Linear Algebra- Linear Transformation

Question

Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ defined by

$T(x_1,x_2,x_3) = (x_2,x_3-cx_1,-bx_2 - ax_3)$, where $a,b,c$
are fixed real numbers. Show that $T$ is a linear
transformation of $\mathbb{R}^3$ and that $A^3 + aA^2 + bA + cI = 0$,
where $A$ is the matrix of $T$ with respect to standard
basis of $\mathbb{R}^3$.

The characteristic polynomial I got is

$$A^3+aA^2+bA+cA+acI=0$$

Please help.

Answer 1

Yes, you are right. The char. polynomial is $x^3+ax^2+(b+c)x+ac$.

By Cayley - Hamilton we get $A^3+aA^2+(b+c)A+acI=0$.

A missprint occurs in your book.