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How can an inverse matrix help solve a system of linear equations?

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modern algebra A simple algebraic curve, showing the graph of the algebraic equation y² = x³ + 1.

In modern algebra, a system of linear equations can be represented as a matrix equation of the form $A X = B$ , where $A$ is the matrix of coefficients, $X$ is the column vector of unknowns, and $B$ is the column vector of constants.12

If the matrix $A$ is a square, nonsingular matrix (meaning its determinant is not zero), it possesses an inverse matrix, denoted as $A^{- 1}$ , such that $A^{- 1} A = I$ , where $I$ is the identity matrix.13

To solve for the unknowns, you can multiply both sides of the equation by the inverse matrix from the left:
$A^{- 1} (A X) = A^{- 1} B$

Because matrix multiplication is associative, this simplifies to:
$(A^{- 1} A) X = A^{- 1} B$
$I X = A^{- 1} B$
$X = A^{- 1} B$

This method provides a direct algebraic solution for $X$ by analogy to solving a simple linear equation $a x = b$ by multiplying by the reciprocal $a^{- 1}$ .2