- #1

- 926

- 56

Surprisingly or not, I haven't found an answer to the above question.

- B
- Thread starter kent davidge
- Start date

- #1

- 926

- 56

Surprisingly or not, I haven't found an answer to the above question.

- #2

fresh_42

Mentor

- 13,457

- 10,515

https://en.wikipedia.org/wiki/System_of_linear_equations#Matrix_solution

https://en.wikipedia.org/wiki/Invertible_matrix#In_relation_to_its_adjugate

- #3

- #4

- 589

- 387

For the system [itex]Ax = b[/itex] with [itex]A\in\mathrm{Mat}_{m, n}(K)[/itex], there are [itex]n-\mathrm{rank}(A)[/itex] free variables. If [itex]K=\mathbb R[/itex] (or any other infinite field), then having at least one variable free means there are infinitely many solutions. Also, a solution can be unique only if the matrix [itex]A[/itex] is a square matrix (i.e only if number of variables = number of equations).

Of course, if the underlying field is finite, there can't be infinitely many solutions. That's something a lot of my students get wrong. They remember from school that if number of eqs is smaller than number of variables, then there are automatically infinitely many solutions. Not necessarily.

Of course, if the underlying field is finite, there can't be infinitely many solutions. That's something a lot of my students get wrong. They remember from school that if number of eqs is smaller than number of variables, then there are automatically infinitely many solutions. Not necessarily.

Last edited:

- #5

mathwonk

Science Advisor

Homework Helper

- 10,996

- 1,189

- #6

- 589

- 387

- Replies
- 3

- Views
- 5K

- Replies
- 2

- Views
- 977

- Last Post

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 5

- Views
- 8K

- Replies
- 5

- Views
- 925

- Replies
- 8

- Views
- 3K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 6

- Views
- 3K

- Replies
- 12

- Views
- 804

- Replies
- 4

- Views
- 3K