determinants of matrix with adjoints of order 2 
Let $A$ be a square matrix of order $2$ with $\lvert A \rvert\ne 0$ such that $\big\lvert A+\lvert A \rvert \operatorname{adj} (A)\big\rvert=0$, then the value of $$\big\lvert A-\lvert A \rvert \operatorname{adj} (A)\big\rvert$$ is :

My attempt
I took the matrix as
$$
        \begin{bmatrix}
        a & b  \\
        c & d  \\
        \end{bmatrix}
$$

and its adjoint to be
$$
        \begin{bmatrix}
        d & -b  \\
        -c & a  \\
        \end{bmatrix}
$$
and substituted in above given equation,took its determinant to be = 0
Yet I could not solve it
(Background) I am 12th grader and I know about adjoint,inverse,determinant and the other basics.
However I do NOT know about eigenvalues and eigenvectors.
 A: $\newcommand{\adj}{\operatorname{adj} (A)}$
Let $\det(A) = x$. Then we have that:
$$B = A + \det(A) * \adj = \begin{bmatrix} a+xd & b-xb \\ c-xc & d+xa\end{bmatrix},$$
which implies 
$$\det(B) = a^2x + adx^2 + ad + d^2x - bcx^2 + 2bcx - bc \tag 1.$$
We also know that $\det(B) =0 $.
Similarly, 
$$C = A - \det(A) * \adj = \begin{bmatrix} a-xd & b+xb \\ c+xc & d-xa\end{bmatrix},$$
which implies 
$$\det(C) = - a^2x + adx^2 + ad - d^2x - bcx^2 - 2bcx - bc
\tag 2.$$
Adding $(1),(2)$ yields:
$$\det(B) + \det(  C )\begin{array}[t]{l}= 2ad - 2bc + 2adx^2 - 2bcx^2 = 2(ad-bc)+2x^2(ad-bc)
\\= 2\det(A)+2\det^3(A).  
\end{array}$$ 
But $$\det(B) = 0 \implies \det( C )  = 2\det(A) + 2(\det(A))^3 = 2\det(A) [1+\det^2(A)]\tag{3}.
$$

Some more findings:
Subtracting $(1),(2)$ and taking advantage that $\det(B) = 0$ yields:
$$\det( C ) = 2\det(A) (-a^2-d^2-2bc)\tag{4}$$
Combining $(3),(4)$ yields:
$$ \det^2(A) = -a^2-d^2-2bc -1.$$
(This means that $bc<0$).
Also, we might need the equation:
$$\begin{aligned}(ad-bc)^2 &= -a^2-d^2-2bc-1\\
a^2d^2+a^2+d^2+(bc+1)^2 &=2abcd
\end{aligned}$$
Because the LHS is non - negative and $bc$ is negative, it must hold that $ad$ is negative as well.  
