Lets say we have:


Where A and B are known matrices, X is unknown. In case B was square, a solution can be found by $\mathbf{B^{-1}A=X}$.

But how do you attempt to solve for X when B is not square, i.e. $n\neq m$?

  • 5
    $\begingroup$ What you mentioned works only if $B$ is invertible, by the way. $\endgroup$ – InterestedGuest Feb 18 '11 at 2:54

Let $\mathbf{a}_1,\ldots,\mathbf{a}_k$ be the columns of $\mathbf{A}$ (so $\mathbf{A}$ is $n\times k$ for some $k$). Notice that we will have $\mathbf{B}$ is $n\times m$ and $\mathbf{X}$ is $m\times k$ for some integer $m$ (for $\mathbf{BX}=\mathbf{A}$ to work out). Let $\mathbf{x}_1,\ldots,\mathbf{x}_k$ be the columns of $\mathbf{X}$.

Notice that $\mathbf{a}_i$ depends only on $\mathbf{B}$ and $\mathbf{x}_i$, since $\mathbf{Bx}_i = \mathbf{a}_i$. So to determine the $i$th column of $\mathbf{X}$, it suffices to solve the system of linear equations $$\mathbf{B}\mathbf{x}_i = \mathbf{a}_i.$$ So finding $\mathbf{X}$ is equivalent to solving $k$ systems of linear equations.

In fact, you can just do them all at the same time. Simply take a matrix that is made up of $\mathbf{B}$ followed by $\mathbf{A}$: $$\left(\mathbf{B}|\mathbf{A}\right)$$ and use Gauss-Jordan elimination on $\mathbf{B}$. The solutions you find for each column corresponding to $\mathbf{A}$ yield the columns of $\mathbf{X}$.

  • $\begingroup$ Can you elaborate on a procedure if you were attempting to solve for B, instead of X? $\endgroup$ – user63228 Feb 21 '13 at 1:33

In addition to Arturo's answer, maybe you could take a look at mine to this question.


The answer is given by the Penrose pseudoinverse:

$z=B^{pseudo}A $. While an exact solution exists only in the case $B^{pseudo}=B^{-1}$ the solution always guarantees that $||A-Bz||=minimum$


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