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Then show that $Ax=0$ also has non trivial solution. Is there a solution considering idea of rank and rref.

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    Hint: $\det A^2 = (\det A)^2\,$. – 2017-02-04
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    Simpler hint: If $z$ is a nontrivial solution of $A^2x = 0$, then either $Az = 0$ or $Az \neq 0$. What can you do in the former case, what in the latter? – 2017-02-04
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    Thanks I think there is a small problem here. Since non trivial solution ,the inverse does not exist. Then how do we write that det a^2 =(det a )^2 the material I am referring to does not invoke concept of determinant. It is using row reduced form and rank ideas – 2017-02-04
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    The determinant is defined for non-invertible matrices. In fact, the determinant will be exactly zero if $A$ is non-invertible. This is one way to proceed. – 2017-02-04
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    Please include those observations in the body of your Question, which should be as self-contained as possible and not wholly reliant on the title to pose the problem's setup. – 2017-02-04

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A linear map $T:X \to Y$ is injective if and only if its kernel is trivial. Hence, if $\ker (A^2) \neq \{0\}$, then $A^2$ is not injective. If $A$ were injective (its kernel would be trivial), then $A^2=A \circ A$ would be injective as well, a contradiction.