Evaluate $|A – |A|\operatorname{adj}A|$ if $|A + |A|\operatorname{adj}A| = 0$ where A is a non-singular matrix of order 2 
$A$ is a non-singular square matrix of order $2$ such that $$|A + |A|\operatorname{adj}A| = 0$$ where $\operatorname{adj}A$ represents adjoint of matrix $A$, and $|A|$ represents $\det(A)$ .
Evaluate $$|A – |A|\operatorname{adj}A|.$$

The answer given is $4$.
My Attempt:
$|A + |A|\operatorname{adj}A| = 0$
$|A+|A|^2A^{-1}|=0$
$|A+|A|^2A^{-1}||A|=0$
$|A^2+|A|^2I|=0$
From here I was not able to proceed further. Is there a solution without the use of concepts of eigenvalues or Cayley-Hamilton Theorem.
 A: As you have correctly determined, we have $|A^2 + |A|^2I| = 0$. That is, the matrix $A^2 + |A|^2 I$ is singular.  It follows that there exists a vector $x \in \Bbb R^2$ such that $(A^2 + |A|^2I)x = 0$, which is to say that $A^2 x = -|A|^2x$.
Note that if $Ax = \lambda x$ (for $\lambda \in \Bbb R$), then it would follow that $A^2x = -|A|^2x = \lambda^2 x$; this is impossible since $\lambda^2 \geq 0$. Thus, the vectors $x$ and $Ax$ must be linearly independent.  
We now note that for $v = x$ and $v = Ax$, we have $A^2 v = -|A|^2 v$. Since $x,Ax$ form a basis of $\Bbb R^2$, we can conclude that $A^2 = -|A|^2I$. 
We note that
$$
\det(A^2)
 = \det[-|A|^2I] \implies |A|^2 = |A|^4 \implies |A|=1 \text{ or } |A| = -1.
$$
From there, we can deduce that $|A| > 0$  (so that $|A| = 1$) as follows. Suppose for the purpose of contradiction that $|A| < 0$. Consider the polynomial $p(t) = |tI - A|$; we see that $p(0) < 0$. On the other hand, $\lim_{t \to \infty} \frac{p(t)}{t^2} = \lim_{t \to \infty} |I - \frac 1t A| = |I| = 1$. So, $\lim_{t \to \infty}p(t) = \infty$. By the intermediate value theorem, there exists a $\lambda$ such that $p(\lambda) = 0$. However, as we established in the second paragraph, this is impossible.
It follows that 
$$
\begin{align}
|A - |A|\operatorname{adj}(A)| &= |A - |A|^2 A^{-1}| 
\\ & = |A^{-1}|\cdot \big|A^2 - |A|^2I\big |
\\ & =|A|^{-1}\cdot \big |A^2 - |A|^2I\big |
\\ & = |A|^{-1}\cdot \big |-|A|^2 I - |A|^2I\big | 
\\ & = 
|-2I| = 4.
\end{align}
$$

Alternative approach: as we have established, $A^2 = -|A|^2 I$. In particular, we see that $A$ is a multiple of the identity. We compute
$$
A^2 = \pmatrix{a&b\\ c&d}^2 = \pmatrix{a^2 + bc & ab + bd\\ ac + cd & bc + d^2}.
$$
Since $A^2$ is diagonal, we must have $ab + bd = (a+d)b = 0$ and $ac + cd = c(a+d) = 0$.  We must either have $a + d = 0$ or $b = c = 0$.  The second case is impossible since $A^2$ needs to have negative entries.  That is, we have $d = -a$.
Note that if $c = 0$ or $b = 0$, then $A^2$ must have non-negative diagonal entries. So, 
We find that
$$
|A| = ad - bc = -a^2 - bc = -(a^2 + bc).
$$
So, $-|A|$ is equal to the upper-left entry of $A^2$. Since $A^2 = -|A|^2I$, this tells us that $-|A| = -|A|^2 I$. Since $|A| \neq 0$, we can conclude that $|A| = 1$, as was desired.  We can now follow the sequence of equations from the end of the previous proof.

A proof using complex eigenvalues:
We see that 
$$
\begin{align}
0 &= \det(A^2 + \det(A)^2 I) = \det(A - i\det(A) I)\det(A + i\det(A) I) 
\\ & = \det(A - i\det(A) I) \overline{\det(A - i\det(A) I)} = 
|\det(A - i\det(A) I)|^2.
\end{align}
$$
So, $i\det(A)$ is an eigenvalue of $A$. By a symmetric argument, $-i\det(A)$ is an eigenvalue of $A$.  However, the determinant is the product of eigenvalues, so we have
$$
\det(A) = -i\det(A) \cdot i \det(A) = \det(A)^2,
$$
from which it follows that $\det(A) = 1$ (since we have excluded the possibility that $\det(A) = 0$).  From there, the proof proceeds as above.
A: Presumably $A$ is real. Otherwise the answer is incorrect. E.g. $\det\left(A-\operatorname{adj}(A)I\right)=0$ when $A=iI$ over $\mathbb C$.
Let $B=A^2$. Then
$$
\det\left(A^2+\det(A^2)I\right)
=\det(A)\det\left(A+\det(A)\operatorname{adj}(A)\right)=0.
$$
Therefore $\lambda_1=-\det(B)$ is an eigenvalue of $B$ and the other eigenvalue is $\lambda_2=-1$ (because $\lambda_1\lambda_2=\det(B)$).
Since $A$ is real and $-1$ is an eigenvalue of $A^2$, $A$ must possess a conjugate pair of eigenvalues. Therefore the eigenvalues of $A$ are $i$ and $-i$. It follows that $\det(A)=1$ and
\begin{aligned}
\det\left(A-\det(A)\operatorname{adj}(A)I\right)
&=\frac{\det\left(A^2-\det(A^2)I\right)}{\det(A)}\\
&=\det(A^2-I)\\
&=(i^2-1)((-i)^2-1)=4.
\end{aligned}
