Let $A\in\mathbb{R}^{2\times 2}$ be such that $\det(A)=d\ne0$ and $\det(A+d\cdot\text{Adj}(A))=0$. Evaluate $\det(A-d\cdot\text{Adj}(A))$. 
Let $A$ be a $2\times2$ matrix with real entries such that $\det(A)=d\ne0$ and $\det(A+d\cdot\text{Adj}(A))=0$. Evaluate $\det(A-d\cdot\text{Adj}(A))$.

My Attempt
I am multiplying the matrices.
$(A+d\cdot\text{Adj}(A))\cdot(A-d\cdot\text{Adj}(A))$
Can I say this matrix multiplication is equal to zero?
That is, I am asking if $AB=C$ and $\det(A)=0$ and $\det(B)\ne0$ then can $C$ matrix be a non-null matrix?
After matrix multiplication, I am getting $A^2-d^2\text{Adj}^2(A)$. I am just not sure if this is zero.
This question already has an answer here. More than the answer, I liked the OP's approach. I wish someone could finish that.
I hope to get an answer that uses just the basic properties of matrices and determinants, as eigen values/Cayley Hamilton are not in syllabus.
 A: A solution that uses eigenvalues (I know the OP asked one without that, but I think that this is much more elegant that a brute force approach, which seems the be solution asked here).
Since $A$ is invertible, then we have that $A^{-1}=\mathrm{Adj}(A)/d$. It is also known that the same basis change make $A$ in Jordan form (or in diagonal form when possible) and its inverse in upper-triangular form. We also note that the determinant is invariant by basis change.
So, let $P$ be this basis change such that $A$ is the Jordan form
$$J=\begin{bmatrix}\alpha & 1\\0 & \alpha\end{bmatrix},$$
where we have that $d=\alpha^2\ne0$ by assumption. This yields
$$\det(J+d\cdot\text{Adj}(J))=0.$$
Expanding this yields the expression $(1+d)^2\alpha^2=0$. Since $\alpha\ne0$, by assumption, then we must have that $(1+d)^2=0$. So, we need $d=-1$, which means that $\alpha=\pm i $, which contradicts the fact that the matrix is real. This means that the matrix $A$ must be diagonalizable.
Therefore, let us consider the diagonal form of $A$ given by
$$D=\begin{bmatrix}d_1 & 0\\0 & d_2\end{bmatrix}$$
and evaluating $\det(D+d\cdot\text{Adj}(D))=0$ yields the condition
$$d_1d_2(1+d_1^2)(1+d_2^2)=0.$$
This implies that $d_1=i$ and $d_2=-i$ (or the other way round) since the matrix $A$ is real (so its eigenvalues must be complex conjugate). This implies that $d=d_1d_2=1$.
We then have that
$$D-d\cdot\text{Adj}(D)=\begin{bmatrix}2i & 0\\0 & -2i\end{bmatrix}$$
and we finally obtain that $\det(D-d\cdot\text{Adj}(D))=4$.
A: 
That is, I am asking if $AB=C$ and $det(A)=0$ and $det(B)≠0$ then can $C$ matrix be a non-null matrix?

It will definitely be a non null matrix (assuming all of $A, B, C$ are square matrices). Because $B$ is invertible, we can write $A=CB^{-1}$ and if $C$ is a null matrix then $A$ will be a null matrix which is not necessarily true. So $C$ can't be a null matrix in this particular case.
My Solution:
Using very basic concepts of matrix algebra this is one approach for $\mathbb{R}^{2×2}$ matrices
Let $A= \begin{bmatrix}x & y\\z & w\end{bmatrix}$
Then $det(A)=wx-yz=d$
It can be verified easily that $$Adj(A)=\begin{bmatrix}w & -y\\-z & x\end{bmatrix}$$
Now $|A+dAdj(A)|=0$ implies that
$$det(\begin{bmatrix}x+dw & y-dy\\z-dz & w+dx\end{bmatrix})=0$$
If we do the algebra we get
$$(x+dw)(w+dx)-(z-dz)(y-dy)=0$$
On expanding and rearranging terms we get
$$\implies (xw-yz)(1+d^2)+d(w^2+x^2+2yz)=0$$
Identifying $(xw-yz)$ as $d$ and noting that $d\neq0$, our equation simplifies to
$$1+d^2+w^2+x^2+2yz=0$$
Now adding and subtracting $2wx$ and substituting $d$ for $(wx-yz)$ reduces the equation to
$$(d-1)^2+(w+x)^2=0$$
Sum of squares is zero implies that all squares are themselves zero (real number case)
$$\implies d=1; w=-x$$
Finally we have to calculate $det(A-dAdj(A))$, using $d=1$ we can write
$$det(A-dAdj(A))=det(A-Adj(A))$$
$$=det(\begin{bmatrix}x & y\\z & w\end{bmatrix}-\begin{bmatrix}w & -y\\-z & x\end{bmatrix})$$
As we know that $x=-w$ the determinant reduces to
$$det(\begin{bmatrix}2x & 2y\\2z & 2w\end{bmatrix})=det(2A)=2^2det(A)$$
$$=4$$
That's it.
Really nice question.
A: Observe that $A^2+d^2I $ is singular, because $\det(A^2+d^2I)=\det(A)\det(A+d\operatorname{adj}(A))=0$. Hence there exists some nonzero vector $x$ such that
\begin{align}
A^2x&=-d^2x,\tag{1}\\
A^2(Ax)&=A(A^2x)=-d^2Ax.\tag{2}
\end{align}
The two vectors $x$ and $Ax$ are not parallel to each other, otherwise, we would have $Ax=cx$ for some real scalar $c$ and hence $c^2x=A^2x=-d^2x$, which is impossible. Therefore $B=\pmatrix{x&Ax}$ is a nonsingular matrix. But then $(1)$ and $(2)$ implies that $A^2B=-d^2B$. Hence
\begin{align}
A^2&=-d^2I,
\ \text{ i.e., }
\operatorname{adj}(A)=-\frac{A}{d}.
\end{align}
Now $d^2=\det(A^2)=\det(-d^2I)=d^4$ and in turn $d=\pm1$. If $d=-1$, then $\operatorname{adj}(A)=A$, but no $A$ will satisfy these two conditions because
$$
A=\pmatrix{\alpha&\beta\\ \gamma&\delta}=\pmatrix{\delta&-\beta\\ -\gamma&\alpha}=\operatorname{adj}(A)
\ \text{ and }\ \alpha\delta-\beta\gamma=d=-1
$$
give the insolvable equation that $\alpha^2=-1$.
Therefore $d=1$, $\operatorname{adj}(A)=-A$ and $\det\left(A-d\operatorname{adj}(A)\right)=\det(2A)=4d=4$.
