# Prove that A is reversible matrix if B is reversible

1. May 6, 2013

### 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?

2. May 6, 2013

### Staff: Mentor

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

3. May 6, 2013

### skrat

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).

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

4. May 6, 2013

### HallsofIvy

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

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

5. May 6, 2013

### skrat

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 definiton?

Does that sound right?

6. May 6, 2013

### micromass

Staff Emeritus
Right

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!

7. May 6, 2013

### skrat

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...