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.
$\newcommand{\adj}{\operatorname{adj} (A)}$ 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.