When A is square and invertible, there's a one-line solution to A\vec{x} = \vec{b}. Multiply both sides on the left by A^{-1}: since A^{-1}A = I, the left side becomes plain \vec{x}, and we're left with

Apply the undo-matrix to \vec{b} and out drops the answer. It's the matrix version of "divide both sides by A."

One inverse, many right-hand sides

Here A^{-1} is fixed (it's the inverse of \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}). Slide the entries of \vec{b} and the solution \vec{x} = A^{-1}\vec{b} updates instantly. That's the quiet advantage of the inverse: compute it once, then solve for any \vec{b} with a single multiply.

Elegant on paper, careful in practice

\vec{x} = A^{-1}\vec{b} is the cleanest statement of the solution, and it proves at a glance that an invertible system has exactly one answer. In real computation, though, forming the full inverse is slower and less accurate than just running Gaussian elimination on that one \vec{b} — so the formula is loved for understanding, while elimination does the heavy lifting. The same closed form reappears as the normal equation that fits a regression line.