# 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.

$$\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}$$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.

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?