Solving Matrix Equations with Inverses

Two Equations, Two Sides

In a scalar equation such as \(ax=b\), multiplication can be rearranged without changing its order. Matrix multiplication does not generally commute: \(\mathbf{A}\mathbf{B}\) and \(\mathbf{B}\mathbf{A}\) may be different, and one product may not even be defined. So solving a matrix equation means paying attention not only to which matrix is unknown, but also to which side each factor occupies.

We will consider square invertible coefficient matrices \(\mathbf{A}\in\mathbb{R}^{n\times n}\). An inverse of \(\mathbf{A}\) is a matrix \(\mathbf{A}^{-1}\) satisfying \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{A}^{-1}\mathbf{A}=\mathbf{I}_n\). For \(\mathbf{A}\mathbf{X}=\mathbf{B}\), if \(\mathbf{B}\in\mathbb{R}^{n\times p}\), then \(\mathbf{X}\) must be in \(\mathbb{R}^{n\times p}\). For \(\mathbf{X}\mathbf{A}=\mathbf{B}\), if \(\mathbf{B}\in\mathbb{R}^{m\times n}\), then \(\mathbf{X}\) must be in \(\mathbb{R}^{m\times n}\). These dimensions make the products and the proposed solutions well-defined.

The side is determined by the original equation. In the first equation, \(\mathbf{A}\) stands immediately to the left of \(\mathbf{X}\), so its inverse must act on the left to cancel it. In the second, \(\mathbf{A}\) stands to the right of \(\mathbf{X}\), so its inverse must act on the right. The order of the other factors is not free to change.

Solving \(\mathbf{A}\mathbf{X}=\mathbf{B}\)

Proof. For existence, set \(\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}\), which has size \(n\times p\). Then \(\mathbf{A}\mathbf{X}=\mathbf{A}(\mathbf{A}^{-1}\mathbf{B})=(\mathbf{A}\mathbf{A}^{-1})\mathbf{B}=\mathbf{I}_n\mathbf{B}=\mathbf{B}\), using associativity and the identity law. For uniqueness, suppose \(\mathbf{Y}\) is any solution, so \(\mathbf{A}\mathbf{Y}=\mathbf{B}\). Multiplying this equality on the left by \(\mathbf{A}^{-1}\) gives \(\mathbf{A}^{-1}(\mathbf{A}\mathbf{Y})=\mathbf{A}^{-1}\mathbf{B}\), hence \((\mathbf{A}^{-1}\mathbf{A})\mathbf{Y}=\mathbf{A}^{-1}\mathbf{B}\) and \(\mathbf{Y}=\mathbf{A}^{-1}\mathbf{B}\). Thus every solution equals the one already found. \(\square\)

Solving \(\mathbf{X}\mathbf{A}=\mathbf{B}\)

Proof. For existence, set \(\mathbf{X}=\mathbf{B}\mathbf{A}^{-1}\), which has size \(m\times n\). Then \(\mathbf{X}\mathbf{A}=(\mathbf{B}\mathbf{A}^{-1})\mathbf{A}=\mathbf{B}(\mathbf{A}^{-1}\mathbf{A})=\mathbf{B}\mathbf{I}_n=\mathbf{B}\). For uniqueness, suppose \(\mathbf{Y}\mathbf{A}=\mathbf{B}\). Multiply on the right by \(\mathbf{A}^{-1}\): \((\mathbf{Y}\mathbf{A})\mathbf{A}^{-1}=\mathbf{B}\mathbf{A}^{-1}\). Associativity gives \(\mathbf{Y}(\mathbf{A}\mathbf{A}^{-1})=\mathbf{B}\mathbf{A}^{-1}\), so \(\mathbf{Y}\mathbf{I}_n=\mathbf{B}\mathbf{A}^{-1}\) and therefore \(\mathbf{Y}=\mathbf{B}\mathbf{A}^{-1}\). Every solution must be this matrix. \(\square\)

Why the Two Answers Can Differ

When \(\mathbf{A}\) and \(\mathbf{B}\) are square, both forms may be written with the same matrices, but they still ask different questions. The equation \(\mathbf{A}\mathbf{X}=\mathbf{B}\) has answer \(\mathbf{A}^{-1}\mathbf{B}\), whereas \(\mathbf{X}\mathbf{A}=\mathbf{B}\) has answer \(\mathbf{B}\mathbf{A}^{-1}\). These expressions need not be equal. They would be equal only in cases where \(\mathbf{A}^{-1}\mathbf{B}=\mathbf{B}\mathbf{A}^{-1}\); no such equality is guaranteed for arbitrary matrices.

Common Mistake: Multiplying on the Wrong Side

A frequent error is to remember “multiply by the inverse” but ignore which side it must be on. In the first equation of the example above, a wrong step would be to write \(\mathbf{X}=\mathbf{B}\mathbf{A}^{-1}\). That expression actually solves \(\mathbf{X}\mathbf{A}=\mathbf{B}\), not \(\mathbf{A}\mathbf{X}=\mathbf{B}\). Substituting it into the left-multiplication equation exposes the error:

The correction follows the original order: from \(\mathbf{A}\mathbf{X}=\mathbf{B}\), multiply both sides on the left by \(\mathbf{A}^{-1}\), obtaining \(\mathbf{A}^{-1}\mathbf{A}\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}\), and hence \(\mathbf{X}=\mathbf{A}^{-1}\mathbf{B}\). For \(\mathbf{X}\mathbf{A}=\mathbf{B}\), multiply both sides on the right instead. Always substitute the proposed answer into the original equation; checking a product in a different order is not a valid verification.

When the Inverse Method Does Not Apply

The two theorems require that \(\mathbf{A}\) be square and invertible. If \(\mathbf{A}\) is singular, the inverse formulas are unavailable, but that does not by itself tell us whether an equation has no solution or many solutions. For example, let \(\mathbf{A}=\begin{bmatrix}1&0\\0&0\end{bmatrix}\). For \(\mathbf{A}\mathbf{X}=\mathbf{O}\), both \(\mathbf{X}=\mathbf{O}\) and \(\mathbf{X}=\begin{bmatrix}0&0\\1&0\end{bmatrix}\) are solutions, since multiplying either on the left by \(\mathbf{A}\) gives \(\mathbf{O}\). In contrast, \(\mathbf{A}\mathbf{X}=\mathbf{I}_2\) has no solution: every product \(\mathbf{A}\mathbf{X}\) has a zero second row, but \(\mathbf{I}_2\) has second row \((0,1)\). Thus invertibility guarantees one solution for every right-hand side, while a singular coefficient requires a separate analysis.

This method matters because matrix equations often encode several ordinary linear systems at once. The rules above handle all the columns of a left-multiplication equation or all the rows of a right-multiplication equation while preserving the order of matrix products. Keeping dimensions, side, and verification together prevents a plausible-looking inverse formula from solving the wrong equation.