Question

2. Use Cayley-Hamilton theorem to find A6 − 5A5 + 8A4 − 2A3 − 9A2 + 31A − 36I,

when A=





1 0 3

2 1 −1

1 −1 1





Answer 1

Find the characteristic polynomial of the matrix A:

$p_A(\lambda)=\det\begin{vmatrix} 1-\lambda & 0 & 3\\ 2 & 1-\lambda & -1\\ 1 & -1 & 1-\lambda \end{vmatrix}=(1-\lambda)^3-6-4(1-\lambda)$

$-p_A(\lambda)=\lambda^3-3\lambda^2-\lambda+9$

Cayley-Hamilton theorem claims that $p_A(A)=0$, that is, $A^3-3A^2-A+9I=0$.

Let's divide $A^6 − 5A^5 + 8A^4 − 2A^3 − 9A^2 + 31A − 36I$ by $A^3-3A^2-A+9I$ with remainder. We obtain:

$A^6 − 5A^5 + 8A^4 − 2A^3 − 9A^2 + 31A − 36I=$

$=(A^3-3A^2-A+9I)(A^3 − 2^2 + 3A − 4I)=0$

Answer. 0.