Proof of Theorem MISM

As with the proof of Theorem SS, we see if $\frac{1}{\alpha}\inverse{A}$ is a suitable inverse for $\alpha A$. \begin{align*} \left(\frac{1}{\alpha}\inverse{A}\right)\left(\alpha A\right) &=\left(\frac{1}{\alpha}\alpha\right)\left(\inverse{A}A\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=1I_n&&\text{Scalar multiplicative inverses}\\ &=I_n&&\knowl{./knowls/property.OM.knowl}{\text{Property OM}} \end{align*} and \begin{align*} \left(\alpha A\right)\left(\frac{1}{\alpha}\inverse{A}\right)&= \left(\alpha\frac{1}{\alpha}\right)\left(A\inverse{A}\right)&&\knowl{./knowls/theorem.MMSMM.knowl}{\text{Theorem MMSMM}}\\ &=1I_n&&\text{Scalar multiplicative inverses}\\ &=I_n&&\knowl{./knowls/property.OM.knowl}{\text{Property OM}} \end{align*}

The matrix $\frac{1}{\alpha}\inverse{A}$ has met all the requirements to be the inverse of $\alpha A$, so we can write $\inverse{\left(\alpha A\right)}=\frac{1}{\alpha}\inverse{A}$.