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