Question Number 54644 by gunawan last updated on 08/Feb/19
![A= [(3,7),((−1),(−2)) ] A^(27) +A^(31) +A^(40) =...](https://www.tinkutara.com/question/Q54644.png)
$$\mathrm{A}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{7}}\\{−\mathrm{1}}&{−\mathrm{2}}\end{bmatrix} \\ $$$${A}^{\mathrm{27}} +{A}^{\mathrm{31}} +{A}^{\mathrm{40}} =… \\ $$
Answered by kaivan.ahmadi last updated on 10/Feb/19
![A^2 = [((3 7)),((−1 −2)) ] [((3 7)),((−1 −2)) ]= [((2 7)),((−1 −3)) ] A^3 = [((−1 0)),((0 −1)) ]=−I but we know that I^2 = [((−1 0)),((0 −1)) ] [((−1 0)),((0 −1)) ]= [((1 0)),((0 1)) ]=I I^3 =I^2 I=I,…,I^n =I so we have A^4 =A^3 A=−IA A^5 =A^4 A=−IA^2 A^6 =A^5 A=−IA^3 =I A^7 =A^6 A=IA=A ⋮ A^(27) =(A^7 )^3 A^6 =A^3 I=−I^2 =−I A^(31) =(A^7 )^4 A^3 =A^4 A^3 =A^7 =A A^(40) =(A^7 )^5 A^5 =A^5 A^5 =A^(10) =A^3 A^7 =−IA so we have A^(27) +A^(31) +A^(40) = [((−1 0)),((0 −1)) ]+ [((3 7)),((−1 −2)) ]+ [((−3 −7)),((1 2)) ]= [((−1 0)),((0 −1)) ]=−I](https://www.tinkutara.com/question/Q54734.png)
$$\mathrm{A}^{\mathrm{2}} =\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{7}}\\{−\mathrm{1}\:−\mathrm{2}}\end{bmatrix}\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{7}}\\{−\mathrm{1}\:−\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\mathrm{2}\:\:\:\:\:\:\:\mathrm{7}}\\{−\mathrm{1}\:\:−\mathrm{3}}\end{bmatrix} \\ $$$$\mathrm{A}^{\mathrm{3}} =\begin{bmatrix}{−\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=−\mathrm{I} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{know}\:\mathrm{that} \\ $$$$\mathrm{I}^{\mathrm{2}} =\begin{bmatrix}{−\mathrm{1}\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:−\mathrm{1}}\end{bmatrix}\begin{bmatrix}{−\mathrm{1}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{bmatrix}=\mathrm{I} \\ $$$$\mathrm{I}^{\mathrm{3}} =\mathrm{I}^{\mathrm{2}} \mathrm{I}=\mathrm{I},\ldots,\mathrm{I}^{\mathrm{n}} =\mathrm{I} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{A}^{\mathrm{4}} =\mathrm{A}^{\mathrm{3}} \mathrm{A}=−\mathrm{IA} \\ $$$$\mathrm{A}^{\mathrm{5}} =\mathrm{A}^{\mathrm{4}} \mathrm{A}=−\mathrm{IA}^{\mathrm{2}} \\ $$$$\mathrm{A}^{\mathrm{6}} =\mathrm{A}^{\mathrm{5}} \mathrm{A}=−\mathrm{IA}^{\mathrm{3}} =\mathrm{I} \\ $$$$\mathrm{A}^{\mathrm{7}} =\mathrm{A}^{\mathrm{6}} \mathrm{A}=\mathrm{IA}=\mathrm{A} \\ $$$$\vdots \\ $$$$\mathrm{A}^{\mathrm{27}} =\left(\mathrm{A}^{\mathrm{7}} \right)^{\mathrm{3}} \mathrm{A}^{\mathrm{6}} =\mathrm{A}^{\mathrm{3}} \mathrm{I}=−\mathrm{I}^{\mathrm{2}} =−\mathrm{I} \\ $$$$\mathrm{A}^{\mathrm{31}} =\left(\mathrm{A}^{\mathrm{7}} \right)^{\mathrm{4}} \mathrm{A}^{\mathrm{3}} =\mathrm{A}^{\mathrm{4}} \mathrm{A}^{\mathrm{3}} =\mathrm{A}^{\mathrm{7}} =\mathrm{A} \\ $$$$\mathrm{A}^{\mathrm{40}} =\left(\mathrm{A}^{\mathrm{7}} \right)^{\mathrm{5}} \mathrm{A}^{\mathrm{5}} =\mathrm{A}^{\mathrm{5}} \mathrm{A}^{\mathrm{5}} =\mathrm{A}^{\mathrm{10}} =\mathrm{A}^{\mathrm{3}} \mathrm{A}^{\mathrm{7}} =−\mathrm{IA} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{A}^{\mathrm{27}} +\mathrm{A}^{\mathrm{31}} +\mathrm{A}^{\mathrm{40}} = \\ $$$$\begin{bmatrix}{−\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:−\mathrm{1}}\end{bmatrix}+\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\:\mathrm{7}}\\{−\mathrm{1}\:\:−\mathrm{2}}\end{bmatrix}+\begin{bmatrix}{−\mathrm{3}\:\:\:−\mathrm{7}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{2}}\end{bmatrix}= \\ $$$$\begin{bmatrix}{−\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:−\mathrm{1}}\end{bmatrix}=−\mathrm{I} \\ $$$$ \\ $$$$ \\ $$