> [!NOTE] Proposition (Existence of inverse) > Given a square matrix $A\in \mathbb{R}^{n\times n}$ let $(R\mid P)$ denote $(A\mid I_{n})$ in [[Reduced Row Echelon Form for Real Matrix#^9f9c22|RREF]]. Then > 1. if $R=I_{n}$ then its [[Inverse of Real Square Matrix|inverse]] $A^{-1}$ exists with $A^{-1}=P$; > 2. if $R\neq I_{n}$ then $A$ is singular. *Proof*. Let $E_{1},\dots,E_{n}$ be a sequence of elementary matrices that reduce $A$ to $R$. So $(A\mid I_{n })$ becomes $(E_{k}E_{k-1}\dots E_{1}A\mid E_{k}E_{k-1}\dots E_{1})=(R\mid P)$Hence $P=E_{k}E_{k-1}\dots E_{1}$ and $R=PA$. If $R=I_{n}$ then $(E_{k}E_{k-1}\dots E_{1})A=I_{n}\implies A^{-1} = E_{k}E_{k-1}\dots E_{1} = P.$ If $R\neq I_{n}$ then, since $R$ is in RREF and is square, it must have at least one zero row. WLOG if row $1$ of $R$ is $\underline{0}$ then $(1,0,\dots ,0)(PA)=\underline{0}$Since $P$ is invertible ([[Elementary Row Operation is Equivalent to Pre-Multiplying by Elementary Matrix|elementary matrices are invertible]]), if $A$ were invertible then we could post-multiply by $A^{-1}P^{-1}$ to obtain $(1,0,\dots,0) = \underline{0}$ which is a contradiction. Hence $A$ is singular.