Solution:

Given matrices $A=\left(\begin{array}{lll}1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1\end{array}\right), B=\left(\begin{array}{lll}1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1\end{array}\right)$ and
$C=\left(\begin{array}{ccc}
1+p q & p & 0 \\
q & 1+p q & p \\
0 & q & 1
\end{array}\right)$
We now need to find the inverses of given matrices.

(i) We now have $A=\left(\begin{array}{lll}1 & p & 0 \\ 0 & 1 & p \\ 0 & 0 & 1\end{array}\right)$
Let’s now calculate the determinant of the given matrix first.
$|A|=\left|\begin{array}{lll}
1 & p & 0 \\
0 & 1 & p \\
0 & 0 & 1
\end{array}\right|=1(1-0)-p(0)+0=1 \neq 0$
So, the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of $A$.
$M_{A}=\left(\begin{array}{lll}
\left|\begin{array}{ll}
1 & p \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
0 & p \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right| \\
\left|\begin{array}{ll}
p & 0 \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
1 & p \\
0 & 0
\end{array}\right| \\
\left|\begin{array}{ll}
p & 0 \\
1 & p
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
0 & p
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
0 & p
\end{array}\right|
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & 0 \\
p & 1 & 0 \\
p^{2} & p & 1
\end{array}\right)$

Step:2 Find the co-factor matrix of $A$.
$\begin{array}{l}
C_{A}=\left(\begin{array}{ccc}
(-1)^{1+1}(1) & (-1)^{1+2}(0) & (-1)^{1+3}(0) \\
(-1)^{2+1}(p) & (-1)^{2+2}(1) & (-1)^{2+3}(0) \\
(-1)^{3+1}\left(p^{2}\right) & (-1)^{3+2}(p) & (-1)^{3+3}(1)
\end{array}\right) \\
=\left(\begin{array}{ccc}
1 & 0 & 0 \\
-p & 1 & 0 \\
p^{2} & -p & 1
\end{array}\right)
\end{array}$

Step: 3 By transpose of $C_{A}$ we will have $\operatorname{adj} A$.
$\operatorname{adj} A=C_{A}^{T}=\left(\begin{array}{ccc}1 & 0 & 0 \\ -p & 1 & 0 \\ p^{2} & -p & 1\end{array}\right)^{T}=\left(\begin{array}{ccc}1 & -p & p^{2} \\ 0 & 1 & -p \\ 0 & 0 & 1\end{array}\right)$
Finally the inverse of the matrix is
$A^{-1}=\frac{1}{|A|} \text { adj } A=\frac{1}{1}\left(\begin{array}{ccc}
1 & -p & p^{2} \\
0 & 1 & -p \\
0 & 0 & 1
\end{array}\right)=\left(\begin{array}{ccc}
1 & -p & p^{2} \\
0 & 1 & -p \\
0 & 0 & 1
\end{array}\right)$

(ii) We now have $B=\left(\begin{array}{lll}1 & 0 & 0 \\ q & 1 & 0 \\ 0 & q & 1\end{array}\right)$
Let’s now calculate the determinant of the given matrix first.
$|B|=\left|\begin{array}{lll}
1 & 0 & 0 \\
q & 1 & 0 \\
0 & q & 1
\end{array}\right|=1(1-0)-0+0=1 \neq 0$
So, the given matrix has inverse.
Find: the adjoint of the given matrix.

Step: 1 Find the minor matrix of $B$.
$M_{B}=\left(\begin{array}{ll}
\left|\begin{array}{ll}
1 & 0 \\
q & 1
\end{array}\right| & \left|\begin{array}{ll}
q & 0 \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
q & 1 \\
0 & q
\end{array}\right| \\
\left|\begin{array}{ll}
0 & 0 \\
q & 1
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
0 & q
\end{array}\right| \\
\left|\begin{array}{ll}
0 & 0 \\
1 & 0
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
q & 0
\end{array}\right| & \left|\begin{array}{ll}
1 & 0 \\
q & 1
\end{array}\right|
\end{array}\right)=\left(\begin{array}{ccc}
1 & q & q^{2} \\
0 & 1 & q \\
0 & 0 & 1
\end{array}\right)$

Step: 2 Find the co-factor matrix of $B$.
$\begin{array}{l}
C_{B}=\left(\begin{array}{ccc}
(-1)^{1+1}(1) & (-1)^{1+2}(q) & (-1)^{1+3}\left(q^{2}\right) \\
(-1)^{2+1}(0) & (-1)^{2+2}(1) & (-1)^{2+3}(q) \\
(-1)^{3+1}(0) & (-1)^{3+2}(0) & (-1)^{3+3}(1)
\end{array}\right) \\
=\left(\begin{array}{ccc}
1 & -q & q^{2} \\
0 & 1 & -q \\
0 & 0 & 1
\end{array}\right)
\end{array}$

Step:3 By transpose of $C_{B}$ we will get $\operatorname{adj} B$.
$\operatorname{adj} B=C_{B}^{T}=\left(\begin{array}{ccc}1 & -q & q^{2} \\ 0 & 1 & -q \\ 0 & 0 & 1\end{array}\right)^{T}=\left(\begin{array}{ccc}1 & 0 & 0 \\ -q & 1 & 0 \\ q^{2} & -q & 1\end{array}\right)$
Finally the inverse of the matrix is
$B^{-1}=\frac{1}{|B|} \text { adj } B=\frac{1}{1}\left(\begin{array}{ccc}
1 & 0 & 0 \\
-q & 1 & 0 \\
q^{2} & -q & 1
\end{array}\right)=\left(\begin{array}{ccc}
1 & 0 & 0 \\
-q & 1 & 0 \\
q^{2} & -q & 1
\end{array}\right)$
Finally the inverse of $C$ is calculated as follow.
$C=\left(\begin{array}{ccc}
1+p q & p & 0 \\
q & 1+p q & p \\
0 & q & 1
\end{array}\right)=\left(\begin{array}{lll}
1 & p & 0 \\
0 & 1 & p \\
0 & 0 & 1
\end{array}\right)\left(\begin{array}{lll}
1 & 0 & 0 \\
q & 1 & 0 \\
0 & q & 1
\end{array}\right)=A B$
Hence,
$\begin{array}{l}
\quad C^{-1}=(A B)^{-1}=B^{-1} A^{-1} \\
=\left(\begin{array}{ccc}
1 & 0 & 0 \\
-q & 1 & 0 \\
q^{2} & -q & 1
\end{array}\right)\left(\begin{array}{ccc}
1 & -p & p^{2} \\
0 & 1 & -p \\
0 & 0 & 1
\end{array}\right) \\
=\left(\begin{array}{ccc}
1 & -p & p^{2} \\
-q & p q+1 & -p^{2} q-p \\
q^{2} & -q^{2} p-q & p^{2} q^{2}+p q+1
\end{array}\right)
\end{array}$