Existence $a,b \in \mathbb{R}$ such that $\det(aA+bB) <0 $? Problem: Let $A,B \in \mathcal{M}_n(\mathbb{R})$, and $AB=BA, \det(AB) \not=0$

Existence $a,b \in \mathbb{R}$ such that  $\det(aA+bB) &lt;0 $?

Attempt: if $\det A &lt;0$ or $\det B&gt;0$, we chose $a=1,b=0$ or $b=1,a=0$.

If $\det A&gt;0$ and $\det B&gt;0$ then $\det A^{-1}&gt;0$ and $\det B^{-1}&gt;0$

*) if $n$ is odd number then $\det(aA+bB) &lt;0 \Leftrightarrow \det A^{-1}\det(aA+bB)=\det(a+bA^{-1}B)&lt;0$, we chose $b=1$ and $\det(a+X)$ with ($X=A^{-1}B \in \mathcal{M}(\mathbb{R})$) is a polynomial has deg is an odd. Hence, there exists $a$ such that $\det (a+X)&lt;0$

**) if $n$ is even, I can not go further :(
If $n=2k$ pick $A=B=I_n$. Then
$$\det(aA+bB)=(a+b)^{2k} \geq 0$$
Thank you very much, so that it's not true if $n=2k$.

If this problem has changed by

Problem: Let $A,B \in \mathcal{M}_n(\mathbb{R})$, and $AB=BA, det(AB) \not=0$

Existence $a,b \in \mathbb{R}, a^2+b^2 \not= 0$ such that  $det(aA+bB) =0 $?