Question

Questão envolvendo matrizes. Dadas as matrizes A = $\left[\begin{array}{ccc}8&8\\8&8\end{array}\right]$ e $A^{k}$= $\left[\begin{array}{ccc} 2^{111} & 2^{111} \\ 2^{111} & 2^{111} \end{array}\right]$ . Então, o valor de k é igual a:

a) 26
b) 27
c) 28
d) 29
e) 30

Answer 1

$A^{1}=\left[\begin{array}{cc}8&8\\8&8\end{array}\right]=\left[\begin{array}{cc}2^{3}&2^{3}\\2^{3}&2^{3}\end{array}\right]$

Vamos tentar achar uma lei para as "potências" de A:

Achando A²:

$A^{2}=A\cdot A=\left[\begin{array}{cc}8&8\\8&8\end{array}\right]\cdot\left[\begin{array}{cc}8&8\\8&8\end{array}\right]=\left[\begin{array}{cc}(8\cdot8+8\cdot8)&(8\cdot8+8\cdot8)\\(8\cdot8+8\cdot8)&(8\cdot8+8\cdot8)\end{array}\right]\\\\\\A^{2}=\left[\begin{array}{cc}128&128\\128&128\end{array}\right]=\left[\begin{array}{cc}2^{7}&2^{7}\\2^{7}&2^{7}\end{array}\right]$

Achando A³:

$A^{3}=A^{2}\cdot A=\left[\begin{array}{cc}2^{7}&2^{7}\\2^{7}&2^{7}\end{array}\right]\cdot\left[\begin{array}{cc}2^{3}&2^{3}\\2^{3}&2^{3}\end{array}\right]\\\\\\A^{3}=\left[\begin{array}{cc}(2^{7}\cdot2^{3}+2^{7}\cdot2^{3})&(2^{7}\cdot2^{3}+2^{7}\cdot2^{3})\\(2^{7}\cdot2^{3}+2^{7}\cdot2^{3})&(2^{7}\cdot2^{3}+2^{7}\cdot2^{3})\end{array}\right]\\\\\\A^{3}=\left[\begin{array}{cc}2^{11}&2^{11}\\2^{11}&2^{11}\end{array}\right]$

Achando A⁴:

$A^{4}=A^{3}\cdot A=\left[\begin{array}{cc}2^{11}&2^{11}\\2^{11}&2^{11}\end{array}\right]\cdot\left[\begin{array}{cc}2^{3}&2^{3}\\2^{3}&2^{3}\end{array}\right]\\\\\\A^{4}=\left[\begin{array}{cc}(2^{11}\cdot2^{3}+2^{11}\cdot2^{3})&(2^{11}\cdot2^{3}+2^{11}\cdot2^{3})\\(2^{11}\cdot2^{3}+2^{11}\cdot2^{3})&(2^{11}\cdot2^{3}+2^{11}\cdot2^{3})\end{array}\right]\\\\\\A^{4}=\left[\begin{array}{cc}2^{15}&2^{15}\\2^{15}&2^{15}\end{array}\right]$

Podemos perceber que os expoentes das entradas dessas matrizes estão em P.A:

$P.A~(3,~7,~11,~15,~...)$

A razão dessa P.A é 4. Achando o termo geral:

$a_{n}=a_{1}+(n-1)\cdot r\\a_{n}=3+(n-1)\cdot4\\a_{n}=3+4n-4\\a_{n}=4n-1$

Veja que n representa o expoente da matriz. Sendo n = k, temos:

$a_{k}=4k-1$

Sabemos que 'ak' = 111 (expoente de 2 na matriz A^k). Achando k:

$111=4k-1\\111+1=4k\\4k=112\\k=112/4\\k=28$

Letra C