Suppose I have a square matrix $A$ (e.g., $n\times n$) that multiplies some vector $y$ ($1\times n$) into a vector of the same arrangement $z$ ($1\times n$), such that $Ay = z$, where both $y$ and $z$ are known but not $A$. How many matrices can solve the equation, and how can I calculate that number?
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If $\bf y$ is the zero vector and $\bf z$ is not, there are no solutions for $A$. Otherwise, there are infinitely many solutions. (Assuming we are dealing with real or complex matrices and vectors say). – Minus One-Twelfth Mar 19 '19 at 23:33
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Welcome to MSE! Please see https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for information on how to format expressions and equations on this site like I did in my edit to your question. It's not that hard to pick up! – Robert Howard Mar 19 '19 at 23:44