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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?

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  • $\begingroup$ I saw what you did there in the mathjax. Next time to insert the $\times$ symbol use the \times mathjax symbol ;) $\endgroup$ – Gaurang Tandon Feb 11 '18 at 16:00
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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.

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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.

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  • $\begingroup$ is there a general proof for this or should we prove it by using your example as a counter example $\endgroup$ – Karen Feb 11 '18 at 15:57
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    $\begingroup$ @Karen It's easily proven by writing a general $n\times n$ matrix in which you choose one row to be like $a_1, a_2 \ldots a_n$ and another row to be $\beta a_1, \beta a_2 \ldots \beta a_n$ then just subtract the latter row from $\beta$ times the first, and you will get a zero row. $\endgroup$ – Von Neumann Feb 11 '18 at 15:58
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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 $0$.

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  • $\begingroup$ That's a nice notation for large matrices, substituting one symbol for entire columns! $\endgroup$ – Gaurang Tandon Feb 11 '18 at 16:02
  • $\begingroup$ @GaurangTandon I don't know why I don't see it more often. $\endgroup$ – José Carlos Santos Feb 11 '18 at 16:03
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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.

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