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A-1-2-0-1-find-A-n-




Question Number 69954 by 20190927 last updated on 29/Sep/19
A= (((1       2)),((0        1)) )    find A^n
$$\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:\:\mathrm{find}\:\mathrm{A}^{\mathrm{n}} \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
A = (((1       0)),((0        1)) ) + (((0       2)),((0         0)) )  =I +J  we have J^2 = (((0        2)),((0        0)) ) . (((0        2)),((0         0)) ) = (((0           0)),((0            0)) )  A^n =(J +I)^n  =Σ_(k=0) ^n  C_n ^k  J^k  I^(n−k)  =Σ_(k=0) ^n  C_n ^k  J^k   =I  +nJ  = (((1       0)),((0        1)) )  + (((0     2n)),((0        0)) ) = (((1      2n)),((0         1)) )
$${A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:={I}\:+{J} \\ $$$${we}\:{have}\:{J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$${A}^{{n}} =\left({J}\:+{I}\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{J}^{{k}} \:{I}^{{n}−{k}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{J}^{{k}} \\ $$$$={I}\:\:+{nJ}\:\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{2}{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{2}{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$
Commented by kaivan.ahmadi last updated on 29/Sep/19
A^2 =A.A= (((1   2)),((0     1)) ) (((1    2)),((0     1)) )= (((1    4)),((0     1)) )  A^3 =A.A^2 = (((1   2)),((0    1)) ) (((1    4)),((0     1)) )= (((1    6)),((0     1)) )  A^4 =A.A^3 = (((1    2)),((0      1)) ) (((1     6)),((0      1)) )= (((1     8)),((0     1)) )  ⋮  ⇒A^n = (((1    2n)),((0      1)) )
$${A}^{\mathrm{2}} ={A}.{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{4}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{\mathrm{3}} ={A}.{A}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{4}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{6}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$${A}^{\mathrm{4}} ={A}.{A}^{\mathrm{3}} =\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{6}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{8}}\\{\mathrm{0}\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\vdots \\ $$$$\Rightarrow{A}^{{n}} =\begin{pmatrix}{\mathrm{1}\:\:\:\:\mathrm{2}{n}}\\{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$
Commented by 20190927 last updated on 29/Sep/19
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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