# Homework Help: Matrices and determinant meaning

1. May 7, 2014

### rzn972

1. The problem statement, all variables and given/known data
For a system Ax= 0, suppose det (A)= .0001. Which of the following describes the solutions to system?
There is exactly one solution, but the system is close to having infinity many.
There is exactly one solution, but the system is close to having none.

2. Relevant equations

3. The attempt at a solution
Since the determinant is not zero, that means that there is one and only solution right? I don't get what it means when the determinant is very small.

2. May 7, 2014

### LCKurtz

Yes there is only one solution. But the determinant is close to 0. If it were actually zero which of the two options is correct?

3. May 7, 2014

### AlephZero

This is a very poor question IMO.

The numerical size of the determinant is the wrong criterion. If all the elements $A$ are approximately equal to $a$, the determinant will be about $a^n$ where $n$ is the size of the matrix. So if all the elements were about 0.1 and n = 10, you could have a perfectly well conditioned set of equations with determinant of about $10^{-10}$.

If all the numbers were about 100 and n = 10, you could have a very poorly conditioned set of equations with a determinant of $10^{10}$ - compared with the $10^{20}$ that you would expect from a well conditioned matrix.

The correct criterion is the condition number of the matrix, which compares the numerical properties of A and its inverse, and is independent of the size of the individual elements in $A$.
http://en.wikipedia.org/wiki/Condition_number

Determinants are a very useful concept in pure mathematics, but there are very few situations were finding the numerical value of a determinant has any value, especially for a large matrix.

4. May 8, 2014

### rzn972

If the determinant is zero, doesn't that mean there is no solution so that "There is exactly one solution, but the system is close to having none?".
The answer is supposed to be the other one, "There is exactly one solution, but the system is close to having infinity many".

5. May 8, 2014

### xiavatar

It should be "There is exactly one solution, but the system is close to having infinitely many". Now I'm going to assume that you already know that you can think of A as a set of column vectors. If $det(A)$ then this implies that the set of vectors in A is linearly dependent. Now what do you know about linearly dependent vectors and the number of linear combinations that combine to equal zero?

6. May 8, 2014

### pasmith

$\vec x = \vec 0$ is always a solution of $A\vec x = \vec 0$.

7. May 8, 2014

### rzn972

Since the det is not equal to zero, it means that the vectors are independent. If they are independent it means that there are infinite solutions?

8. May 8, 2014

### HallsofIvy

No, it means there is exactly one solution! You were told that in the original problem:
"There is exactly one solution, but the system is close to having infinity many.
There is exactly one solution, but the system is close to having none. "
Both of those say "there is exactly one solution". All you need to distinguish between is "the system is close to having infinity many" and "the system is close to having none".

And, as you were told. "$A\vec{x}= \vec{0}$" always has the solution $\vec{x}= \vec{0}$.

9. May 8, 2014

### rzn972

Okay, I think I get it. I am still a bit hazy on one part though . If the det=0, it implies that it is linearly independent meaning there are infinite solutions. I thought that if det=0 it could mean there are infinite or no solutions. When is there a case of no solutions?

10. May 8, 2014

### Ray Vickson

You can have no solution for $Ax = b$ with the wrong kind of $b$. For example, the system
$$x + y = 2\\ 2x + 2y = 3$$
has no solution.

This type of thing cannot happen if all right-hand-sides = 0.

Last edited: May 8, 2014
11. May 8, 2014

### xiavatar

No, when det(A)=0 it implies linear dependence.

12. May 8, 2014

### rzn972

Right, my mistake. Det(A)=0 implies linear dependence. Thanks everyone!