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

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

   =>2×1-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:

     $A^{2020}=P.D^{2020}.P^{-1}$ 

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

    

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