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Question Number 54938 by gunawan last updated on 14/Feb/19
![If A= [(2,1),(1,2) ],then A^(2006) =...](Q54938.png)
$$\mathrm{If}\:{A}=\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}}\end{bmatrix},\mathrm{then}\:{A}^{\mathrm{2006}} =... \\ $$
Commented by Abdo msup. last updated on 15/Feb/19
![we have A = (((2 0)),((0 2)) ) + (((0 1)),((1 0)) ) =2I +J J^2 = (((0 1)),((1 0)) ) . (((0 1)),((1 0)) ) = (((1 0)),((0 1)) ) =I ⇒ J^(2n) =I and J^(2n+1) =J and A^n =(2I +J)^n =Σ_(k=0) ^n C_n ^k J^k (2I)^(n−k) =Σ_(p=0) ^([(n/2)]) C_n ^(2p) j^(2p) (2I)^(n−2p) +Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) j^(2p+1) (2I)^(n−2p−1) =Σ_(p=0) ^([(n/2)]) C_n ^(2p) 2^(n−2p) I +Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) 2^(n−2p−1) J ⇒ A^(2006) =Σ_(p=0) ^(1003) C_(2006) ^(2p) 2^(2006−2p) +Σ_(p=0) ^(2002) C_(2006) ^(2p+1) 2^(2005−2p)](Q54942.png)
$${we}\:{have}\:{A}\:=\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{2}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:=\mathrm{2}{I}\:+{J} \\ $$$${J}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:={I}\:\Rightarrow \\ $$$${J}^{\mathrm{2}{n}} ={I}\:\:{and}\:{J}^{\mathrm{2}{n}+\mathrm{1}} ={J}\:\:\:{and}\:{A}^{{n}} =\left(\mathrm{2}{I}\:+{J}\right)^{{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{J}^{{k}} \:\left(\mathrm{2}{I}\right)^{{n}−{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \:\:{j}^{\mathrm{2}{p}} \left(\mathrm{2}{I}\right)^{{n}−\mathrm{2}{p}} \:\:\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{j}^{\mathrm{2}{p}+\mathrm{1}} \left(\mathrm{2}{I}\right)^{{n}−\mathrm{2}{p}−\mathrm{1}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}} \mathrm{2}^{{n}−\mathrm{2}{p}} \:{I}\:\:\:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\mathrm{2}^{{n}−\mathrm{2}{p}−\mathrm{1}} \:{J}\:\Rightarrow \\ $$$${A}^{\mathrm{2006}} \:=\sum_{{p}=\mathrm{0}} ^{\mathrm{1003}} {C}_{\mathrm{2006}} ^{\mathrm{2}{p}} \:\:\mathrm{2}^{\mathrm{2006}−\mathrm{2}{p}} \:+\sum_{{p}=\mathrm{0}} ^{\mathrm{2002}} \:{C}_{\mathrm{2006}} ^{\mathrm{2}{p}+\mathrm{1}} \:\mathrm{2}^{\mathrm{2005}−\mathrm{2}{p}} \\ $$$$ \\ $$