# the number of possible matrix of order $3$ and whose polynomial equation of adjoint is given

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

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.