Cho ma trận A sau: $\left[\begin{array}{ccc}3&0\\8&-1\\\end{array}\right]$ Tính $A^{2020}$

Đáp án:

 $A^{2020}$=$\left(\begin{array}{ccc}3^{2020}&0\\2+2.3^{2020}&-1\end{array}\right)$ 

Giải thích các bước giải:

Ta có:

$\left|\begin{array}{ccc}3-lamda&0\\8&-1-lamda\end{array}\right|=0$

<=>\(\left[ \begin{array}{l}lamda=-1\\lamda=3\end{array} \right.\) 

=>A chéo hoá được.

lamda=-1:A-lamda.I=$\left(\begin{array}{ccc}4&0\\8&0\end{array}\right)$ 

=>x1=0=>u1=(0;1)

lamda=3:A-lamda.I=$\left(\begin{array}{ccc}0&0\\8&-4\end{array}\right)$ 

=>2x1-x2=0=>u2=(1;2)

Suy ra:

P=$\left(\begin{array}{ccc}0&1\\1&2\end{array}\right)$ 

D=$\left(\begin{array}{ccc}-1&0\\0&3\end{array}\right)$ 

và $P^{-1}AP=D$

Vậy ta có:

  $A^{2020}=P.D^{2020}.P^{-1}$ 

=$\left(\begin{array}{ccc}0&1\\1&2\end{array}\right)$.$\left(\begin{array}{ccc}-1&0\\0&3^{2020}\end{array}\right)$.$\left(\begin{array}{ccc}-2&1\\1&0\end{array}\right)$=$\left(\begin{array}{ccc}3^{2020}&0\\2+2.3^{2020}&-1\end{array}\right)$