Q: If a matrix of order $9$ has a column that is a multiple of another,
does the system $Ax = 0$ has infinite solutions?
A: Yes, because the matrix would not be invertible.
Can you explain this answer further?
I understand that the matrix needs to be invertible in order to have an unique solution, but how do you know that having a column that is a multiple of another causes the matrix to be singular? Does the same happen with rows?
Best Answer
Assume that the column $c_i$ of $A$ is equal to the columns $c_j$ times $\alpha$ so we have
$$Ae_i=\alpha A e_j$$ where $(e_1,\ldots,e_9)$ is the standard basis, hence $$A(\beta(e_i-\alpha e_j))=0,\quad \forall \beta\in\Bbb R $$