If $A$ is invertible, $Ax=b$ has the unique solution $x=A^{-1}b$. Multiplication verifies existence, and applying $A^{-1}$ proves uniqueness. For $A=\begin{bmatrix}2&1\\1&1\end{bmatrix}$ and $b=(3,2)^T$, the solution is $(1,1)^T$. In computation, use a factorization and triangular solves.
Parents
Like how this is organised?
Build your own knowledge tree on knowen.md: notes, decisions and tasks in one connected markdown graph, shared with your team and your AI agents.