What are the units of an inverse matrix? As the title suggests. For example if I have a matrix $A = \begin{pmatrix}
 a & b\\ 
 c& d 
\end{pmatrix}$ and all elements consist of variables with units $kg$ and then I take the inverse of the matrix is the resulting units simply $kg^{-1}$? How can this be the case if not all matrices have inverses?
Somewhat related to my other question about unit quantities in other matrix equations.
 A: $
\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}
$The wonderful book $\,$Multidimensional Analysis (by George W Hart)$\,$ addresses this question in detail. It is surprisingly difficult to get it right.
For example, here is the dimensional sketch of a rectangular matrix and that of its pseudoinverse
$$\eqalign{
A &= \m{
(m\cdot C^{-1})  & (m\cdot s\cdot K^{-1}) \\
(kg\cdot C^{-1}) & (kg\cdot s\cdot K^{-1}) \\
(m\cdot s^{-1}\cdot C^{-1})&(m\cdot K^{-1})} 
\\\\
A^{+} &= \m{
(C\cdot m^{-1}) & (C\cdot kg^{-1}) & (C\cdot s\cdot m^{-1}) \\
(K\cdot s^{-1}\cdot m^{-1}) & (K\cdot s^{-1}\cdot kg^{-1}) & (K\cdot m^{-1}) }
\\\\
}$$
Matrices which are squareable must have a special dimensional structure, e.g.
$$\eqalign{
B &= \m{
({\tt1}) & (m\cdot s^{-1}) \\
(s\cdot m^{-1}) & ({\tt1}) }
\quad\implies\quad 
B^2 &\overset{\Delta}{\;=\;} B \\
}$$
In this case, the diagonal elements carry no units, while the units of the other elements are the reciprocal of those in the transposed position. All powers of $B$ carry the same units.
Likewise, functions of $B$ such as the square root or exponential, carry the same units as $B$.
Note that for the rectangular matrix above
$$\eqalign{
AA^+ &= \m{
({\tt1}) & (m\cdot kg^{-1}) & (s) \\
(kg\cdot m^{-1}) & ({\tt1}) & (kg\cdot s\cdot m^{-1}) \\
(s^{-1}) & (m\cdot s^{-1}\cdot kg^{-1}) & ({\tt1})
}
\\\\
A^+A &= \m{
({\tt1}) & (C\cdot s\cdot kg^{-1}) \\
(kg\cdot s^{-1}\cdot C^{-1}) & ({\tt1}) \\
}
\\
}$$
So these projection matrices are squareable.
Also, an identity matrix has no dimensions on its diagonal elements, but does carry dimensions in its off-diagonal elements.
An equation like $(I+B)$ only makes sense if $I$ carries the same units as $B$.
So there is not one, but an infinite number of $2\times 2$ identity matrices when units are included.
A: As you know from the cofactor formula, the entries of the inverse matrix are homogeneous rational functions of degree $-1$ in the entries of the original matrix, so their unit will be $\mathrm{kg}^{-1}$.
As for your objection, I don't see your point. Are you fazed by the fact that $\sqrt{\bullet}$ is only defined for positive real numbers when you solve for time in, say, a uniformly accelerated motion?
