The document discusses using matrices to represent systems of linear equations. It introduces the concept of an augmented matrix, which writes the coefficients of the variables and constants of a linear system as a matrix. The document also covers row operations that can be performed on matrices, such as adding a multiple of one row to another. It defines what makes a matrix be in row-echelon form and provides an example. Finally, it works through using Gaussian elimination with an augmented matrix to solve a sample system of three linear equations.

1. 9.4 Systems of Linear Equations
Matrices
Matthew 6:14 "For if you forgive others their trespasses, your
heavenly Father will also forgive you,"

2. Let’s take a closer look at using a matrix ...

3. Let’s take a closer look at using a matrix ...
⎡ a11 a12 a13 a14 ⎤
⎢ ⎥
A = ⎢ a21 a22 a23 a24 ⎥
⎢ a a32 a33 a34 ⎥
31
⎣ ⎦

4. Let’s take a closer look at using a matrix ...
⎡ a11 a12 a13 a14 ⎤
⎢ ⎥
A = ⎢ a21 a22 a23 a24 ⎥
⎢ a a32 a33 a34 ⎥
31
⎣ ⎦
So, b25 is the value of the element in the B matrix
occupying the 2nd row & 5th column position.

5. Here is a slightly different way of representing a
linear system with a matrix ... called putting it into
Augmented Form

6. Here is a slightly different way of representing a
linear system with a matrix ... called putting it into
Augmented Form
⎧ 7x − 2y − z = 4
⎪
Given this system: ⎨ x + 3z = 6
⎪ 4y + z = 7
⎩

7. Here is a slightly different way of representing a
linear system with a matrix ... called putting it into
Augmented Form
⎧ 7x − 2y − z = 4
⎪
Given this system: ⎨ x + 3z = 6
⎪ 4y + z = 7
⎩
⎡ 7 −2 −1 4 ⎤
written in Augmented Form: ⎢ ⎥
⎢ 1 0 3 6 ⎥
⎢ 0 4 1 7 ⎥
⎣ ⎦

8. Here is a slightly different way of representing a
linear system with a matrix ... called putting it into
Augmented Form
⎧ 7x − 2y − z = 4
⎪
Given this system: ⎨ x + 3z = 6
⎪ 4y + z = 7
⎩
⎡ 7 −2 −1 4 ⎤
written in Augmented Form: ⎢ ⎥
⎢ 1 0 3 6 ⎥
⎢ 0 4 1 7 ⎥
⎣ ⎦
it simply includes the constants as well ...

9. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows

10. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:

11. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1

12. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1
The result of adding Row 1 to 3 times Row 3 is placed
into Row 1

13. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1
The result of adding Row 1 to 3 times Row 3 is placed
into Row 1
−2R3 → R3

14. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1
The result of adding Row 1 to 3 times Row 3 is placed
into Row 1
−2R3 → R3
Replace Row 3 with -2 times Row 3

15. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1
The result of adding Row 1 to 3 times Row 3 is placed
into Row 1
−2R3 → R3
Replace Row 3 with -2 times Row 3
R2 ↔ R3

16. Matrix Row Operations: We can ...
1) Add a multiple of one row to another
2) Multiply a row by a nonzero constant
3) Interchange two rows
Matrix Row Operation Notation:
R1 + 3R3 → R1
The result of adding Row 1 to 3 times Row 3 is placed
into Row 1
−2R3 → R3
Replace Row 3 with -2 times Row 3
R2 ↔ R3
Interchange rows 2 and 3

17. Use Matrix Row Operation Notation when transforming
a matrix in order to document what you are doing.

18. A matrix is in Row-Echelon Form if:

19. A matrix is in Row-Echelon Form if:
1) The leading entry in each row is 1

20. A matrix is in Row-Echelon Form if:
1) The leading entry in each row is 1
2) The leading entry (of 1) in each row is to the right
of the leading entry in the row above it

21. A matrix is in Row-Echelon Form if:
1) The leading entry in each row is 1
2) The leading entry (of 1) in each row is to the right
of the leading entry in the row above it
3) All rows consisting only of zeros is at the bottom

22. A matrix is in Row-Echelon Form if:
1) The leading entry in each row is 1
2) The leading entry (of 1) in each row is to the right
of the leading entry in the row above it
3) All rows consisting only of zeros is at the bottom
⎡ 1 2 −4 6 ⎤
Example: ⎢ ⎥
⎢ 0 1 3 −7 ⎥
⎢ 0 0 1
⎣ 2 ⎥
⎦

