Question

Using Cayley — Hamilton theorem, evaluate A^(7) if A=[ 3 2
-1. 0 ]

Answer 1

The characteristics equation is $|A-\lambda I|=0$

$\begin{vmatrix} 3-\lambda & 2\\ -1 & \lambda \end{vmatrix}=0\\ -\lambda(3-\lambda)+2=0\\ -3\lambda+\lambda^2+2=0\\ \lambda^2-3\lambda+2=0$

By Cayley-Hamilton Theorem,

$A^2-3A+2I=0\\ \implies A^2=3A-2I$

$A^7=A^4.A^3\\ A^7=(A^2)^2.A^2.A\\ A^7=(3A-2I) ^2.(3A-2I).A\\ A^7=(9A^2-12A+4I)(3A^2-2A)\\ A^7=(9(3A-2I)-12A+4I)(3(3A-3I)-2A) \\ A^7=(27A-18I-12A+4I)(9A-6I-2A) \\ A^7=(15A+4I)(7A-6I) \\ A^7=(105A^2-188A+84I)\\ A^7=(105(3A-2I)-188A+84I)\\ A^7=315A-210A-188A+84I\\ A^7=127A-126I\\ A^7=127\begin{pmatrix} 3&2 \\ -1&0 \end{pmatrix}-126\begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix}\\ A^7=\begin{pmatrix} 381&254\\ -127&0 \end{pmatrix}-\begin{pmatrix} 126&0 \\ 0&126 \end{pmatrix}\\ A^7=\begin{pmatrix} 255&254 \\ -127&-126 \end{pmatrix}$