If a multiple of one row of a matrix is added to another row, its determinant is equal to original determinant : $$\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|=\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}+ka_{31}& a_{22}+ka_{32}&a_{23}+ka_{33}\\a_{31}& a_{32}&a_{33}\end{array}\right|$$
Proof
We prove when \(n=3\).
Since the property of (4), we have
$$\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{31}& a_{32}&a_{33}\\a_{31}& a_{32}&a_{33}\end{array}\right|=0$$
Therefore, we obtain
\begin{eqnarray*}\text{(RHS)}&=&\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|+\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ ka_{31}& ka_{32}&ka_{33}\\a_{31}& a_{32}&a_{33}\end{array}\right|\ \ \ \text{( ∵ Property of Determinant (1) )}\\ && \\&=&\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|+k\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{31}& a_{32}&a_{33}\\a_{31}& a_{32}&a_{33}\end{array}\right| \ \ \ \text{( ∵ Property of Determinant (2) )}\\ && \\ &=&\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right| \ \ \ \text{( ∵ Property of Determinant (4) )}\\ && \\&=&\text{(LHS)}\end{eqnarray*}