Suppose that a matrix A is \(n\times n\) matrix. Then if a matrix is multiplied by k, the determinant is multiplied by \(k^{n}\) $$\left|k\left(\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right)\right|=k^{3}\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|$$
Proof
We prove when \(n=3\).
Since the property of (2), we have
$$\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\ka_{31}& ka_{32}&ka_{33}\end{array}\right|=k\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\a_{31}& a_{32}&a_{33}\end{array}\right|$$
Therefore, we obtain
\begin{eqnarray*}\text{(LHS)}&=&\left|k\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)\right|\\ && \\&=&\left|\begin{array}{ccc}ka_{11}& ka_{12}&ka_{13}\\ ka_{21}& ka_{22}&ka_{23}\\ka_{31}& ka_{32}&ka_{33}\end{array}\right|\\ && \\ &=&k\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ ka_{21}& ka_{22}&ka_{23}\\ka_{31}& ka_{32}&ka_{33}\end{array}\right| \ \ \ \text{( ∵ Property of Determinant (2) )}\\ && \\&=&k^{2}\left|\begin{array}{ccc}a_{11}& a_{12}&a_{13}\\ a_{21}& a_{22}&a_{23}\\ka_{31}& ka_{32}&ka_{33}\end{array}\right| \ \ \ \text{( ∵ Property of Determinant (2) )}\\ && \\&=&k^{3}\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 (2) )}\\ && \\&=&\text{(RHS)}\end{eqnarray*}