Proof about rank of a matrix depending on the value of its determinant

Question

I've got the following exercise: $A \in M(2 \times 2; \mathbb{R})$. Show that $$\text{rank}(A)= \begin{cases} 2, \ \ \ \ \ \ \text{if det(A) ≠ 0} \\ \leq 1, \ \ \text{if det(A) = 0} \end{cases}$$

I am not sure how to start this. I am grateful for every hint.

Answer 1

only just an HINT

rank(A) = #numer of linearly independent columns

thus

1. rank(A)=2 $\iff det(A) ≠ 0$
2. rank(A)<2 $\iff det(A) = 0$

NOTE take a look at a good book on linear algebra