Infinit number of solutions (lenear akgebra)

[nhrock3]

question:
there are two systems of MXN non homogeneus equations:
they both are from Ax=b type
and their redused A is the same.

the first system c' has infinit number of solution
does the other one(c) has infinit number of solutions too .

?

i know that its wrong but in this course i have a list of laws regarding systems of equations
and if the answer was "yes" then i should have prove it.

but i don't know why by the laws its impossible?(so i need to look for contradicting example)

 

[HallsofIvy]

No, that doesn't follow. If A is invertible (det(A) not 0) then Ax= b has a unique solution for any b. If A is not invertible (det(A)= 0) then Ax= b either has an infinite number of solutions or no solution, depending on b.

For a simple counter example, look at
\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}2 \\ 2\end{bmatrix}
and
\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}2 \\ 1\end{bmatrix}

 

[nhrock3]

yes but invertibility holds of square martrices
i was talled MXN