<s>[INST] For any $2$ x $2$ matrix $A$, does there always exist a $2$ x $2$ matrix $B$ such that det($A+B$) = det($A$) + det($B$)? For each invertible $2$ x $2$ matrix $A$, does there exist an invertible $2$ x $2$ matrix $B$ such that the following conditions hold? 
(1) $A + B$ is invertible
(2) det($A+B$) = det($A$) + det($B$)
I know that for $2$ x $2$ matrices det($A+B$) = det($A$) + det($B$) + tr($A$)tr($B$) - tr($AB$). So this means tr($A$)tr($B$) = tr($AB$). Right now I am having trouble proving that there exists a $B$ that satisfies this equation as well as condition (1). 
 [/INST] Since we aretalking of $2\times2$ matrices, its slightly easier to write down explicitly.
So $\det(A+B)=\det(A)+\det(B)$ happens when $(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})=(a_{11}a_{22}-a_{12}a_{21})+(b_{11}b_{22}-b_{12}b_{21})$
$\implies a_{11}b_{22}+b_{11}a_{22}-a_{12}b_{21}-b_{12}a_{21}=0=\det\begin{bmatrix}a_{11}\,a_{12}\\b_{21}\,b_{22}\end{bmatrix}+\det\begin{bmatrix}b_{11}\,b_{12}\\a_{21}\,a_{22}\end{bmatrix}$
Choosing $b_{11} = -a_{21}, b_{12} = -a_{22}, b_{21} = a_{11}$ and $b_{22} = a_{12}$ we get our matrix B.
The key point here is to choose $b_{ij}$ such that the two determinants are zero. Part 1 follows by considering matrix B such that $\det(A+B)\neq0$
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Example for $\det(A+B)\neq0$ consider the matrix $A=\begin{bmatrix}4\: 5 \\7\: 9\end{bmatrix}$. We can choose matrix B as $\begin{bmatrix}-7 -9 \\4 \:\: 5\end{bmatrix}$.
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