## Precalculus (6th Edition) Blitzer

The easiest way to evaluate $\left| \begin{matrix} 3 & 2 & 8 \\ 5 & -4 & 0 \\ -6 & 7 & 0 \\ \end{matrix} \right|$ is to expand about the elements in column 3.
To calculate the value for a given determinant, choose any row or column and calculate 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: $A={{a}_{11}}C\left( {{a}_{11}} \right)+{{a}_{12}}C\left( {{a}_{12}} \right)+{{a}_{13}}C\left( {{a}_{13}} \right)$ Therefore, column 3 consists of two zeroes, so we need to calculate only one co-factor. Thus, it is easier to calculate.