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Consider the Difference Equation xk+1=Axk\mathbf { x } _ { \mathbf { k } + 1 } = \mathrm { Ax } _ { \mathbf { k } }

Question 1

Multiple Choice

Consider the difference equation xk+1=Axk\mathbf { x } _ { \mathbf { k } + 1 } = \mathrm { Ax } _ { \mathbf { k } } , where A\mathrm { A } has eigenvalues and corresponding eigenvectors v1\mathbf { v } _ { 1 } , v2\mathbf { v } _ { 2 } , and v3\mathbf { v } _ { 3 } given below. Find the general solution of this difference equation if x0x _ { 0 } is given as below.

- λ1=1.3,λ2=0.8,λ3=0.6,v1=[661],v2=[212],v3=[122]\lambda _ { 1 } = 1.3 , \lambda _ { 2 } = 0.8 , \lambda _ { 3 } = 0.6 , \mathbf { v } _ { 1 } = \left[ \begin{array} { r } - 6 \\ 6 \\ 1 \end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { l } 2 \\ 1 \\ 2 \end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } 1 \\ 2 \\ - 2 \end{array} \right] , and x0=[33337]\mathbf { x } _ { 0 } = \left[ \begin{array} { r } - 33 \\ 33 \\ - 7 \end{array} \right]


A)
xk=(1.3) kv13(0.8) kv2+3(0.6) kv3\mathrm{x}_{\mathrm{k}}=(1.3) ^{\mathrm{k}_{\mathbf{v1}}}-3(0.8) ^{\mathrm{k}_{\mathbf{}}} \mathbf{v}_{2}+3(0.6) ^{\mathrm{k}_{\mathbf{v3}}}
B)
xk=(1.3) kv1+(0.8) kv2+(0.6) kv3\mathrm{x}_{\mathrm{k}}=(1.3) ^{\mathrm{k}_{\mathbf{v1}}}+(0.8) ^{\mathrm{k}_{\mathbf{v2}}}+(0.6) ^{\mathrm{k}_{\mathbf{v3}}}


C) xk=5(1.3) kv13(0.8) kv2+(0.6) kv3\mathbf { x } _ { \mathrm { k } } = 5 ( 1.3 ) ^ { \mathrm { k } _ { \mathbf { v1 } } } - 3 ( 0.8 ) ^ { \mathrm { k } _ { \mathbf {v2} } } + ( 0.6 ) ^ { \mathrm { k } _ { \mathbf { v3 } } }

D) xk=5(1.3) kv13(0.8) kv2+3(0.6) kv3\mathbf { x } _ { \mathrm { k } } = 5 ( 1.3 ) ^ { \mathrm { k } _ { \mathbf { v1 } } } - 3 ( 0.8 ) ^ { \mathrm { k } _ { \mathbf { v2 } } } + 3 ( 0.6 ) ^ { \mathrm { k } _ { \mathbf { v3 } } }

Correct Answer:

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