The eigen values and eigen vectors of the system of equations. The eigen values of the system of equations are λ = 3.0 , λ = − 0.6 _ . The eigen vectors of the system of equations is [ 0 1 ] T and [ 1 0 ] T _ . 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 [ 3.0 0 0 − 0.6 ] . Let A = [ 3.0 0 0 − 0.6 ] . Substitute A = [ 3.0 0 0 − 0.6 ] in det ( A − λ I ) = 0 and obtain the characteristic equation as shown below: | [ 3.0 0 0 − 0.6 ] − λ [ 1 0 0 1 ] | = 0 | 3.0 − λ 0 0 − 0.6 − λ | = 0 ⋯ ⋯ ( 1 ) ( 3.0 − λ ) ( − 0.6 − λ ) = 0 Solve the and obtain the eigen values as follows: ( 3.0 − λ ) ( − 0.6 − λ ) = 0 λ = 3.0 , λ = − 0.6 Thus, the eigen values are λ = 3.0 , λ = − 0.6 _ . Substitute λ = 3.0 in equation (1) as, [ A − λ I ] λ = 3 = [ 3 − 3 0 0 − 0.6 − 3 ] = [ 0 0 0 − 3.6 ] 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 ⇒ [ 0 0 0 − 3.6 ] [ x 1 x 2 ] = 0 ⇒ { 0 x 1 + 0 x 2 = 0 0 x 1 − 3.6 x 2 = 0 ⇒ x 2 = 0 So, x 1 can be taken as 1. Therefore, the eigen vector for λ = 3.0 is [ 1 0 ] T _ Substitute λ = − 0.6 in equation (1) as, [ A − λ I ] λ = − 0.6 = [ 3 − ( − 0.6 ) 0 0 − 0.6 − ( − 0.6 ) ] = [ 3.6 0 0 0 ] 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 ⇒ [ 3.6 0 0 0 ] [ x 1 x 2 ] = 0 ⇒ { 3.6 x 1 + 0 x 2 = 0 0 x 1 + 0 x 2 = 0 ⇒ x 1 = 0 So, x 2 can be taken as 1. Therefore, the eigen vector for λ = − 0.6 is [ 0 1 ] T _

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