Matrix equalation system. Which statement is wrong and why?

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LilaSan

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Hello, I have two statements, one of them is wrong but I just CANT figure it out :

If an equalation system Ax = b has exactly one solution, then the matrix can only be regular
If a matrix is regular, then the equalation system Ax = b has always one solution

Which one of them can be wrong and can it be please explained to me? :) Best would be with a small, easy example so I would understand, because to be dearly honest..those questions seem just the same for me :( but actually there HAS to be something wrong.

Thanks for the help!
 
I don't think I've heard of a "regular matrix" before, can you please provide the definition that you've been given? One of the options on this Wikipedia page says that a regular matrix is one that is invertible. But can it be non-square (so long as a right or left inverse exists?), if so then this might provide an insight?

If A must be square:- then if x=[ p ; q^2 ] and A is the identity matrix and b=[1 ; 1] would you say that the equation system has one or two solutions? It seems to me that x=[1;1] which is a single solution, but this implies that ( p=1 and q=1 ) OR ( p=1 and q=-1 ) which is obviously two solutions. I guess it depends on the exact definition of "the number of solutions to an equation system".
 
Yes, thank you!
I mean the invertible Matrix, yes, the "opposite" of the singular Matrix.

In the first statements it says that if it has one solution, it can only be a square, invertible matrix.
The second statement says that if the matrix is invertible(what means it needs to be square, right?) it has to have ONE solution.

What I understood:
When It doesnt have one solution that means it should be a singular matrix, not the invertible one. So for me both were correct XD
 
LilaSan said:
Yes, thank you!
I mean the invertible Matrix, yes, the "opposite" of the singular Matrix.

In the first statements it says that if it has one solution, it can only be a square, invertible matrix.
The second statement says that if the matrix is invertible(what means it needs to be square, right?) it has to have ONE solution.

What I understood:
When It doesnt have one solution that means it should be a singular matrix, not the invertible one. So for me both were correct XD
Click to expand...
When It doesnt have one solution:

It can have infinite number of solutions

or

NO solution
 
The two statements are converse to one another.

The converse of if A then B is if B then A.

Both statements are NOT the same (but some converses are)!

Example: If you are a student then you are smart. If you are smart then you are a student.

Assume the first statement is true. The 2nd statement is not true as there are some (at least one) people who are smart that are not students (maybe they already finished their schooling or they are quite young and haven't started school yet).

Now it can be that both statements are the same. If you are an integer of the form 2r, then you are an even integer. If you are an even integer then you are of the form 2x. These two statements are the same.


Think about this. Suppose you have 4 equations with unknowns. Can this yield a unique solution? Is the matrix regular?
 
Are there not only "regular/inverse" and singular Matrix?
If the second statement would be wrong, I would need to "proof" that if I get an equalation with one solution it should be something else than a regular matrix.
Means a singular?
I was looking into it and it can't be singular then if it has one solution.
Is there another kind of Matrix that can actually have only ONE solution without being regular or singular?

What I found was this:
If the rank of the matrix is equal to the number of unknown n, there is ONE solution. For systems of equations with n equations, this is the case if they are linearly independent.

All this is true for a regular matrix tho
 
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