What Does ##ad-bc=0## Imply About Solutions to ##AX=0##?

  • Thread starter Thread starter Bashyboy
  • Start date Start date
Bashyboy
Messages
1,419
Reaction score
5

Homework Statement


Consider the system ##AX=0##, where
$$A = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$$
is a 2x2 matrix over the field F. Prove that if ##ad-bc=0## and some entry of ##A## is different from ##0##, then there is a solution ##(p,q)## such that ##(x,y)## is a solution if and only if there is some scalar ##t## such that ##x=pt## and ##y = qt##.

Homework Equations

The Attempt at a Solution


Am I asked to find the vector ##(p,q)## for which the statement "##(x,y)## is a solution if and only if there is some scalar ##t## such that ##x=pt## and ##y = qt##" holds, or am I assuming that I have such a solution ##(p,q)## and proving that this statement?
 
Physics news on Phys.org
Bashyboy said:
Am I asked to find the vector ##(p,q)## for which the statement "##(x,y)## is a solution if and only if there is some scalar ##t## such that ##x=pt## and ##y = qt##" holds, or am I assuming that I have such a solution ##(p,q)## and proving that this statement?

Well, the problem doesn't ask you to find (p,q), only to show that if (p,q) is a solution, and (x,y) is a second solution, then there is some t such that x=pt and y=qt.

For a 2x2 matrix, it's easy enough to convert \left( \begin{array}\\ a & b \\ c & d \end{array} \right) \left( \begin{array} \\ x \\ y \end{array} \right) = 0 into two ordinary (non-matrix) equations for x and y to see how the possible values of x and y must be related.
 
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