<s>[INST] Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible. If $Det(A)=k$,and $Det(A+kadj(A))=0$ Let $A$ be a $2\times2$ matrix with real entries such that $A$ is invertible. 
If $Det(A)=k$,and  
$Det(A+kadj(A))=0$,
then find the value of $Det(A-kadj(A))$
My attempt:
$Det(A+kadj(A))=0$
$Det(A)Det(A+kadj(A))=0$
$Det(A^2+kAadj(A))=0$
$Det(A^2+k^2I)=0$
Using same steps,I get $Det(A-kadj(A))=\frac{1}{k}Det(A^2-k^2I)$ 
but can't understand what to do next
 [/INST] Since you're given that $A$ is a $2\times 2$ you could brute force it...
Calculating the determinant of
$$
\begin{align*}
A^2 + k^2I = 
\begin{pmatrix}
a + k^2 & b \\
c & d + k^2
\end{pmatrix}
\end{align*}
$$
and setting it equal to zero gives that $a+d = -(1+k^2)$. Now solving for the determinant of
$$
\begin{align*}
A^2 - k^2I = 
\begin{pmatrix}
a - k^2 & b \\
c & d - k^2
\end{pmatrix}
\end{align*}
$$
gives $\frac 1k \det(A^2 -k^2I) = 2k(1 + k^2)$.
</s>