Solving a System with the Matrix Inverse

Writing a linear system as AX = B, why multiplying by the inverse replaces division, and the four steps that lead to X = A inverse times B with a full check.

--

Any system of linear equations can be packed into the single matrix equation AX = B . You cannot divide by a matrix — but you can multiply by its inverse, and that turns solving the system into one multiplication.

Concept The three matrices
A — the coefficient matrix, the numbers in front of the unknowns.
X — the column of unknowns.
B — the column of constants on the right-hand side.

Together they form AX = B , whatever the size of the system.

Theorem Multiply by the inverse

There is no division for matrices. Instead, multiply both sides on the left by A^{-1} :

A^{-1}AX = A^{-1}B
Since A^{-1}A = I and IX = X :
X = A^{-1}B

Order matters: matrix multiplication is not commutative, so A^{-1} must go on the left of both sides.

Example Step 1 — write it as AX = B

Take the system:

x + y = 100
2x + 3y = 220
In matrix form:
\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 100 \\ 220 \end{pmatrix}
Example Step 2 — find A⁻¹
The determinant:
\det(A) = (1)(3) - (1)(2) = 3 - 2 = 1
Swap the main diagonal, flip the other signs:
\begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}
Divide by the determinant 1, which changes nothing:
A^{-1} = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}
Example Step 3 — compute X = A⁻¹B
3 −1 −2 1 100 220 80 20 = A⁻¹ × B = X
Each entry of the answer is a row of A^{-1} multiplied against the column B .
x = (3)(100) + (-1)(220) = 300 - 220 = 80
y = (-2)(100) + (1)(220) = -200 + 220 = 20
⟹ x = 80 and y = 20
Example Step 4 — check the solution

Substitute both values back into the original equations.

80 + 20 = 100 ✓
2(80) + 3(20) = 160 + 60 = 220 ✓
⟹ both equations hold, so the solution is correct
Note Points to watch
If \det(A) = 0 there is no inverse, so this method cannot be used.
Multiply by A^{-1} on the left of both sides, never on the right.
Keep the equations in the same variable order when building A .
The method extends to larger systems, not just 2×2.
Summary
  1. Every linear system can be written as AX = B.
  2. There is no matrix division; multiply by A⁻¹ on the left instead.
  3. This gives the solution rule X = A⁻¹B.
  4. The worked system gave x = 80 and y = 20.
  5. A⁻¹ exists only when det(A) ≠ 0; always substitute back to check.
231. Arithmetic Sequences and Series
← Back to Semester