To verify: The matrix equation x = P y for the matrices A = [ − 5 0 15 3 4 − 9 − 5 0 15 ] and P = [ 0 1 0 1 0 0 0 0 1 ] if x = P y is eigenvectors of A and y is an eigenvector of A ^ = P − 1 A P . Given that, the matrices are A = [ − 5 0 15 3 4 − 9 − 5 0 15 ] and P = [ 0 1 0 1 0 0 0 0 1 ] . In order to verify the given statement, first find the matrix A ^ as follows. The inverse matrix of P = [ 0 1 0 1 0 0 0 0 1 ] is computed as follows. P − 1 = 1 | P − 1 | a d j [ 0 1 0 1 0 0 0 0 1 ] = 1 − 1 a d j [ 0 1 0 1 0 0 0 0 1 ] = [ 0 1 0 1 0 0 0 0 1 ] Now compute the value of P − 1 A P as follows. P − 1 A P = [ 0 1 0 1 0 0 0 0 1 ] [ − 5 0 15 3 4 − 9 − 5 0 15 ] [ 0 1 0 1 0 0 0 0 1 ] = [ 3 4 − 9 − 5 0 15 − 5 0 15 ] [ 0 1 0 1 0 0 0 0 1 ] = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] Thus, the value of the matrix A ^ = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] . In order to obtain y , compute the eigen vectors for the matrix A ^ = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] as follows. The characteristic equation of A ^ = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] is | A ^ − λ I | = 0 . | [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | 4 − λ 3 − 9 0 − 5 − λ 15 0 − 5 15 − λ | = 0 Simplify further and obtain the values of λ as follows. [ ( 4 − λ ) ( ( − 5 − λ ) ( − 15 − λ ) + 75 ) − 3 ( 0 − 0 ) − 9 ( 0 − 0 ) ] = 0 [ ( 4 − λ ) ( λ 2 − 10 λ ) − 0 + 0 ] = 0 − λ 3 + 14 λ 2 − 40 λ = 0 λ ( − λ 2 + 14 λ − 40 ) = 0 Simplify further, λ ( − λ 2 + 14 λ − 40 ) = 0 λ ( − λ 2 + 4 λ + 10 λ − 40 ) = 0 λ ( λ − 10 ) ( λ − 4 ) = 0 λ = 0 or λ = 4 or λ = 10 Now obtain eigen vector for λ = 0 as follows. A ^ − λ I = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] + 0 [ 1 0 0 0 1 0 0 0 1 ] = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] + [ 0 0 0 0 0 0 0 0 0 ] = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] Use row reduction to obtain the eigen vectors as follows. A ^ − λ I = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] ∼ [ 4 3 − 9 0 − 5 15 0 0 0 ] ( R 3 → R 3 − R 2 ) ∼ [ 4 0 0 0 − 5 15 0 0 0 ] ( R 1 → 3 R 1 − R 2 ) ∼ [ 1 0 0 0 1 − 3 0 0 0 ] ( R 1 → 1 4 R 1 R 2 → − 1 5 R 2 ) From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 + 0 x 3 = 0 0 x 1 + x 2 − 3 x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the values of x 1 = 0 and x 2 = 3 x 3 . Let x 3 = 1 , then the value of x 2 = 3 . Thus, the eigen vector is [ 0 3 1 ] . Now obtain eigen vector for λ = 4 as follows. A ^ − λ I = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − 4 [ 1 0 0 0 1 0 0 0 1 ] = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − [ 4 0 0 0 4 0 0 0 4 ] = [ 0 3 − 9 0 − 9 15 0 − 5 11 ] Use row reduction to obtain the eigen vectors as follows. A ^ − λ I = [ 0 3 − 9 0 − 9 15 0 − 5 11 ] ∼ [ 0 3 − 9 0 − 9 15 0 0 8 3 ] ( R 3 → R 3 − 5 9 R 2 ) ∼ [ 0 3 − 9 0 − 9 0 0 0 1 ] ( R 3 → 3 8 R 3 R 2 → R 2 − 15 R 3 ) ∼ [ 0 3 0 0 1 0 0 0 1 ] ( R 2 → − 1 9 R 2 R 1 → R 1 + 9 R 3 ) Simplify further, A ^ − λ I = [ 0 3 0 0 1 0 0 0 1 ] ∼ [ 0 3 0 0 0 0 0 0 1 ] ( R 2 → R 1 − 3 R 2 ) ∼ [ 0 1 0 0 0 0 0 0 1 ] From above matrix, it is observed that the system of equation is { 0 x 1 + x 2 + 0 x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 0 x 1 + 0 x 2 + x 3 = 0 . So, the values of x 2 = 0 and x 3 = 0 . Consider the value of x 1 = 1 . Thus, the eigen vector is [ 1 0 0 ] . Now obtain eigen vector for λ = 10 as follows. A ^ − λ I = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − 10 [ 1 0 0 0 1 0 0 0 1 ] = [ 4 3 − 9 0 − 5 15 0 − 5 15 ] − [ 10 