The eigen values and eigen vectors of the system of equations. The eigen values of the system of equations are λ = 4 + 3 i 5 , λ = 4 − 3 i 5 _ . The eigen vectors of the system of equations is λ = cos θ + i sin θ , λ = cos θ − i sin θ _ . Definition used: Let A be a n × n matrix. The eigen values of the matrix A is obtained by solving the characteristic equation D ( λ ) = 0 that is det ( A − λ I ) = 0 , where λ is the eigen value. Calculation: The given matrix is [ cos θ − sin θ sin θ cos θ ] . Let A = [ cos θ − sin θ sin θ cos θ ] . Substitute A = [ cos θ − sin θ sin θ cos θ ] in det ( A − λ I ) = 0 and obtain the characteristic equation as shown below: | [ cos θ − sin θ sin θ cos θ ] − λ [ 1 0 0 1 ] | = 0 | cos θ − λ − sin θ sin θ cos θ − λ | = 0 ( cos θ − λ ) ( cos θ − λ ) + sin 2 θ = 0 Solve the above equation and obtain the eigen values as follows: ( cos θ − λ ) ( cos θ − λ ) + sin 2 θ = 0 cos 2 θ − 2 λ cos θ + λ 2 + sin 2 θ = 0 λ = cos θ + i sin θ , λ = cos θ − i sin θ Thus, the eigen values are λ = cos θ + i sin θ , λ = cos θ − i sin θ _ . Substitute λ = cos θ + i sin θ in [ A − λ I ] as, [ A − λ I ] λ = cos θ + i sin θ = [ cos θ − ( cos θ + i sin θ ) − sin θ sin θ cos θ − ( cos θ + i sin θ ) ] = [ − i sin θ − sin θ sin θ − i sin θ ] Let x = [ x 1 x 2 ] be the eigen vector. Solve the equation [ A − λ I ] x = 0 as follows: [ A − λ I ] x = 0 ⇒ [ − i sin θ − sin θ sin θ − i sin θ ] [ x 1 x 2 ] = 0 ⇒ { − i sin θ x 1 − sin θ x 2 = 0 sin θ x 1 − i sin θ x 2 = 0 ⇒ { − i x 1 − x 2 = 0 x 1 − i x 2 = 0 ⇒ − i x 1 = x 2 Thus, the vector x can be written as follows: x = [ x 1 x 2 ] = [ x 1 − i x 1 ] = x 1 [ 1 − i ] Therefore, the eigen vector for λ = 4 + 3 i 5 is [ 1 − i ] T _ . Substitute λ = cos θ − i sin θ in [ A − λ I ] as, [ A − λ I ] λ = cos θ − i sin θ = [ cos θ − ( cos θ − i sin θ ) − sin θ sin θ cos θ − ( cos θ − i sin θ ) ] = [ i sin θ − sin θ sin θ i sin θ ] Let x = [ x 1 x 2 ] be the eigen vector. Solve the equation [ A − λ I ] x = 0 as follows: [ A − λ I ] x = 0 ⇒ [ i sin θ − sin θ sin θ i sin θ ] [ x 1 x 2 ] = 0 ⇒ { i sin θ x 1 − sin θ x 2 = 0 sin θ x 1 + i sin θ x 2 = 0 ⇒ { i x 1 − x 2 = 0 x 1 + i x 2 = 0 ⇒ i x 1 = x 2 Let x = [ x 1 x 2 ] be the eigen vector. Solve the equation [ A − λ I ] x = 0 as follows: [ A − λ I ] x = 0 ⇒ [ 3 i 5 − 0.6 0.6 3 i 5 ] [ x 1 x 2 ] = 0 ⇒ { 3 i 5 x 1 − 0.6 x 2 = 0 0.6 x 1 3 i 5 x 2 = 0 ⇒ { i x 1 − x 2 = 0 x 1 + i x 2 = 0 ⇒ i x 1 = x 2 Thus, the vector x can be written as follows: x = [ x 1 x 2 ] = [ x 1 i x 1 ] = x 1 [ 1 i ] Therefore, the eigen vector for λ = cos θ − i sin θ is [ 1 i ] T _ .

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