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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).

Since we aretalking of $2times2$ matrices, its slightly easier to write down explicitly.

So $det(A+B)=det(A)+det(B)$ happens when $(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})=(a_{11}a_{22}-a_{12}a_{21})+(b_{11}b_{22}-b_{12}b_{21})$

$implies a_{11}b_{22}+b_{11}a_{22}-a_{12}b_{21}-b_{12}a_{21}=0=detbegin{bmatrix}a_{11},a_{12}\b_{21},b_{22}end{bmatrix}+detbegin{bmatrix}b_{11},b_{12}\a_{21},a_{22}end{bmatrix}$

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 $det(A+B)neq0$

$rule{17cm}{1pt}$

Example for $det(A+B)neq0$ consider the matrix $A=begin{bmatrix}4: 5 \7: 9end{bmatrix}$. We can choose matrix B as $begin{bmatrix}-7 -9 \4 :: 5end{bmatrix}$.