq 1+ Eq 2 → Eq 2 leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎪
⎨ 4y − 4z = 12
⎪ 8y − 13z = 9
⎩
−2 ⋅ Eq 2 + Eq 3 → Eq 3

48. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1 ⎧ x − 2y + 3z = 1
⎪ ⎪
⎨ x + 2y − z = 13 ⎨ 4y − 4z = 12
⎪2x + 4y − 7z = 11 ⎪ − 5z = −15
⎩ ⎩
−1⋅ Eq 1+ Eq 2 → Eq 2 leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎧ x − 2y + 3z = 1 ⎪
⎪ ⎨ y−z = 3
⎨ 4y − 4z = 12 ⎪
⎪ 8y − 13z = 9 ⎩ z=3
⎩
−2 ⋅ Eq 2 + Eq 3 → Eq 3

49. Solve this linear system using Gaussian Elimination.
Verify with matrices.
⎧ x − 2y + 3z = 1 ⎧ x − 2y + 3z = 1
⎪ ⎪
⎨ x + 2y − z = 13 ⎨ 4y − 4z = 12
⎪2x + 4y − 7z = 11 ⎪ − 5z = −15
⎩ ⎩
−1⋅ Eq 1+ Eq 2 → Eq 2 leading coefficients to 1
−2 ⋅ Eq 1+ Eq 3 → Eq 3
⎧ x − 2y + 3z = 1
⎧ x − 2y + 3z = 1 ⎪
⎪ ⎨ y−z = 3
⎨ 4y − 4z = 12 ⎪
⎪ 8y − 13z = 9 ⎩ z=3
⎩
−2 ⋅ Eq 2 + Eq 3 → Eq 3
( 4, 6, 3) and verify

50. HW #5
Opportunities multiply as they are seized.
Sun Tzu