Consider the map $f : M_{2\times3}(\Bbb R) \to \Bbb R^3$ given by sending a matrix to the triple of its $2\times 2$ minors.

The set of rank $2$ matrices is the inverse image of the set $$\{f(x_1, x_2, x_3)\in \Bbb R^3 \mid (x_1,x_2,x_3) \neq (0,0,0)\}$$ This set is open in $\Bbb R^3$, hence the set of rank $2$ matrices is open if the map $f$ is continuous.

But how to prove that $f$ is continuous?

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    $\begingroup$ I have MathJax-ed up your question. For next time, please refer to, for instance, this guide on how to write readable math on this site. $\endgroup$ – Arthur May 6 '19 at 7:36
  • $\begingroup$ As to the problem itself, before you can talk about continuous, you have to be aware of the topology on both your spaces. Presumably you have the standard topology on $\Bbb R^3$. Is the topology on $M_{2\times3}(\Bbb R)$ given basically by the standard topology on $\Bbb R^6$ by way of the six entries of the matrices? $\endgroup$ – Arthur May 6 '19 at 7:38
  • $\begingroup$ @Arthur, I suspect, the OP may not yet be familiar with the concept of topology. But I understand you wanted to ask them, what "continuous" means in that context. $\endgroup$ – avs May 6 '19 at 7:50

How about the map $ f : M_{2\times3}(\mathbb{R}) \to \mathbb{R} $ that maps a matrix A to sum of determinant squares of all the possible $2\times2$ minors. Then, set of matrices of rank 2 is just the inverse image of $\mathbb{R}\setminus{0} $, which is open in $\mathbb{R}$. Note that this map is continuous because it is a polynomial in the entries of the matrix.

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