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

Proof

The relation between rank, invertibility and linear independence was proven in fact. Suppose now that the rows are linearly dependent. After exchanging rows, we may assume that ${}v_{n}=\sum _{i=1}^{n-1}s_{i}v_{i}$. Then, due to fact and fact, we get

${}\det M=\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\\sum _{i=1}^{n-1}s_{i}v_{i}\end{pmatrix}}=\sum _{i=1}^{n-1}s_{i}\det {\begin{pmatrix}v_{1}\\\vdots \\v_{n-1}\\v_{i}\end{pmatrix}}=0\,.$

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 ${}\neq 0$. Since the determinant of the identity matrix is ${}1$, the determinant of the initial matrix is ${}\neq 0$.

To fact