Recipe

Solve Ax = b using a known inverse

Use when: you already have A^{-1} in hand for an invertible matrix A, and need to solve A\mathbf x=\mathbf b

Goal: the solution vector \mathbf x, as one matrix-vector multiply instead of a fresh elimination

Taught in Solving Ax = b with an inverse -- and why you usually should not

1. 1

   This recipe only pays off when A^{-1} is already sitting in front of you. If all you have is A itself and this is a one-off system, skip ahead to row-reducing the augmented matrix [\,A\mid\mathbf b\,] directly instead — computing A^{-1} just to throw it away after one solve costs roughly three times the work.

   Example

   A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix} with A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} already given — the inverse is the input to this recipe, not something to derive here.

   Why: multiplying by A^{-1} is only the cheap move when computing A^{-1} isn't the actual bottleneck of this particular solve.

2. 2

   Left-multiply both sides of A\mathbf x=\mathbf b by A^{-1}, writing the product with A^{-1} on the left and \mathbf b on the right.

   Example

   For \mathbf b = \begin{bmatrix} 4 \\ 7 \end{bmatrix}: \mathbf{x} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\begin{bmatrix} 4 \\ 7 \end{bmatrix}

   Why: A^{-1}A=I_n collapses the left side of A^{-1}(A\mathbf x) down to \mathbf x alone, which is exactly the rearrangement that makes the inverse useful here.

3. 3

   Each entry of \mathbf x is one row of A^{-1} dotted with \mathbf b — multiply matching entries and add.

   Example

   Top entry: 2(4)+(-1)(7) = 8-7 = 1. Bottom entry: (-5)(4)+3(7) = -20+21=1, so \mathbf x = \begin{bmatrix}1\\1\end{bmatrix}.

   Why: this is nothing more than the ordinary matrix-vector product definition, applied to A^{-1} and \mathbf b.

4. 4

   Plug the computed \mathbf x into the original equation and confirm both rows come out right — don't just trust the multiplication.

   Example

   Row 1: 3(1)+1(1)=4. Row 2: 5(1)+2(1)=7. Both match \mathbf b, so \mathbf x=\begin{bmatrix}1\\1\end{bmatrix} is confirmed.

   ⚡ Shortcut

   Solving for several different right-hand sides against the same A? Do step 1 exactly once — A^{-1} doesn't change — then repeat only steps 2-3 for each new \mathbf b. That's the entire payoff of having A^{-1} on hand: the expensive part happens once, and every additional solve afterward is one cheap multiply, never a fresh elimination.

   Why: this catches an arithmetic slip in the row-by-row multiplication before it's mistaken for a final answer.