Evaluating Determinant using Cayley-Hamilton Theorem A is a non singular square matrix of order 2 such that |A + |A|adjA| = 0, where adjA represents adjoint of matrix A, and |A| represents det(A) (determinant of matrix A) 
Evaluate |A – |A|adjA|. 
I believe this can be solved using Cayley Hamilton Theorem, but I'm not sure exactly how one would proceed to find the value of the required determinant. The motivation behind my thought was appearance of |A - kI| = 0 in the determinant to be evaluated, where k is a constant. 
#1 Of course, |A|adjA can be reduced to |A|² A-1, leaving us with |A + |A|² A-1|. I don't know how to take it from here. A detailed solution using Cayley Hamilton theorem would be appreciated. 
#2 I guess that for a square matrix of order n, the required determinant would be equal to n². Could someone please help prove or disprove the result, by generalising for a nxn matrix? 
Thanks a lot! 
 A: $\text{adj A}=|A|A^{-1}$, at least when $A$ is non-singular, so
$$A-|A|\text{adj A}=A-|A|^2A^{-1}=A^{-1}(A-|A|I)(A+|A|I).$$
The determinant of this is
$$|A|^{-1}\chi(|A|)\chi(-|A|)$$
where $\chi(x)=\det(A-xI)$ is the characteristic polynomial of $A$.
If we write $\chi(x)=x^2-tx+|A|$ (where $t$ is the trace of $A$)
then $\chi(\pm |A|)=|A|(1\mp t+|A|)$
etc.
A: We can take your expression $|A + |A|^2A^{-1}| = 0$, noting that we have assumed that $A$ is invertible, and find that
$$
|A| \cdot |A + |A|^2A^{-1}| = 0 \implies\\
|(A + |A|^2A^{-1})A| = 0 \implies\\
|A^2 + |A|^2I| = 0
$$
In other words, if $A$ is invertible, then the condition given is equivalent to stating that $-|A|^2$ is an eigenvalue of $A^2$, which is true if and only if either $i|A|$ or $-i|A|$ is an eigenvalue of $A$ (if $A$ is a real matrix, then both must be eigenvalues).  Since the product of the eigenvalues of $A$ is $|A|$, we can determine both eigenvalues of $A$, which allows us to evaluate
$$
|A - |A|\operatorname{adj}(A)| = \frac{|A^2 - |A|^2I|}{|A|}
$$
In the $n \times n$ case, we should not expect that knowing $|A + |A|^2A^{-1}| = 0$ is enough information to determine $|A - |A|\operatorname{adj}(A)|$.  The case in which $|A| = 0$ should be handled separately, but perhaps continuity arguments suffice.

So, we have determined that the eigenvalues of $A$ are either $i|A|$ and $-i$ or $-iA$ and $i$.  We compute
$$
|A^2 - |A|^2 I| = \prod_{i=1}^2(\lambda_i^2 - |A|^2) = (-1 - |A|^2)(-|A|^2 - |A|^2) = 2|A|^2(1 + |A|^2)
$$
