Consider the linear system
$$A \vec{x}=\vec{b},$$
where $A$ is an invertible matrix. We can solve this system in two different ways:
? By finding the reduced row-echelon form of the augmented matrix $[A | \vec{b}]$
?By computing $A^{-1}$ and using the formula $\vec{x}=A^{-1} \vec{b}$. In general, which approach requires fewer multiplicative operations? See Exercise 45.