The document discusses solving a system of linear equations using Gauss-Jordan elimination. It begins with a 4x4 coefficient matrix representing a system with 4 unknowns and 3 equations, making it dependent. The method shows row operations that transform the matrix into reduced row echelon form, revealing the system is dependent. Therefore, the solution is expressed with parametric values for two of the unknowns.

1. 9.4 Systems of Linear Equations
Matrices
(Day Two)
Mark 11:25 "And whenever you stand praying, forgive, if you
have anything against anyone, so that your Father also who is
in heaven may forgive you your trespasses.”"

2. Inconsistent and Dependent systems work the same
way we’ve done previously.

3. Inconsistent and Dependent systems work the same
way we’ve done previously.
⎧ x + 2y − 4z − 5w = 14
Use Gauss - Jordan ⎪
to solve: ⎨ x + 3y − 4z − 5w = 21
⎪2x + 2y − 8z − 10w = 14
⎩

4. Inconsistent and Dependent systems work the same
way we’ve done previously.
⎧ x + 2y − 4z − 5w = 14
Use Gauss - Jordan ⎪
to solve: ⎨ x + 3y − 4z − 5w = 21
⎪2x + 2y − 8z − 10w = 14
⎩
Ok ... so this is weird ...
4 unknowns and 3 equations ... ????
Note the order of the variables: x, y, z, w

5. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦

6. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

7. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

8. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

9. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ ⎥
⎣ ⎦

10. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦

11. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

12. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
−2R2 + R3 → R3 ⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

13. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
−2R2 + R3 → R3 ⎢ ⎥
⎢ ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦

14. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
−2R2 + R3 → R3 ⎢ ⎥
⎢ ⎥
−1R2 → R2 ⎢ 0 0 0 0 0 ⎥
⎣ ⎦

15. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
⎢ 1 3 −4 −5 21 ⎥
⎢ 2 2 −8 −10 14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
R1 − R2 → R2 ⎢ 0 −1 0 0 −7 ⎥
⎢ ⎥
−2R1 + R3 → R3 ⎢ 0 −2 0 0 −14 ⎥
⎣ ⎦
⎡ 1 2 −4 −5 14 ⎤
−2R2 + R3 → R3 ⎢ ⎥
⎢ 0 1 0 0 7 ⎥
−1R2 → R2 ⎢ 0 0 0 0 0 ⎥
⎣ ⎦

16. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦

17. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!

18. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!
y=7 z=a w=b

19. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!
y=7 z=a w=b
x + 2 ( 7 ) − 4 ( a ) − 5 ( b ) = 14

20. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!
y=7 z=a w=b
x + 2 ( 7 ) − 4 ( a ) − 5 ( b ) = 14
x − 4a − 5b = 0

21. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!
y=7 z=a w=b
x + 2 ( 7 ) − 4 ( a ) − 5 ( b ) = 14
x − 4a − 5b = 0
x = 4a + 5b

22. ⎡ 1 2 −4 −5 14 ⎤
⎢ ⎥
From previous slide ⎢ 0 1 0 0 7 ⎥
⎢ 0
⎣ 0 0 0 0 ⎥ ⎦
All zeros means Dependent (0=0 which is true)
Use parametric values!
y=7 z=a w=b
x + 2 ( 7 ) − 4 ( a ) − 5 ( b ) = 14
x − 4a − 5b = 0
x = 4a + 5b
( 4a + 5b, 7, a, b )

23. Ok ... so you should now be really good at doing
these by hand!
Let’s explore more deeply what our fancy-schmancy
graphing calculators can do for us!!

24. ref ([ A ]) will take Matrix A and put it into
Row-Echelon Form (ref)

25. ref ([ A ]) will take Matrix A and put it into
Row-Echelon Form (ref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦

26. ref ([ A ]) will take Matrix A and put it into
Row-Echelon Form (ref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦
⎡ 1 2 −1 1 ⎤
⎢
ref ([ A ]) = 0 1 2 −3 ⎥
⎢ ⎥
⎢ 0 0 1 −2 ⎥
⎣ ⎦

27. ref ([ A ]) will take Matrix A and put it into
Row-Echelon Form (ref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦
⎡ 1 2 −1 1 ⎤
⎢
ref ([ A ]) = 0 1 2 −3 ⎥
⎢ ⎥
⎢ 0 0 1 −2 ⎥
⎣ ⎦
and we could use back-substitution to
finish solving the system ...

28. rref ([ A ]) will take Matrix A and put it into
Reduced Row-Echelon Form (rref)

29. rref ([ A ]) will take Matrix A and put it into
Reduced Row-Echelon Form (rref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦

30. rref ([ A ]) will take Matrix A and put it into
Reduced Row-Echelon Form (rref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦
⎡ 1 0 0 −3 ⎤
⎢
rref ([ A ]) = 0 1 0 1 ⎥
⎢ ⎥
⎢ 0 0 1 −2 ⎥
⎣ ⎦

31. rref ([ A ]) will take Matrix A and put it into
Reduced Row-Echelon Form (rref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦
⎡ 1 0 0 −3 ⎤
⎢
rref ([ A ]) = 0 1 0 1 ⎥
⎢ ⎥
⎢ 0 0 1 −2 ⎥
⎣ ⎦
and our system is solved!!

32. rref ([ A ]) will take Matrix A and put it into
Reduced Row-Echelon Form (rref)
⎡ 4 8 −4 4 ⎤
⎢ ⎥
let A = ⎢ 3 8 5 −11 ⎥
⎢ −2 1 12 −17 ⎥
⎣ ⎦
⎡ 1 0 0 −3 ⎤
⎢
rref ([ A ]) = 0 1 0 1 ⎥
⎢ ⎥
⎢ 0 0 1 −2 ⎥
⎣ ⎦
and our system is solved!!
(we still need to handle the dependent cases by hand ...)

33. Personally, RREF is my preferred method of solution
for a system of linear equations. How about you?

34. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 3z = 1
⎪
⎨ x − 3y − z = 0
⎪2x − 6z = 6
⎩

35. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 3z = 1
⎪
⎨ x − 3y − z = 0
⎪2x − 6z = 6
⎩
⎡ 1 −2 3 1 ⎤
⎢
rref ([ A ]) = 1 −3 −1 0 ⎥
⎢ ⎥
⎢ 2 0 −6 6 ⎥
⎣ ⎦
( 3, 1, 0 )
Sweet!!

36. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩

37. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣

38. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣

39. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
z=t

40. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
t 7
z=t y− =
2 2

41. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
t 7
z=t y− =
2 2
t+7
y=
2

42. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
t 7
z=t y− = x + 4t = 10
2 2
t+7
y=
2

43. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
t 7
z=t y− = x + 4t = 10
2 2
t+7 x = 10 − 4t
y=
2

44. Use rref ([ A ]) to solve this system:
⎧ x − 2y + 5z = 3
⎪
⎨−2x + 6y − 11z = 1
⎪ 3x − 16y + 20z = −26
⎩
⎡ 1 0 4 10 ⎤
⎢ ⎥
1 7 ⎥ Dependent!
rref ([ A ]) = ⎢ 0 1 −
⎢ 2 2 ⎥
⎢ 0 0 0 ⎥
0 ⎦
⎣
t 7
z=t y− = x + 4t = 10
2 2 ⎛ t + 7 ⎞
⎜ 10 − 4t,
⎝
, t ⎟
⎠
t+7 x = 10 − 4t 2
y=
2

45. HW #8
Whether you think you can or whether you think you
can’t, you’re right.
Henry Ford