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I have read that the product of two nonsingular matrices is nonsingular. So if $A$ and $B$ are nonsingular, then so is $AB$.

But what if we have $AC$ is nonsingular. Is there a rule in factorisation such that if we have a nonsingular matrix $AC$, then both $A$ and $C$ have to be nonsingular?

From my understanding you can have an $n \times n$ nonsingular matrix $A$ and a $n \times n$ nonsingular matrix $D$. So the multiplication would work, but because we have a nonsingular $\times$ singular, do we know what the result would be in terms of invertibility?

All I know is that a nonsingular $\times$ nonsingular $=$ nonsingular

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  • $\begingroup$ But what if all I have is $AC$ is nonsingular and I know that $A$ is nonsingular. Is there any way to draw conclusions about $C$ without knowing any entries of the matrices, ie, cannot use det(C). $\endgroup$ – Bucephalus Jun 30 '17 at 8:40
  • $\begingroup$ oopps, sorry, there is one more detailed I missed. $ACx=0$ homogenous. $\endgroup$ – Bucephalus Jun 30 '17 at 8:42
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A matrix R is singular if $\det$(R)=$0$.

A matrix Q is invertible if $\det$(Q)$\neq0$.

We also have that det (S) x det(T) = det(ST).

Therefore, if both S and T have nonzero determinants, so does ST and therefore is invertible.

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  • $\begingroup$ Oh yes, I see @SakethMalyala... $\endgroup$ – Bucephalus Jun 30 '17 at 18:40
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If $AB$ is non-singular, then there exists $C$ such that $(AB)C=I$

This implies that $A(BC)=I$, which means that $A$ is invertible and it's inverse is $BC$.

Same works for $B$.

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