An application of two 3x3 matrices identity

Question

What is the proper context or the physical meaning for the following problem? In particular, is it related to a classical equation?

Let $A$ and $B$ be two $3\times3$ matrices, such that $\det A=1$ and $B=A+A^{-1}$, then: $$ 4+\det B=Tr(B^2). $$

Answer 1

The equation appears rather bizarre, and I would be surprised to see a "physical meaning" in any reasonable sense of the word "physical". You might note, though, that if $A$ has nonzero eigenvalues $\lambda_1, \lambda_2, \lambda_3$ (counted by algebraic multiplicity), so that $\det(A) = \lambda_1 \lambda_2 \lambda_3$, then $B$ has eigenvalues $\lambda_i + \lambda_i^{-1}$, and the equation becomes

$$ 4 + \prod_{i=1}^3 (\lambda_i + \lambda_i^{-1}) = \sum_{i=1}^3 (\lambda_i + \lambda_i^{-1})^2 \tag{1}$$

Now the difference between the two sides turns out to be $$ \dfrac{(\lambda_1 - \lambda_2 \lambda_3)(\lambda_2 - \lambda_1 \lambda_3)(\lambda_3 - \lambda_1 - \lambda_2)(1 - \lambda_1 \lambda_2 \lambda_3)}{ \lambda_1^2 \lambda_2^2 \lambda_3^2} $$

So in particular this is $0$ if $\det(A) = 1$.

EDIT: Now, how could one come up with such a bizarre equation? I'll try some reverse engineering. I think the appropriate context is the elementary symmetric functions.

Multiplying by the common denominator $\lambda_1^2 \lambda_2^2 \lambda_3^2$, equation (1) becomes an equation in symmetric polynomials:

$$ 4 \lambda_1^2 \lambda_2^2 \lambda_3^2 + \lambda_1 \lambda_2 \lambda_3 \prod_{i=1}^3 (\lambda_i^2+1) = \lambda_1^2 \lambda_2^2 \lambda_3^2 \sum_{i=1}^3 (\lambda_i^2 + 1)^2\lambda_i^{-2}$$

We can express each of these symmetric polynomials in terms of the elementary symmetric polynomials $s_1 = \lambda_1 + \lambda_2 + \lambda_3$, $s_2 = \lambda_1 \lambda_2 + \lambda_1 \lambda_3 + \lambda_2 \lambda_3$, $s_3 = \lambda_1 \lambda_2 \lambda_3$:

$$ \eqalign{ q_1 = \lambda_1^2 \lambda_2^2 \lambda_3^2 &= s_3^2 \cr q_2 = \lambda_1 \lambda_2 \lambda_3 \prod_{i=1}^3 (\lambda_i^2+1) &= s_3 (s_3^2-2 s_1 s_3+s_1^2+(s_2-1)^2)\cr q_3 = \lambda_1^2 \lambda_2^2 \lambda_3^2 \sum_{i=1}^3 (\lambda_i^2 + 1)^2\lambda_i^{-2} &= s_2^2-2 s_1 s_3+(s_1^2-2 s_2+6) s_3^2}$$ How do we get a linear combination of these that is $0$ when $s_3 = 1$? Take the equation $a q_1 + b q_2 + c q_3 = 0$, and substitute $s_3 = 1$:

$$\eqalign{ 0 = a &+ b (1 - 2 s_1 + s_1^2 + (s_2-1)^2) + c (s_2^2 - 2 s_1 + s_1^2 - 2 s_2 + 6\cr & =(b+c) s_1^2 - 2 (b+c) s_ 1+(b+c) s_2^2 - 2(b+c)s_2+a+2 b+6 c} $$ To make this $0$ for all $s_1$ and $s_2$, we take $c=-b$ and $a=4b$. With $b=1$ this corresponds to equation (1).