Answer
$\left| \begin{matrix}
3 & 2 & 1 \\
4 & 3 & 1 \\
5 & 1 & 1 \\
\end{matrix} \right|=3\left| \begin{matrix}
\underline{3} & \underline{1} \\
\underline{1} & \underline{1} \\
\end{matrix} \right|-4\left| \begin{matrix}
\underline{2} & \underline{1} \\
\underline{1} & \underline{1} \\
\end{matrix} \right|+5\left| \begin{matrix}
\underline{2} & \underline{1} \\
\underline{3} & \underline{1} \\
\end{matrix} \right|$
Work Step by Step
To calculate the value for a given determinant, choose any row or column and find the co-factors. The co-factor of a determinant $A$ is:
$A=\left| \begin{matrix}
{{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\
{{b}_{21}} & {{b}_{22}} & {{b}_{23}} \\
{{c}_{31}} & {{c}_{32}} & {{c}_{33}} \\
\end{matrix} \right|$
$C\left( {{a}_{11}} \right)={{\left( -1 \right)}^{i+j}}\left| \begin{matrix}
{{b}_{22}} & {{b}_{23}} \\
{{c}_{32}} & {{c}_{33}} \\
\end{matrix} \right|$, where $i,j$ are row and column.
Now multiply the co-factors with the same number:
$A={{a}_{11}}C\left( {{a}_{11}} \right)+{{a}_{12}}C\left( {{a}_{12}} \right)+{{a}_{13}}C\left( {{a}_{13}} \right)$
From the given information, the above result can be given as here:
$\left| \begin{matrix}
3 & 2 & 1 \\
4 & 3 & 1 \\
5 & 1 & 1 \\
\end{matrix} \right|=3\left| \begin{matrix}
3 & 1 \\
1 & 1 \\
\end{matrix} \right|-4\left| \begin{matrix}
2 & 1 \\
1 & 1 \\
\end{matrix} \right|+5\left| \begin{matrix}
2 & 1 \\
3 & 1 \\
\end{matrix} \right|$
