Solution:
Suppose the numbers as $x$, $y$, $z$
$\begin{array}{l}
x+y+z=2 \\
\ldots \cdots(i)
\end{array}$
Also, $2 y+(x+z)+1$
$x+2 y+z=1 \ldots \ldots \text { (ii) }$
Again,
$\begin{array}{l}
x+z+5(x)=6 \\
5 x+y+z=6 \ldots \ldots \text { (iii) }
\end{array}$
$\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
5 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{l}
2 \\
1 \\
6
\end{array}\right]$
$\begin{array}{l}
A X=B \\
|A|=1(1)-1(-4)+1(-9) \\
=1+4-9 \\
=-4
\end{array}$
As a result, the unique solution given by $x=A^{-1} B$
$C_{11}=(-1)^{1+1}(2-1)=1$
$C_{12}=(-1)^{1+2}(1-5)=4$
$C_{13}=(-1)^{1+3}(1-10)=-9$
$C_{21}=(-1)^{2+1}(1-1)=0$
$\mathrm{C}_{22}=(-1)^{2+2}(1-5)=-4$
$C_{23}=(-1)^{2+3}(1-5)=4$
$C_{31}=(-1)^{3+1}(1-2)=-1$
$C_{32}=(-1)^{3+2}(1-1)=0$
$\mathrm{C}_{33}=(-1)^{3+3}(2-1)=1$
$X=A^{-1} B=\frac{1}{|A|}($ adj $A) B$
$\operatorname{Adj} A=\left[\begin{array}{ccc}1 & 4 & -9 \\ 0 & -4 & 4 \\ -1 & 0 & 1\end{array}\right]^{T}=\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1\end{array}\right]$
$\mathrm{X}=\frac{1}{-4}\left[\begin{array}{ccc}1 & 0 & -1 \\ 4 & -4 & 0 \\ -9 & 4 & 1\end{array}\right]\left[\begin{array}{l}2 \\ 1 \\ 6\end{array}\right]$
$X=\frac{1}{-4}\left[\begin{array}{c}2-6 \\ 8-4 \\ -18+4+6\end{array}\right]$
$=\frac{1}{-4}\left[\begin{array}{c}
-4 \\
4 \\
-8
\end{array}\right]$
As a result, $\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{Z}\end{array}\right]=\left[\begin{array}{c}1 \\ -2 \\ 2\end{array}\right]$