Find total number of possible square matrix of order $3$ with real entries, whose adjoint matrix $B$ has characteristics polynomial equation $\lambda^3-\lambda^2+\lambda+1=0$ is

Plan $$A=\begin{pmatrix}a& b& c\\ d&e &f\\g&h& i\end{pmatrix}$$

Then characteristic polynomial matrix is $|A-\lambda I|=0$

$$A=\begin{vmatrix}a-\lambda& b& c\\ d&e-\lambda &f\\g&h& i-\lambda\end{vmatrix}=0$$

$$(a-\lambda)(ei-e\lambda-i\lambda+\lambda^2-fh)-b(di-d\lambda-gf)+c(dh-eg+\lambda g)=0$$

$$(\lambda-a)(\lambda^2-(e+i)\lambda+fh-ei)+b(di-d\lambda-gf)-e(dh-eg+\lambda g)=0$$

$$\lambda^3-(\cdots +e+i)\lambda^2+\cdots \cdots =0$$

How do i solve it help me please


Hints. (I suppose that "adjoint" means classical adjoint rather than Hermitian adjoint.)

  1. Determine $\det(B)$ from the characteristic polynomial of $B$.
  2. Express $\det(B)$ in terms of $\det(A)$ using the identity $AB=\det(A)I_3$ (care should be taken to deal with the possibility that $\det(A)=0$).
  3. Using the fact that $A$ is real, argue that the results from (1) and (2) are contradicting each other. Hence the required number is zero.
  • $\begingroup$ It seems that you are mistaking "adjugate" for "adjoint". $\endgroup$ – Alex M. Jun 2 at 12:31
  • $\begingroup$ @AlexM. Please read the first line of my answer. $\endgroup$ – user1551 Jun 2 at 19:13
  • $\begingroup$ I have, and what I am saying is that you make a wrong assumption: "adjoint" means "Hermitian adjoint", because this is the only meaning of the word. What you are working with is called "adjugate", and is not what the OP is asking about. Unless, of course, the OP is not a native English speaker and is making some confusion. $\endgroup$ – Alex M. Jun 2 at 20:14
  • $\begingroup$ @AlexM. It's up to the OP to clarify what "adjoint" means. Based on the problem statement, it makes more sense to interpret it as classical adjoint. $\endgroup$ – user1551 Jun 2 at 20:36

Infinitely many, since any two similar matrices have the same characteristic polynomial.


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