# Determinant/Zero, linear dependent and rank property/Fact/Proof

Suppose now that the rows are linearly dependent. After exchanging rows, we may assume that $\displaystyle {{}} v_n = \sum_{i [[Category:Wikiversity soft redirects|Determinant/Zero, linear dependent and rank property/Fact/Proof]] __NOINDEX__ 1}^{n-1} s_i v_i$ . Then, due to
$pmatrix}"): {\displaystyle {{}} \det M = \det \begin{pmatrix} v_1 \\ \vdots \\ v_{n-1} \\ \sum_{i [[Category:Wikiversity soft redirects|Determinant/Zero, linear dependent and rank property/Fact/Proof]] __NOINDEX__ 1}^{n-1} s_i v_i \end{pmatrix} = \sum_{i [[Category:Wikiversity soft redirects|Determinant/Zero, linear dependent and rank property/Fact/Proof]] __NOINDEX__ 1}^{n-1} s_i \det \begin{pmatrix} v_1 \\ \vdots \\ v_{n-1} \\ v_i \end{pmatrix} = 0 \, .$
Now suppose that the rows are linearly independent. Then, by exchanging of rows, scaling and addition of a row to another row, we can transform the matrix successively into the identity matrix. During these manipulations, the determinant is multiplied with some factor ${\displaystyle {}\neq 0}$. Since the determinant of the identity matrix is ${\displaystyle {}1}$, the determinant of the initial matrix is ${\displaystyle {}\neq 0}$.