Let $A, B \in \mathbb{M}_3(\mathbb{C})$ such that $\det(A) = \det(B) = 1.$ Show that $\det(A + \sqrt{2}\cdot B) \not= 0.$ Let $A, B \in \mathbb{M}_3(\mathbb{C})$ such that $\det(A) = \det(B) = 1.$ Show that $\det(A + \sqrt{2}\cdot B) \not = 0.$
I tried a bunch of techniques but nothing seems to work. Usually one can factorize expressions using complex numbers but there the matrices do not commute. Perhaps solutions using eigenvalues can be used. Since the determinant is non-zero we can claim that the eigenvalues are non-zero. But what does that say about the sum, I am not sure. I also tried to show that the columns of the matrix $A+\sqrt{2}B$ are independent but that get's messy. And lastly I tried to construct the following linear polynomial: 
$$|A+xB|=p(x)$$ then $p(x)=ax+b$ and since $p(0)=1$ we have that $p(x)=ax+1.$ But what should I do after this, I am again not sure. Any ideas/hints will be much appreciated. 
 A: This is not true. There exist matrices with determinant $1$ such that the first row of $A$ is $(-\sqrt 2,0,0)$ and the first row of $B$ is $(1,0,0)$.
Are you sure the problem said  $A, B \in \mathbb{M}_3(\mathbb{C})$ ? It's been hinted that the problem is right for  $A, B \in \mathbb{M}_3(\mathbb{Z})$. That seemed totally obvious at first, then thinking about it it wasn't so obvious for a minute. I think the following's correct - there's probably a less clumsy way:
If we had a counterexample with integer entries then $M=AB^{-1}$ would be a matrix with integer entries, determinant $1$, with an eigenvalue $-\sqrt 2$. Since the minimal polynomial for $-\sqrt 2$ over $\mathbb Q$ is $t^2-2$ it follows that $t^2-2$ divides the characteristic polynomial for $M$, so $\sqrt 2$ is also an eigenvalue. Since $\det(M)=1$ the third eigenvalue is $-1/2$, but now the sum of the eigenvalues is not an integer.
In fact there is a matrix $M$ with rational entries having determinant $1$ and $-\sqrt 2$ for an eigenvalue, namely $$M=\begin{bmatrix}0&2&0
\\1&0&0
\\0&0&-1/2\end{bmatrix}.$$And sure enough if $A=M$, $B=I$ then $A$ and $B$ have determinant $1$ while $\det(A+\sqrt 2B)=0$.
A: It is better to replace $\mathbb{C}$ by $\mathbb{Q}$ and $\sqrt{2}$ by $\pi$.
Proposition. Let $K$ be a field and $u$ be transcendental over $K$. Let $A,B\in M_n(K)$ s.t. $\det(B)\not=0$. Then $\det(A+uB)\not= 0$. 
Proof. $\det(A+uB)=u^n \det(B)+\sum_{i<n}a_ix^i$ where $\det(B)\in K\setminus 0$ and $a_i\in K$. If $\det(A+uB)=0$, then $u$ is algebraic over $K$, a contradiction.
