Possible Solutions for Square Matrices: How Many Solutions Are There?

  • Thread starter Thread starter ivanyo
  • Start date Start date
  • Tags Tags
    Matrix Square
ivanyo
Messages
5
Reaction score
0
Hi everyone,

I've been reading through my textbook for some guidance on this, but I have yet to find anything. The question asks to "describe the set of possible solutions when the following occur (assuming Ax=b). I've written my own knowledge in italics - however, I think they require an answer in terms of "no solution"/"# of unique solutions" or "infinite solutions".

(i) detA = 0, b = 0 (no inverse)
(ii) detA ≠ 0, b = 0 (inverse possible)
(iii) detA = 0, b ≠ 0 (no inverse)
(iv) detA ≠ 0, b ≠ 0 (inverse possible)

The way I learned was that if there was 3 solutions, there had to be 3 variables/pieces of information, otherwise it was redundant or inconsistent depending on what you were given. I've yet to learn using determinants and the b-vector.

Any insight would be helpful - Thanks.
 
Physics news on Phys.org
ivanyo said:
Hi everyone,

I've been reading through my textbook for some guidance on this, but I have yet to find anything. The question asks to "describe the set of possible solutions when the following occur (assuming Ax=b). I've written my own knowledge in italics - however, I think they require an answer in terms of "no solution"/"# of unique solutions" or "infinite solutions".

(i) detA = 0, b = 0 (no inverse)
It's true that the matrix A doesn't have an inverse, but they're asking about the equation Ax = 0. Does this equation have a) no solutions, b) exactly 1 solutions, c) an infinite number of solutions?
ivanyo said:
(ii) detA ≠ 0, b = 0 (inverse possible)
If det(A) ≠ 0, then A definitely has an inverse, so how many solutions does the equation Ax = 0 have?
ivanyo said:
(iii) detA = 0, b ≠ 0 (no inverse)
Again, how many solutions does the equation Ax = b have?
ivanyo said:
(iv) detA ≠ 0, b ≠ 0 (inverse possible)
As before, if det(A) ≠ 0, then A definitely has an inverse, so how many solutions does the equation Ax = b have?
ivanyo said:
The way I learned was that if there was 3 solutions, there had to be 3 variables/pieces of information, otherwise it was redundant or inconsistent depending on what you were given. I've yet to learn using determinants and the b-vector.
If you have 3 equations and 3 unknowns, such a system might have no solutions, 1 solution, or an infinite number of solutions. It can never have exactly three solutions. As a really simple example, here is a very simple system of 3 equation in 3 variables:
x = 2
y = 3
z = 5
This can be represented as a matrix equation in the form Ax = b, where A is
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}

and b is
\begin{bmatrix} 2 \\ 3 \\ 5 \end{bmatrix}

This system has one solution, not three; namely x = 2, y = 3, z = 5.

ivanyo said:
Any insight would be helpful - Thanks.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top