23. A matrix is in Row-Echelon Form if:
1) The leading entry in each row is 1
2) The leading entry (of 1) in each row is to the right
of the leading entry in the row above it
3) All rows consisting only of zeros is at the bottom
⎡ 1 2 −4 6 ⎤
Example: ⎢ ⎥
⎢ 0 1 3 −7 ⎥
⎢ 0 0 1
⎣ 2 ⎥
⎦
Notice this is just like Triangular Form ...
in essence, z=2 and we could use back substitution
to find y and x

24. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3
⎪
⎨ x + 2y − 6z = 7
⎪ 4x − y + 8z = 15
⎩

25. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦

26. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

27. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

28. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
⎢
⎣ ⎥
⎦

29. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
−4R1 + R3 → R3 ⎢
⎣ ⎥
⎦

30. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
−4R1 + R3 → R3 ⎢ 0 3 −12 3 ⎥
⎣ ⎦

31. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
−4R1 + R3 → R3 ⎢ 0 3 −12 3 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ ⎥
⎢
⎣ ⎥
⎦

32. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
−4R1 + R3 → R3 ⎢ 0 3 −12 3 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
R2 ↔ R3 ⎢ ⎥
⎢
⎣ ⎥
⎦

33. Let’s use an Augmented Matrix and Gaussian
Elimination to solve this system:
⎧ x − y + 5z = 3 ⎡ 1 −1 5 3 ⎤
⎪ ⎢ ⎥
⎨ x + 2y − 6z = 7 ⎢ 1 2 −6 7 ⎥
⎪ 4x − y + 8z = 15 ⎢ 4 −1 8 15 ⎥
⎩ ⎣ ⎦
⎡ 1 −1 5 3 ⎤
R1 − R2 → R2 ⎢ ⎥
⎢ 0 −3 11 −4 ⎥
−4R1 + R3 → R3 ⎢ 0 3 −12 3 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
R2 ↔ R3 ⎢ 0 3 −12 3 ⎥
⎢ 0 −3 11 −4 ⎥
⎣ ⎦

34. ⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ ⎥
⎢ 0 −3 11 −4 ⎥
⎣ ⎦

35. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦

36. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦

37. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢
⎣ ⎥
⎦

38. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
3R2 + R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢
⎣ ⎥
⎦

39. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
3R2 + R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 −1 −1 ⎥
⎦

40. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
3R2 + R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 −1 −1 ⎥
⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢
⎣ ⎥
⎦

41. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
3R2 + R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 −1 −1 ⎥
⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
−1R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢
⎣ ⎥
⎦

42. ⎡ 1 −1 5 3 ⎤
R2 ⎢ ⎥
→ R2 ⎢ 0 1 −4 1 ⎥
3 ⎢ 0 −3 11 −4 ⎥
⎣ ⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
3R2 + R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 −1 −1 ⎥
⎦
⎡ 1 −1 5 3 ⎤
⎢ ⎥
−1R3 → R3 ⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 1 1 ⎥
⎦

43. ⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 1 1 ⎥
⎦
This is now in Row-Echelon Form (REF)

44. ⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 1 1 ⎥
⎦
This is now in Row-Echelon Form (REF)
z =1 y − 4 (1) = 1 x − 1(5) + 5(1) = 3
y=5 x=3

45. ⎡ 1 −1 5 3 ⎤
⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0
⎣ 0 1 1 ⎥
⎦
This is now in Row-Echelon Form (REF)
z =1 y − 4 (1) = 1 x − 1(5) + 5(1) = 3
y=5 x=3
( 3, 5, 1)

46. A matrix is in Reduced Row-Echelon Form (RREF) if it
is in REF and every number above and below each
leading term is a zero.

47. A matrix is in Reduced Row-Echelon Form (RREF) if it
is in REF and every number above and below each
leading term is a zero.
⎡ 1 0 0 3 ⎤
RREF: ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

48. A matrix is in Reduced Row-Echelon Form (RREF) if it
is in REF and every number above and below each
leading term is a zero.
⎡ 1 0 0 3 ⎤
RREF: ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
Notice that this system is solved

49. A matrix is in Reduced Row-Echelon Form (RREF) if it
is in REF and every number above and below each
leading term is a zero.
⎡ 1 0 0 3 ⎤
RREF: ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
Notice that this system is solved
x=3 y=5 z =1

50. A matrix is in Reduced Row-Echelon Form (RREF) if it
is in REF and every number above and below each
leading term is a zero.
⎡ 1 0 0 3 ⎤
RREF: ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
Notice that this system is solved
x=3 y=5 z =1
So ... rather than back substitute when we get the
matrix to REF, we could continue to use row operations
in order to get it to RREF.
(start at the bottom and work up)

51. Let’s take the REF of that last problem and
put it into RREF

52. Let’s take the REF of that last problem and
put it into RREF
⎡ 1 −1 5 3 ⎤
REF: ⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

