## Precalculus (6th Edition) Blitzer

$\left| \begin{matrix} 3 & 2 & 1 \\ 4 & 3 & 1 \\ 5 & 1 & 1 \\ \end{matrix} \right|=3\left| \begin{matrix} \underline{3} & \underline{1} \\ \underline{1} & \underline{1} \\ \end{matrix} \right|-4\left| \begin{matrix} \underline{2} & \underline{1} \\ \underline{1} & \underline{1} \\ \end{matrix} \right|+5\left| \begin{matrix} \underline{2} & \underline{1} \\ \underline{3} & \underline{1} \\ \end{matrix} \right|$
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: $\left| \begin{matrix} 3 & 2 & 1 \\ 4 & 3 & 1 \\ 5 & 1 & 1 \\ \end{matrix} \right|=3\left| \begin{matrix} 3 & 1 \\ 1 & 1 \\ \end{matrix} \right|-4\left| \begin{matrix} 2 & 1 \\ 1 & 1 \\ \end{matrix} \right|+5\left| \begin{matrix} 2 & 1 \\ 3 & 1 \\ \end{matrix} \right|$