Determinant of a square matrix is $0$ if one row is the multiple of another row

Question

If we have an $n\times n$ matrix in which one row is the multiple of another row, can we say that the determinant is $0$ because by getting the reduced row echelon form, we can get a row of $0$s?

Answer 1

Yes, a matrix has non-zero determinant if and only if it is invertible, which is equivalent to the set of rows being linearly independant.

Answer 2

If one row is exactly a multiple of another row, then you may perform the row subtraction, and one will become an identically zero row, hence the determinant will be zero.

Trivial example

$$\mathsf{A} = \begin{pmatrix} 2 & 3 \\ 4 & 6 \end{pmatrix}$$

By subtracting the second from twice the first you get

$$\mathsf{A} = \begin{pmatrix} 0 & 0 \\ 4 & 6 \end{pmatrix}$$

Hence $\ \text{det}\mathsf{A} = 0$ by definition.

This holds for every $n\times n$ matrix.

Answer 3

Yes, that argument is correct. But it is not the simplest approach. If your matrix is $\begin{pmatrix}v_1&v_2&\ldots&v_n\end{pmatrix}$ (the $v_k$'s are columns) and if, say, $v_3=\lambda v_1$, then\begin{align}\det\begin{pmatrix}v_1&v_2&v_3&\ldots&v_n\end{pmatrix}&=\det\begin{pmatrix}v_1&v_2&\lambda v_1&\ldots&v_n\end{pmatrix}\\&=\lambda\det\begin{pmatrix}v_1&v_2&v_1&\ldots&v_n\end{pmatrix}\\&=0,\end{align}because $\det$ is multilinear and alternate and therefore when two columns are equal, the determinant is $0$.

Answer 4

Use the fact that adding a multiple of a row to another row doesn't change the determinant of that matrix. If we have a row that is a multiple of the other row, then by one elementary row operation we can get a row consists of all zeros.

Finally, with cofactor expansion along that row, we get 0 as the determinant of the matrix.