Find the value of the determinant using the method of expansion by minors; expand on the third row

For the matrix
[tex]\begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}[/tex]the determinant using the method of expansion by minors, expanding on the third row is:
[tex]\det \begin{bmatrix}{a_{11}} & {a_{12}} & {a_{13}} \\ {a_{21}} & {a_{22}} & {a_{23}} \\ {a_{31}} & {a_{32}} & {a_{33}}\end{bmatrix}=a_{31}\det \begin{bmatrix}{a_{12}} & {a_{13}} & {} \\ {a_{22}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}-a_{32}\det \begin{bmatrix}{a_{11}} & {a_{13}} & {} \\ {a_{21}} & {a_{23}} & {} \\ {} & {} & {}\end{bmatrix}+a_{33}\det \begin{bmatrix}{a_{12}} & {a_{12}} & {} \\ {a_{22}} & {a_{22}} & {} \\ {} & {} & {}\end{bmatrix}[/tex]Answer:
First, we compute the determinants of the minors:
[tex]\begin{gathered} \det \begin{bmatrix}{0} & {4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=0+4=4, \\ \det \begin{bmatrix}{1} & {4} & {} \\ {1} & {3} & {} \\ {} & {} & {}\end{bmatrix}=3-4=-1, \\ \det \begin{bmatrix}{1} & {0} & {} \\ {1} & {-1} & {} \\ {} & {} & {}\end{bmatrix}=-1-0=-1. \end{gathered}[/tex]Therefore:
[tex]\det \begin{bmatrix}{1} & {0} & {4} \\ {1} & {-1} & {3} \\ {0} & {5} & {-2}\end{bmatrix}=0\times4-5\times(-1)+(-2)\times(-1)=5+2=7.[/tex]