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If $A$ is a matrix with $5$ rows and $3$ columns, how many determinants must you compute to be sure the rank is $< 3$?

I read an example at here. They only compute $3$ determinants: Row $1, 2, 3$, $1, 2, 4$ and $1, 2, 5$ only. All of them has determinant $= 0$, so they conclude that the rank is $< 3$.

But what if row $2$ is multiple of row $1$ but row $3, 4$ and $5$ are linearly independent?

Shouldn't they calculate the $5! /( 3!\cdot 2!) = 10$ possible determinants of order $3$?

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  • $\begingroup$ I think your link is wrong? I don't see the specific example you are mentioning. But it does not look like a very good site... $\endgroup$ – Morgan Rodgers Oct 12 '15 at 14:46
  • $\begingroup$ Actually, you need to compute only one determinant: $\det(A^tA)$. If it is zero then $A$ has rank smaller than 3 if it is not zero $A$ has rank 3. $\endgroup$ – Daniel Oct 12 '15 at 19:56
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You are right, the three determinants are not enough. For example the matrix $$ P = \begin{pmatrix} 0 &0 & 0\\ 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} $$ has rank three, but the determinants you listed are all zero. As this example shows, leaving of one determinant can lead to a wrong result, you have indeed to compute all $\binom 53$ minors (of course you can stop after you found a non-zero one).

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