Polynomial form of two variable matrix determinant function

Question

We have the function

$$f(x,y)=\det(A^2+B^2-xAB-yBA)$$

where $x, y$ are real numbers and $A, B$ are $2 \times 2$ matrices with real coefficients.

What are the coefficients of $f(x,y)$ in polynomial form?

Answer 1

$f(x,y)=\det(A^2+B^2)-Trace(ABadj(A^2+B^2))x-Trace(BAadj(A^2+B^2))y$

$+(\det(AB+BA)-2\det(AB))xy+\det(AB)(x^2+y^2).$

Answer 2

Let $C(x,y)=A^2+B^2-xAB-yBA$. Then $f(x,y)=\det(C(x,y))$ is a polynomial in $x$ and $y$, and the coeffcients of $A$ and $B$, considered as constants, because the determinant of a matrix is a polynomial in its entries. For example, if $A=B=I$, then $C(x,y)=2I-xI-yI=(2-x-y)I$, so that $\det(C(x,y))=(2-x-y)^n$, where $n$ is the size of the matrices - here $n=2$.

Explicitly, if $A$ has entries $a_1,a_2,a_3,a_4$ and $B$ has entries $b_1,b_2,b_3,b_4$, we obtain $$ \det(C(x,y))=x^2(a_1a_4b_1b_4 - a_1a_4b_2b_3 - a_2a_3b_1b_4 + a_2a_3b_2b_3) + xy( - a_1^2b_2b_3 + a_1 a_2b_1b_3 - a_1a_2b_3b_4 + a_1a_3b_1b_2 - a_1a_3b_2b_4 + 2a_1a_4b_1b_4 + a_2^2b_3^2 - a_2a_3b_1^2 - a_2a_3b_4^2 - a_2a_4b_1b_3 + a_2a_4b_3b_4 + a_3^2b_2^2 - a_3a_4b_1b_2 + a_3a_4b_2b_4 - a_4^2b_2 b_3) + x( - a_1^2a_4b_4 + a_1a_2a_3b_4 + a_1a_2a_4b_3 + a_1a_3a_4b_2 - a_1a_4^2b_1 - a_1b_1b_4^2 + a_1b_2b_3b_4 - a_2^2a_3b_3 - a_2a_3^2b_2 + a_2a_3a_4b_1 + a_2b_1b_3b_4 - a_2b_2b_3^2 + a_3b_1b_2 b_4 - a_3b_2^2b_3 - a_4b_1^2b_4 + a_4b_1b_2b_3) + y^2(a_1a_4b_1b_4 - a_1a_4b_2b_3 - a_2a_3b_1b_4 + a_2a_3b_2b_3) + y( - a_1^2a_4b_4 + a_1a_2a_3b_4 + a_1a_2a_4b_3 + a_1a_3a_4b_2 - a_1a_4^2b_1 - a_1 b_1b_4^2 + a_1b_2b_3b_4 - a_2^2a_3b_3 - a_2a_3^2b_2 + a_2a_3a_4b_1 + a_2b_1b_3b_4 - a_2b_2b_3^2 + a_3b_1b_2b_4 - a_3b_2^2b_3 - a_4b_1^2b_4 + a_4b_1b_2b_3) + a_1^2a_4^2 + a_1^2b_2b_3 + a_1^2b_4^ 2 - 2a_1a_2a_3a_4 - a_1a_2b_1b_3 - a_1a_2b_3b_4 - a_1a_3b_1b_2 - a_1a_3b_2b_4 + a_2^2a_3^2 + a_2a_3 b_1^2 + 2a_2a_3b_2b_3 + a_2a_3b_4^2 - a_2a_4b_1b_3 - a_2a_4b_3b_4 - a_3a_4b_1b_2 - a_3a_4b_2b_4 + a_4^2b_1^2 + a_4^2b_2b_3 + b_1^2b_4^2 - 2b_1b_2b_3b_4 + b_2^2b_3^2. $$