Problem source

The matrices $\mathbf{A},\mathbf{B}$ and $\mathbf{M}$ are given
by 
\[
\mathbf{A}=\begin{pmatrix}a & 0 & 0\\
b & c & 0\\
d & e & f
\end{pmatrix},\quad\mathbf{B}=\begin{pmatrix}1 & p & q\\
0 & 1 & r\\
0 & 0 & 1
\end{pmatrix},\quad\mathbf{M}=\begin{pmatrix}1 & 3 & 2\\
4 & 13 & 5\\
3 & 8 & 7
\end{pmatrix},
\]
where $a,b,\ldots,r$ are real numbers. Given that $\mathbf{M=AB},$
show that $a=1,b=4,c=1,d=3,e=1,f=-2,p=3,q=2$ and $r=-3$ gives the
\textit{unique }solution for $\mathbf{A}$ and $\mathbf{B}.$ Evaluate
$\mathbf{A}^{-1}$ and $\mathbf{B}^{-1},$

Hence, or otherwise, solve the simultaneous equations 
\begin{alignat*}{1}
x+3y+2z & =7\\
4x+13y+5z & =18\\
3x+8y+7z & =25.
\end{alignat*}

Solution source

\begin{align*}
&& \begin{pmatrix}1 & 3 & 2\\
4 & 13 & 5\\
3 & 8 & 7
\end{pmatrix} &= \begin{pmatrix}a & 0 & 0\\
b & c & 0\\
d & e & f
\end{pmatrix}\begin{pmatrix}1 & p & q\\
0 & 1 & r\\
0 & 0 & 1
\end{pmatrix} \\
&&&= \begin{pmatrix} 
a & ap & aq \\
b & pb + c & qb + cr\\
d & pd  + e & qd + er +f
\end{pmatrix} \\
\Rightarrow && a,b,d,p,q&=1,4,3,3,2\\
&&&= \begin{pmatrix} 
1 & 3 & 2 \\
4 & 12 + c & 8+ cr\\
3 & 9  + e & 6 + er +f
\end{pmatrix} \\
\Rightarrow && c, e&=1,-1\\
&&&= \begin{pmatrix} 
1 & 3 & 2 \\
4 & 13 & 8+ r\\
3 & 8 & 6  -r +f
\end{pmatrix} \\
\Rightarrow && r, f &= -3, -2
\end{align*}

\begin{align*}
\mathbf{A}^{-1} &= \begin{pmatrix} 
1 & 0 & 0 \\
4 & 1 & 0\\
3 & -1 & -2
\end{pmatrix}^{-1} \\
&=\frac{1}{-2} \begin{pmatrix} 
-2 & 0 & 0 \\
8 & -2 & 0\\
-7 & 1 & 1
\end{pmatrix} \\
\\
\mathbf{B}^{-1} &= \begin{pmatrix} 
1 & 3 & 2 \\
0 & 1 & -3\\
0 & 0 & 1
\end{pmatrix}^{-1} \\
&= \begin{pmatrix} 
1 & -3 & -11 \\
0 & 1 & 3\\
0 & 0 & 1
\end{pmatrix} \\
\end{align*}

We want to solve $\mathbf{M}\mathbf{v} = \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix}$, ie 
\begin{align*}
\mathbf{v} &= \mathbf{M}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\
&= \mathbf{B}^{-1} \mathbf{A}^{-1}\begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\
&= \frac{1}{-2}\mathbf{B}^{-1}  \begin{pmatrix} 
-2 & 0 & 0 \\
8 & -2 & 0\\
-7 & 1 & 1
\end{pmatrix} \begin{pmatrix} 7 \\ 18 \\ 25 \end{pmatrix} \\
&= \frac{1}{-2}\mathbf{B}^{-1} \begin{pmatrix} -14 \\ 20 \\ -6 \end{pmatrix} \\
&= \begin{pmatrix} 
1 & -3 & -11 \\
0 & 1 & 3\\
0 & 0 & 1
\end{pmatrix}\begin{pmatrix} 7 \\ -10 \\ 3 \end{pmatrix} \\
&= \begin{pmatrix} 4\\ -1 \\ 3 \end{pmatrix} 
\end{align*}

This algorithm is called the "LU-decomposition"