How to Solve Non-Square Matrix Equations

A rectangular matrix generally has no ordinary inverse, so you cannot solve a matrix equation by multiplying by $A^{-1}$. Instead, identify the dimensions and interpret the equation as one or more systems of linear equations. This approach applies whether you are solving $AX=B$, $XA=B$, or a more complicated equation involving an unknown matrix.

The most common case is $AX=B$, where $A$ and $B$ are known and $X$ is unknown. The equation may have no exact solution, a unique solution, or infinitely many solutions.

Start with dimensions

Suppose $A$ is an $m\times n$ matrix and $B$ is an $m\times p$ matrix. For the product $AX$ to be defined and equal to $B$, $X$ must be an $n\times p$ matrix:

$$A_{m\times n}X_{n\times p}=B_{m\times p}.$$

Write the unknown as columns, $X=[x_1\;x_2\;\cdots\;x_p]$, and write $B=[b_1\;b_2\;\cdots\;b_p]$. Then $AX=B$ means

$$Ax_j=b_j \qquad (j=1,\ldots,p).$$

So you can solve the equation by solving $p$ ordinary systems with the same coefficient matrix $A$, one for each column of $B$.

Decide whether exact solutions exist

A system $Ax=b$ has a solution exactly when $b$ is in the column space of $A$. Therefore, $AX=B$ has an exact solution exactly when every column of $B$ is in the column space of $A$.

You can test this by row-reducing the augmented matrix $[A\mid B]$. If row reduction produces a row whose coefficient entries are all zero but whose entry in $B$ is nonzero, the equation is inconsistent. Otherwise, the solutions can be found by expressing pivot variables in terms of any free variables.

The number of solutions depends on the rank of $A$ and consistency:

• If $A$ has full column rank ($\operatorname{rank}(A)=n$), each consistent column system has a unique solution.
• If $\operatorname{rank}(A)<n$, a consistent system has free variables and therefore infinitely many solutions (over $\mathbb{R}$).
• If any column system is inconsistent, there is no exact matrix $X$ satisfying the equation.

Worked example: a rectangular equation with infinitely many solutions

Let

$$A=\begin{bmatrix}1&0&1\\0&1&1\end{bmatrix}, \qquad B=\begin{bmatrix}1&2\\3&4\end{bmatrix}.$$

Since $A$ is $2\times3$ and $B$ is $2\times2$, the unknown $X$ must be $3\times2$. Write its columns as $x_1$ and $x_2$. Solving $Ax_1=b_1$ gives

$$x_{11}+x_{31}=1,\qquad x_{21}+x_{31}=3.$$

Set the free variable $x_{31}=t_1$. Then $x_{11}=1-t_1$ and $x_{21}=3-t_1$. The second column gives, in the same way, $x_{12}=2-t_2$, $x_{22}=4-t_2$, and $x_{32}=t_2$. Thus every solution is

$$X=\begin{bmatrix}1-t_1&2-t_2\\3-t_1&4-t_2\\t_1&t_2\end{bmatrix}, \qquad t_1,t_2\in\mathbb{R}.$$

For example, taking $t_1=t_2=0$ gives one solution. Substituting it verifies the equation:

$$AX= \begin{bmatrix}1&0&1\\0&1&1\end{bmatrix} \begin{bmatrix}1&2\\3&4\\0&0\end{bmatrix} =\begin{bmatrix}1&2\\3&4\end{bmatrix}=B.$$

The two free parameters arise because $A$ has three columns but rank two. Each column system has one free variable.

If the equation is $XA=B$

The same principle works, but the row space replaces the column space. If $A$ is $m\times n$, then $X$ must have $m$ columns for $XA$ to be defined. Each row of $XA$ is a linear combination of the rows of $A$. Consequently, $XA=B$ has a solution exactly when every row of $B$ lies in the row space of $A$.

You can also transpose the equation: $XA=B$ is equivalent to $A^T X^T=B^T$. Then solve it as a left-multiplication problem, one column at a time.

If there is no exact solution: least squares

In applications, $AX=B$ may be inconsistent because data contain measurement error or because the system is overdetermined. A common alternative is to choose $X$ that minimizes the total squared residual, $\lVert AX-B\rVert_F^2$, where the Frobenius norm sums the squares of all entries.

The Moore–Penrose pseudoinverse gives a standard least-squares choice: $X=A^+B$. When $A$ has full column rank, this simplifies to $X=(A^T A)^{-1}A^T B$. In numerical work, QR factorization or singular value decomposition is generally preferred to explicitly forming that inverse. Least squares gives an approximate fit; it does not turn an inconsistent equation into an exact one.

Common mistakes

• Trying to invert a rectangular matrix. Ordinary matrix inverses are defined only for square matrices, and even a square matrix must be nonsingular. Use elimination or a pseudoinverse for the relevant problem instead.
• Ignoring dimensions. Check the inner dimensions of every product before solving, and state the size of the unknown.
• Assuming consistency. A rectangular system can have no solution. Check the augmented system or the column-space condition first.
• Giving just one solution when there are free variables. A particular solution does not describe all solutions. Include the free parameters, or state clearly that you are giving one example.
• Confusing exact and least-squares solutions. A least-squares answer minimizes error; it satisfies $AX=B$ exactly only if the residual is zero.

For equations with the unknown matrix appearing more than once, such as $AXB=C$, you can often convert the problem into a scalar linear system in the entries of $X$. The dimensions and placement of the unknown determine which system to solve.