Prove statement about determinants.

Question

$A$ is a $3\times 3$ matrix over $\mathbb{R}$, I want to show that if $$\det(A + I_3)=\det(A+2I_3),$$ then $$2\det(A+I_3) + \det(A-I_3) + 6 = 3\det A.$$

Can you help me?

Answer 1

If $a,b,c$ are the eigenvalues of $A$, then $\det(A)=abc$, and $\det(A+dI)=(a+d)(b+d)(c+d)$.

So we are given that $$ (a+1)(b+1)(c+1)=\det (A+I)=\det (A+2I)=(a+2)(b+2)(c+2), $$ and hence $$ (ab+bc+ca)+(a+b+c)+1=2(ab+bc+ca)+4(a+b+c)+8, $$ or simpler $$ (ab+bc+ca)+3(a+b+c)+7=0.\tag{1} $$

We need to show that $$ 2\det (A+I)+\det (A-I)+6=3\det(A) $$ which is equivalent to $$ 2(abc+(ab+bc+ca)+(a+b+c)+1)+(abc-(ab+bc+ca)+(a+b+c)-1)+6=3abc, $$ or simpler $$ (ab+bc+ca)+3(a+b+c)+7=0\tag{2} $$ Clearly $(1)$ and $(2)$ are identical and hence we are done.

Answer 2

First show that you can assume $A$ diagonal. Then the question reduces to proving a basic identity between two polynomials.

More explicitely, you can write $\det (A+ x I_3) = \sigma_3 + x \sigma_2 + x^2 \sigma_1 + x^3$, where $\sigma_3=a_1 a_2 a_3$, $\sigma_2 = a_1 a_2 + a_1 a_3 + a_2 a_3$ and $\sigma_1 = a_1 + a_2 + a_3$, if the $a_i$ are the diagonal elements of $A$. The assumption gives you $\sigma_2 = -3 \sigma_1 - 7$, and the wanted equality follows readily.