202311270959 Exercise 1.4.33

Simplify:

$(AB)^{-1}(AC^{-1})(D^{-1}C^{-1})^{-1}D^{-1}$ .

_{Extracted from H. Anton & C. Rorres. (2010). Elementary Linear Algebra Application Version (10e)}

Answer. $B^{-1}$

Roughwork.

Let

$\begin{aligned} A_{m\times n}&:m\textrm{-by-}n \\ B_{n\times o}&:n\textrm{-by-}o \\ (C^{\mathrm{T}})_{n\times p}}&:n\textrm{-by-}p \\ (D^{\mathrm{T}})_{q\times n}&:q\textrm{-by-}n \\ C_{p\times n}&:p\textrm{-by-}n \\ D_{n\times q}&:n\textrm{-by-}q \\ (AB)_{m\times o}&:m\textrm{-by-}o \\ ((AB)^{\mathrm{T}})_{o\times m}&:o\textrm{-by-}m \\ ((AC)^{\mathrm{T}})_{m\times p}&:m\textrm{-by-}p \\ (D^{\mathrm{T}}C^{\mathrm{T}})_{q\times p}&:q\textrm{-by-}p \\ ((D^{\mathrm{T}}C^{\mathrm{T}})^{\mathrm{T}})_{p\times q}&:p\textrm{-by-}q \\ \end{aligned}$

such that after simplification the original array of matrices (here transpose in place of inverse) should give an $o$ -by- $n$ matrix. For merely satisfying this requirement one candidate can be $B_{n\times o}$ ‘s transpose, i.e., $(B^{\mathrm{T}})_{o\times n}$ , but not much evidence than coincidence here.

Observe, that the simplified expression, in general, should $\textrm{\scriptsize{NOT}}$ necessarily be a span (/linear combination) of matrices in the list above, for none any pair of matrices have necessarily the same dimension for which matrix addition is possible, apart from square matrices.

Hence, assume $AB$ , $C$ , and $D$ to be three square matrices which are invertible.

Note, that the simplified expression should be an arrangement of matrix multiplication amongst some of $8$ matrices $A$ and $A^{-1}$ , $B$ and $B^{-1}$ , $C$ and $C^{-1}$ , and $D$ and $D^{-1}$ .

Consider, by simplification the matrix product should have strictly less than $r=7$ factors, i.e., at most $6$ ; and in adding an extra identity matrix $I$ to the preceding (*such as to avoid terms $AA^{\mathrm{-1}}=A^{\mathrm{-1}}A=I$ to go nearby) we have $n=9$ matrices at hand; the possible outcomes is a permutation