# Thread: Determinant proof

1. ## Determinant proof

Q: Prove that if $|A|=|B|\neq{0}$ and A and B are of the same size, then there exists a matrix C such that $|C|=1$ and A=CB.

A: Since $|A|=|B|\neq{0}$ we know $A^{-1}$ and $B^{-1}$ exists. So,

$|A^{-1}||A|=|A^{-1}||B|$

$1=|A^{-1}||B|$

Now, let $|C|=1$

$|C|=|A^{-1}||B|$

Since $|A|=|B|$ and $|C|=|A^{-1}||A|$

$|A^{-1}||A|=|A^{-1}||B|$

$(|A||A^{-1}|)|A|=(|A||A^{-1}|)|B|$

$1|A|=(|A||A^{-1}|)|B|$

$|A|=|C||B|$

$\therefore$ $A=CB$

Does that work? I guess A=I, B=I, and C=I. Do I need to show more?

2. I presume from the thread title that $|A|$ is the determinant of the matrix $A$. I'm afraid your arguments only allow you to say that if $A,B,C$ are matrices with $|A|=|B|\ne 0$ and $|C|=1$, then $|A|=|C||B|=|B|$ which is vacuously true. That does not imply $A=CB$. Example: $A=B=I, C=\begin{pmatrix}1 &1 \\ 0 & 1\end{pmatrix}$.

You started at the right place: Since $|A|,|B|$ are nonzero, they are invertible. Then $A=AI=A(B^{-1}B)=(AB^{-1})B$, so if we can show $|AB^{-1}|=1$, then we have found our $C$!

Now we need some properties of the determinant, namely that $|A^{-1}|=|A|^{-1}$ and $|AB|=|A||B|$. Then it is easy to see that $|AB^{-1}|=|A||B^{-1}|=|A||B|^{-1}=1$ as required.

3. Then it is easy to see that $|A^{-1}B|=|A^{-1}||B|=|A|^{-1}|A|=1$ as required.
I'm assuming you meant to show $|AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1$ which gives us the $|C|$ we were searching for. Correct? I guess it doesn't really matter what order since A=B and these are square matrices, but it looks cleaner.

4. Originally Posted by Danneedshelp
I'm assuming you meant to show $|AB^{-1}|=|A||B^{-1}|=|A||A|^{-1}=1$ which gives us the $|C|$ we were searching for. Correct? I guess it doesn't really matter what order since A=B and these are square matrices, but it looks cleaner.
Precisely