A question regarding Similar Matrix

Question

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Suppose there exist a $3 \times 3$ matrix $A$ and a $3$-dimensional column-vector $x$ such that the set of vectors $x,Ax,A^2x$ are linearly independent, and $$ A^3x = 3Ax - 2A^2 x $$

1. Let $P = [x,\ Ax, \ A^2 x].$ Find a matrix $B$ such that $A = PBP^{-1}$.
2. Compute the determinant $|A^2 + A + I|$.

Can you please tell me how to proceed ? I can only conclude from the given information that $A$ is singular.

Answer 1

It's usually easier to find $B$ such that $AP=PB$. Notice that $$\begin{eqnarray*} AP &=& A[x,Ax,A^2x]\\ &=&[Ax,A^2x,A^3x]\\ &=& [Ax,A^2x,3Ax-2A^2x]\\ &=&PB \end{eqnarray*}$$ where $B=\begin{pmatrix}0&0&0\\ 1&0&3\\ 0&1&-2\end{pmatrix}$. This means we can say $$\begin{eqnarray*} \det(A^2+A+I)&=&\det(P[B^2+B+I]P^{-1}) \\&=& \det(B^2+B+I) \\ &=& \det\begin{pmatrix}1&0&0\\ 1&4&-3\\ 1&-1&6\end{pmatrix} \\ &=& 21 \end{eqnarray*}$$