Use (9.11)and (9.12) to simplify${\left({\text{AB}}^{\text{T}}\text{C}\right)}^{\mathbf{\text{T}}}\mathbf{,}{\left({\text{C}}^{-1}\text{MC}\right)}^{\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{(}\mathbf{\text{AH}}{\mathbf{)}}^{\mathbf{-}\mathbf{1}}{\left({\text{AHA}}^{-1}\right)}^{\mathbf{3}}{\left({\text{HA}}^{-1}\right)}^{\mathbf{-}\mathbf{1}}$

From the mentioned equation,

$$\begin{gathered} (ABCD{)}^T=D^T{C}^T{B}^T{A}^T \\ (ABCD{)}^{-1}=D^{-1}{C}^{-1}{B}^{-1}{A}^{-1} \end{gathered}$$

A matrix's transpose is determined by converting its rows into columns or columns into rows.

Considering ${\left(A{B}^{T}C\right)}^T$, and using the given equation,

${\left(A{B}^{T}C\right)}^T=C^T{\left(B^T\right)}^T{A}^T$

Since ${\left(B^T\right)}^T=B$,

$$\begin{gathered} C^T{\left(B^T\right)}^T{A}^T=C^{T}B{A}^T \\ {\left(A{B}^{T}C\right)}^T=C^{T}B{A}^T \end{gathered}$$

Considering ${\left(C^{-1}MC\right)}^T$, and using the given equation,

${\left(C^{-1}MC\right)}^{-1}=C^{-1}{M}^{-1}{\left(C^{-1}\right)}^{-1}$

Since ${\left(C^{-1}\right)}^{-1}=C$,

$$\begin{gathered} C^{-1}{M}^{-1}{\left(C^{-1}\right)}^{-1}=C^{-1}{M}^{-1}C \\ {\left(C^{-1}MC\right)}^{-1}=C^{-1}{M}^{-1}C \end{gathered}$$

Now, considering $(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2{\left(H{A}^{-1}\right)}^{-1}$, and using the given equation,

$$\begin{gathered} (AH{)}^{-1}{\left(AH{A}^{-1}\right)}^3{\left(H{A}^{-1}\right)}^{-1}=(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH{A}^{-1}\right){\left(A^{-1}\right)}^{-1}{H}^{-1} \\ (AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH{A}^{-1}\right){\left(A^{-1}\right)}^{-1}{H}^{-1}=(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH{A}^{-1}\right)A{H}^{-1} \\ (AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH{A}^{-1}\right)A{H}^{-1}=(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH\right)\left(A^{-1}A\right)H^{-1} \\ =(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(AH\right)\left(I\right)H^{-1} \end{gathered}$$

Proceed further with the problem.

role="math" localid="1664270555585" $$\begin{gathered} =(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(A\right)\left(H{H}^{-1}\right) \\ =(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(A\right)\left(H{H}^{-1}\right) \\ =(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^2\left(A\right)I \\ =(AH{)}^{-1}\left(AH{A}^{-1}\right)\left(AH{A}^{-1}\right)\left(A\right) \end{gathered}$$

Solve further.

$$\begin{gathered} =(AH{)}^{-1}\left(AH{A}^{-1}\right)\left(AH\right)\left(A^{-1}A\right) \\ =H^{-1}{A}^{-1}AH\left(A^{-1}A\right)H \\ =H^{-1}\left(A^{-1}A\right)H^2 \\ =H^{-1}{H}^2 \end{gathered}$$

Solve further and get,

$=H$

Therefore, $(AH{)}^{-1}{\left(AH{A}^{-1}\right)}^3{\left(H{A}^{-1}\right)}^{-1}=\text{H}$.