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

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 19 '18 at 22:55

• Rank of $A$ is $2$, hence $A^TA$ cannot be rank $5$ and it must be singular.
• Hence the determinant must be $0$.
• We know that $rank(AB) \le \min(rank(A), rank(B))$. – Siong Thye Goh Oct 13 '18 at 4:43