53. Let’s take the REF of that last problem and
put it into RREF
⎡ 1 −1 5 3 ⎤
REF: ⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
We need to get these 3 elements to be 0

54. Let’s take the REF of that last problem and
put it into RREF
⎡ 1 −1 5 3 ⎤
REF: ⎢ ⎥
⎢ 0 1 −4 1 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
We need to get these 3 elements to be 0
⎡ 1 −1 5 3 ⎤
4R3 + R2 → R2 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

55. ⎡ 1 −1 5 3 ⎤
(from previous slide) ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

56. ⎡ 1 −1 5 3 ⎤
(from previous slide) ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
⎡ 1 0 5 8 ⎤
R2 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

57. ⎡ 1 −1 5 3 ⎤
(from previous slide) ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
⎡ 1 0 5 8 ⎤
R2 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
⎡ 1 0 0 3 ⎤
−5R3 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦

58. ⎡ 1 −1 5 3 ⎤
(from previous slide) ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
⎡ 1 0 5 8 ⎤
R2 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
⎡ 1 0 0 3 ⎤
−5R3 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 5 ⎥
⎢ 0 0 1 1 ⎥
⎣ ⎦
( 3, 5, 1)

59. Solving a system of linear equations by putting an
augmented matrix into reduced row-echelon form is called
Gauss - Jordan Elimination

60. Solving a system of linear equations by putting an
augmented matrix into reduced row-echelon form is called
Gauss - Jordan Elimination
⎧6x + 18y − 6z = 24
⎪
Let’s do it with: ⎨5x + 15y + 3z = −20
⎪−3x + y + 33z = −42
⎩

61. Solving a system of linear equations by putting an
augmented matrix into reduced row-echelon form is called
Gauss - Jordan Elimination
⎧6x + 18y − 6z = 24
⎪
Let’s do it with: ⎨5x + 15y + 3z = −20
⎪−3x + y + 33z = −42
⎩
⎡ 6 18 −6 24 ⎤
⎢ ⎥
⎢ 5 15 3 −20 ⎥
⎢ −3 1 33 −42 ⎥
⎣ ⎦

62. ⎡ 1 3 −1 4 ⎤
R1 ⎢ ⎥
→ R1 ⎢ 5 15 3 −20 ⎥
6
⎢ −3 1 33 −42 ⎥
⎣ ⎦

63. ⎡ 1 3 −1 4 ⎤
R1 ⎢ ⎥
→ R1 ⎢ 5 15 3 −20 ⎥
6
⎢ −3 1 33 −42 ⎥
⎣ ⎦
⎡ 1 3 −1 4 ⎤
−5R1 + R2 → R2 ⎢ ⎥
⎢ 0 0 8 −40 ⎥
3R1 + R3 → R3 ⎢ 0 10 30 −30 ⎥
⎣ ⎦

64. ⎡ 1 3 −1 4 ⎤
R1 ⎢ ⎥
→ R1 ⎢ 5 15 3 −20 ⎥
6
⎢ −3 1 33 −42 ⎥
⎣ ⎦
⎡ 1 3 −1 4 ⎤
−5R1 + R2 → R2 ⎢ ⎥
⎢ 0 0 8 −40 ⎥
3R1 + R3 → R3 ⎢ 0 10 30 −30 ⎥
⎣ ⎦
⎡ 1 3 −1 4 ⎤
R2 ↔ R3 ⎢ ⎥
⎢ 0 10 30 −30 ⎥
⎢ 0 0 8 −40 ⎥
⎣ ⎦

65. R2 ⎡ 1 3 −1 4 ⎤
→ R2
10 ⎢ ⎥
⎢ 0 1 3 −3 ⎥ REF
R3
→ R3 ⎢ 0 0 1 −5 ⎥
⎣ ⎦
8

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

67. R2 ⎡ 1 3 −1 4 ⎤
→ R2
10 ⎢ ⎥
⎢ 0 1 3 −3 ⎥ REF
R3
→ R3 ⎢ 0 0 1 −5 ⎥
⎣ ⎦
8
⎡ 1 3 0 −1 ⎤
R3 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 12 ⎥
−3R3 + R2 → R2 ⎢ 0 0 1 −5 ⎥
⎣ ⎦
⎡ 1 0 0 −37 ⎤
−3R2 + R1 → R1 ⎢ ⎥
⎢ 0 1 0 12 ⎥ RREF
⎢ 0 0 1 −5 ⎥
⎣ ⎦
( −37, 12, − 5 )

68. HW #7
And in the end, it’s not the years in your life that count.
It’s the life in your years.
Abraham Lincoln