A system of equations can be written as one matrix equation, $AX = B$. If $A$ has an inverse, you can solve it the way you solve $3x = 12$, as long as you respect the fact that matrix multiplication depends on order. It is an alternative to row reduction.
Key idea
The system $ax + by = e$, $cx + dy = f$ is $$\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} e \\ f \end{bmatrix}, \text{ that is } AX = B.$$ If $A$ is invertible, multiply both sides on the left by $A^{-1}$: $X = A^{-1}B$.
Why it works
$A^{-1}(AX) = (A^{-1}A)X = IX = X$, and the right side becomes $A^{-1}B$. The inverse must go on the left of both sides, because that is where it meets $A$. Putting it on the right would give $AXA^{-1}$, which does not simplify. Once you have $A^{-1}$, a new right side $B$ costs only one multiplication, which is why computers use this idea for many systems with the same coefficients.
Worked examples
Worked example: A two-by-two system
Solve $3x + y = 7$, $5x + 2y = 12$ using an inverse matrix.
- $A = \begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 7 \\ 12 \end{bmatrix}$. $\det A = 1$, so $A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}$.
- $$X = A^{-1}B = \begin{bmatrix} 2 \cdot 7 - 1 \cdot 12 \\ -5 \cdot 7 + 3 \cdot 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}.$$
- Check: $3(2) + 1 = 7$ and $5(2) + 2(1) = 12$. ✓
Answer: $(x, y) = (2, 1)$
Worked example: A given inverse
The coefficient matrix of a system has $A^{-1} = \begin{bmatrix} 1 & 0 & -1 \\ -2 & 1 & 1 \\ 0 & -1 & 2 \end{bmatrix}$, and the constants are $B = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}$. Find $(x, y, z)$.
- $X = A^{-1}B$: row 1 gives $3 + 0 - 2 = 1$.
- Row 2: $-6 + 1 + 2 = -3$. Row 3: $0 - 1 + 4 = 3$.
Answer: $(1, -3, 3)$
Worked example: A new right side
Using the same $A$ as in the first example, solve $3x + y = 1$, $5x + 2y = 3$.
- $$X = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix}\begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}.$$
Answer: $(-1, 4)$
Common mistake
Multiplying by $A^{-1}$ on the right and computing $BA^{-1}$. For a column $B$ that product is not even defined, and for square $B$ it usually gives the wrong matrix. Write the step out: $A^{-1}AX = A^{-1}B$. The inverse sits next to $A$, on the left.