Question 1
Multiple Choice
Consider the difference equation x k + 1 = A x k \mathbf { x } _ { \mathbf { k } + 1 } = \mathrm { Ax } _ { \mathbf { k } } x k + 1 = Ax k , where A \mathrm { A } A has eigenvalues and corresponding eigenvectors v 1 \mathbf { v } _ { 1 } v 1 , v 2 \mathbf { v } _ { 2 } v 2 , and v 3 \mathbf { v } _ { 3 } v 3 given below. Find the general solution of this difference equation if x 0 x _ { 0 } x 0 is given as below. - λ 1 = 1.3 , λ 2 = 0.8 , λ 3 = 0.6 , v 1 = [ − 6 6 1 ] , v 2 = [ 2 1 2 ] , v 3 = [ 1 2 − 2 ] \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] λ 1 = 1.3 , λ 2 = 0.8 , λ 3 = 0.6 , v 1 = − 6 6 1 , v 2 = 2 1 2 , v 3 = 1 2 − 2 , and x 0 = [ − 33 33 − 7 ] \mathbf { x } _ { 0 } = \left[ \begin{array} { r } - 33 \\ 33 \\ - 7 \end{array} \right] x 0 = − 33 33 − 7
A) x k = ( 1.3 ) k v 1 − 3 ( 0.8 ) k v 2 + 3 ( 0.6 ) k v 3 \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}}} x k = ( 1.3 ) k v1 − 3 ( 0.8 ) k v 2 + 3 ( 0.6 ) k v3 B) x k = ( 1.3 ) k v 1 + ( 0.8 ) k v 2 + ( 0.6 ) k v 3 \mathrm{x}_{\mathrm{k}}=(1.3) ^{\mathrm{k}_{\mathbf{v1}}}+(0.8) ^{\mathrm{k}_{\mathbf{v2}}}+(0.6) ^{\mathrm{k}_{\mathbf{v3}}} x k = ( 1.3 ) k v1 + ( 0.8 ) k v2 + ( 0.6 ) k v3 C) x k = 5 ( 1.3 ) k v 1 − 3 ( 0.8 ) k v 2 + ( 0.6 ) k v 3 \mathbf { x } _ { \mathrm { k } } = 5 ( 1.3 ) ^ { \mathrm { k } _ { \mathbf { v1 } } } - 3 ( 0.8 ) ^ { \mathrm { k } _ { \mathbf {v2} } } + ( 0.6 ) ^ { \mathrm { k } _ { \mathbf { v3 } } } x k = 5 ( 1.3 ) k v1 − 3 ( 0.8 ) k v2 + ( 0.6 ) k v3 D) x k = 5 ( 1.3 ) k v 1 − 3 ( 0.8 ) k v 2 + 3 ( 0.6 ) k v 3 \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 } } } x k = 5 ( 1.3 ) k v1 − 3 ( 0.8 ) k v2 + 3 ( 0.6 ) k v3
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