Is the Adjugate of a Singular Matrix Also Singular?

[loli12]

Ques: Show that if A is singular then adj A is also singular. Given that I = [1/ det(A)]*A*adj (A)
I tried to prove it by assuming adj(A) is nonsingular. so, [1/det(A)]*adj(A) is nonsingular, as well as I. Which makes A to be nonsingular. I feel like I am forcing out this proof. So, do you guys have any good suggestion to this?

 

[matt grime]

If A is invertible, then there is a B such that AB=1, that alos implies that B is invertible...

 

[shmoe]

Ques: Show that if A is singular then adj A is also singular. Given that I = [1/ det(A)]*A*adj (A)
I tried to prove it by assuming adj(A) is nonsingular. so, [1/det(A)]*adj(A) is nonsingular, as well as I. Which makes A to be nonsingular. I feel like I am forcing out this proof. So, do you guys have any good suggestion to this?

When you write [1/det(A)]*adj(A) you are assuming that det(A) is not zero and A is non-singular.

In all cases A*adj(A)=det(A)*I. If A is singular, then det(A)=0. In other words, A*adj(A)=0. What does this say about adj(A)?

 

[loli12]

thanks guys, but i still have a ques.. (sure this is easy for you guys) so you are saying if A is singular and A*adj(A)=0 then it implies that adj(A) is also singular? how can that be?

 

[shmoe]

Suppose BC=0, where B and C are both square and non-zero.

If C is non-singular, then multiply by C^(-1) to get:

B*C*C^(-1)=0*C^(-1)

or

B=0, contradicting assumption of B non-zero

If B is nonsingular multiply by B^(-1) on the left and get C=0.

Therefore both B and C must be singular.

 

[TomMe]

Hi,

Maybe a late reply, but since I was also looking for a proof to this (got stuck at the BC = 0 thing), just wanted to say thanks!

Btw, for those interested: the opposite is also true (Adj A is singular => A is singular) and much easier to prove obviously.