Prove that A is reversible matrix if B is reversible

[skrat]

Using A\widetilde{A}=(detA)I prove that \widetilde{A} is reversible matrix if and only if A is reversible. Also prove det(\widetilde{A})=(detA)^{n-1} for any square matrix A.

First part:
1. direction:
Lets say \widetilde{A} is reversible, this means that \widetilde{A}\widetilde{A}^{-1}=I and det\widetilde{A}\neq 0.

A\widetilde{A}=(detA)I can than be written as:
A=(detA)\widetilde{A}^{-1} but now I don't know what else I can do here to prove that A is reversible?

 

[mfb]

but now I don't know what else I can do here to prove that A is reversible?

Try to find a matrix, which, if multiplied by (detA)\widetilde{A}^{-1}, gives I.

 

[skrat]

Try to find a matrix, which, if multiplied by (detA)\widetilde{A}^{-1}, gives I.

Something like ((detA)\widetilde{A}^{-1})^{-1}? But I have absoulutely no reason to believe that (detA)\neq 0 (that's what I am trying to prove).

((detA)\widetilde{A}^{-1})^{-1}
(detA)^{-1}\widetilde{A}

 

[HallsofIvy]

Then do it as two separate cases:
Suppose det(A)= 0. Then what?

Suppose det(A)\ne 0. Then what?

 

[skrat]

Then do it as two separate cases:
Suppose det(A)= 0. Then what?

Suppose det(A)\ne 0. Then what?

Hmm, why don't I have brains like you? That would really help me! :D

Case No.1: suppose det(A)\ne 0
A=(detA)\widetilde{A}^{-1} I can than multiply both sides with (detA)^{-1}\widetilde{A} which gives me
(detA)^{-1}\widetilde{A}A=((detA)\widetilde{A}^{-1})((detA)^{-1}\widetilde{A})
(detA)^{-1}\widetilde{A}A=I
So A is reversible and A^{-1}=(detA)^{-1}\widetilde{A}

Case No.2: suppose detA=0
A=(detA)\widetilde{A}^{-1}
A=0

So if detA=0 than A is not reversible by definition?

Does that sound right?

 

[micromass]

Hmm, why don't I have brains like you? That would really help me! :D

Case No.1: suppose det(A)\ne 0
A=(detA)\widetilde{A}^{-1} I can than multiply both sides with (detA)^{-1}\widetilde{A} which gives me
(detA)^{-1}\widetilde{A}A=((detA)\widetilde{A}^{-1})((detA)^{-1}\widetilde{A})
(detA)^{-1}\widetilde{A}A=I
So A is reversible and A^{-1}=(detA)^{-1}\widetilde{A}

Right

Case No.2: suppose detA=0
A=(detA)\widetilde{A}^{-1}
A=0

So if detA=0 than A is not reversible by definition?

It is correct that $A$ will not be reversible. But it's not good enough. You are proving that if $\tilde{A}$ is invertible, then $A$ is invertible.
You have now found that if $det(A)=0$ and $\tilde{A}$ is invertible, then $A$ is not invertible. So you have found a situation where $\tilde{A}$ is invertible and $A$ is not invertible, but you had to prove that $A$ is always inverible whenever $\tilde{A}$ was. This is an apparent contradiction! To solve it, you must prove that the situation $det(A)=0$ and $\tilde{A}$ invertible never occurs! So you need to prove that if $det(A)=0$ and $\tilde{A}$ is invetible, then some contradiction occurs!

 

[skrat]

To solve it, you must prove that the situation $det(A)=0$ and $\tilde{A}$ invertible never occurs! So you need to prove that if $det(A)=0$ and $\tilde{A}$ is invetible, then some contradiction occurs!

Hmmm, I can see where this is going but up to this moment I haven't got a good idea yet. We often proved this kind of statement using paradox:
Let's say that $A$ is invertible and show that this statement will lead us to paradox and prove that $A$ is NOT invertible.

Hmm, if $A$ is invertible, than $detA\neq 0$ but one of the previous condition says that $detA=0$ and this is paradox, so $A$ is NOT invertible.

I tried using the same idea in equation: (so $detA=0$, $AA^{-1}=I$ and $\tilde{A}\tilde{A}^{-1}=I$)

$A\tilde{A}=(detA)I$ but I didn't get really far this way... all I found out is that $I=(detA)\tilde{A}^{-1}A^{-1}$ where I can see that $A=(detA)\tilde{A}^{-1}$ which was obvious before too...