Suppose $$A=\pmatrix{1&2&3&4&5\\2&3&4&5&6}$$ Find $\det(A^TA)$.

I know exactly how to calculate it by writing it as a $5\times5$ matrix. But how to calculate it smartly?


marked as duplicate by amd, amWhy linear-algebra Aug 19 '18 at 22:55

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  • Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.

  • Hence the determinant must be $0$.

  • $\begingroup$ Can you explain why the first bullet point is true? $\endgroup$ – IntegrateThis Oct 13 '18 at 4:41
  • 1
    $\begingroup$ We know that $rank(AB) \le \min(rank(A), rank(B))$. $\endgroup$ – Siong Thye Goh Oct 13 '18 at 4:43

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