0 0 0 10 0 0 0 10 ] = [ − 6 3 − 9 0 − 15 15 0 − 5 5 ] Use row reduction to obtain the eigen vectors as follows. A ^ − λ I = [ − 6 3 − 9 0 − 15 15 0 − 5 5 ] ∼ [ − 6 3 − 9 0 − 15 15 0 0 0 ] ( R 3 → R 2 − 5 R 3 ) ∼ [ − 6 3 − 9 0 1 − 1 0 0 0 ] ( R 2 → − 1 15 R 2 ) ∼ [ − 6 0 − 6 0 1 − 1 0 0 0 ] ( R 1 → R 1 − 3 R 2 ) Simplify further, A ^ − λ I ∼ [ − 6 0 − 6 0 1 − 1 0 0 0 ] ∼ [ 1 0 1 0 1 − 1 0 0 0 ] ( R 1 → − 1 6 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 + x 3 = 0 0 x 1 + x 2 − x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the values of x 1 = − x 3 and x 2 = x 3 . Let x 3 = 1 , then the values of x 1 = − 1 and x 2 = 1 . Thus, the eigen vector is [ − 1 1 1 ] . For the eigen value λ = 0 the vector Py is computed as follows. [ 0 1 0 1 0 0 0 0 1 ] [ 0 3 1 ] = [ 0 + 3 + 0 0 + 0 + 0 0 + 0 + 1 ] P y = [ 3 0 1 ] ⋯ ⋯ ( 1 ) For the eigen value λ = 4 the vector Py is computed as follows. [ 0 1 0 1 0 0 0 0 1 ] [ 1 0 0 ] = [ 0 + 0 + 0 1 + 0 + 0 0 + 0 + 0 ] P y = [ 0 1 0 ] ⋯ ⋯ ( 2 ) For the eigen value λ = 10 the vector Py is computed as follows. [ 0 1 0 1 0 0 0 0 1 ] [ − 1 1 1 ] = [ 0 + 1 + 0 − 1 + 0 + 0 0 + 0 + 1 ] P y = [ 1 − 1 1 ] ⋯ ⋯ ( 3 ) In order to compute the matrix x , compute the eigen vectors for the matrix A = [ − 5 0 15 3 4 − 9 − 5 0 15 ] as follows. The characteristic equation of A = [ − 5 0 15 3 4 − 9 − 5 0 15 ] is | A − λ I | = 0 . | [ − 5 0 15 3 4 − 9 − 5 0 15 ] − λ [ 1 0 0 0 1 0 0 0 1 ] | = 0 | [ − 5 0 15 3 4 − 9 − 5 0 15 ] − [ λ 0 0 0 λ 0 0 0 λ ] | = 0 | − 5 − λ 0 15 3 4 − λ − 9 − 5 0 15 − λ | = 0 Simplify further and obtain the values of λ as follows. ( − 5 − λ ) ( ( 4 − λ ) ( 15 − λ ) + 0 ) − 0 + 15 ( 0 + 5 ( 4 − λ ) ) = 0 ( − 5 − λ ) ( λ 2 − 19 λ + 60 ) − 0 + 20 − 5 λ = 0 − λ 3 + 14 λ 2 − 40 λ = 0 Simplify further, λ ( − λ 2 + 14 λ − 40 ) = 0 λ ( − λ 2 + 4 λ + 10 λ − 40 ) = 0 λ ( λ − 10 ) ( λ − 4 ) = 0 λ = 0 or λ = 4 or λ = 10 Now obtain eigen vector for λ = 0 as follows. A − λ I = [ − 5 0 15 3 4 − 9 − 5 0 15 ] + 0 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 0 15 3 4 − 9 − 5 0 15 ] + [ 0 0 0 0 0 0 0 0 0 ] = [ − 5 0 15 3 4 − 9 − 5 0 15 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 5 0 15 3 4 − 9 − 5 0 15 ] ∼ [ − 5 0 15 3 4 − 9 0 0 0 ] ( R 3 → R 1 − R 3 ) ∼ [ − 5 0 15 0 4 0 0 0 0 ] ( R 2 → R 2 + 3 5 R 1 ) ∼ [ 1 0 − 3 0 1 0 0 0 0 ] ( R 2 → 1 4 R 2 R 1 → − 1 5 R 2 ) From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 − 3 x 3 = 0 0 x 1 + x 2 + 0 x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the values of x 1 = 3 x 3 and x 2 = 0 . Let x 3 = 1 , then the value of x 1 = 3 . Thus, the eigen vector is x = [ 3 0 1 ] ⋯ ⋯ ( 4 ) . Now obtain eigen vector for λ = 4 as follows. A − λ I = [ − 5 0 15 3 4 − 9 − 5 0 15 ] − 4 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 0 15 3 4 − 9 − 5 0 15 ] − [ 4 0 0 0 4 0 0 0 4 ] = [ − 9 0 15 3 0 − 9 − 5 0 11 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 9 0 15 3 0 − 9 − 5 0 11 ] ∼ [ − 9 0 15 0 0 − 4 0 0 8 3 ] ( R 3 → R 3 − 5 9 R 1 R 2 → R 2 + 1 3 R 1 ) ∼ [ − 9 0 15 0 0 − 4 0 0 1 ] ( R 3 → 3 8 R 3 ) Simplify further, A − λ I ∼ [ − 9 0 15 0 0 1 0 0 1 ] ( R 2 → − 1 4 R 2 ) ∼ [ − 9 0 15 0 0 1 0 0 0 ] ( R 3 → R 3 − R 2 ) ∼ [ − 9 0 0 0 0 1 0 0 0 ] ( R 1 → R 1 − 15 R 2 ) ∼ [ 1 0 0 0 0 1 0 0 0 ] ( R 1 → − 1 9 R 1 ) rom above matrix, it is observed that the system of equation is { x 1 + 0 x 2 + 0 x 3 = 0 0 x 1 + 0 x 2 + x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the values of x 1 = 0 and x 3 = 0 . Consider the value of x 2 = 1 . Thus, the eigen vector is x = [ 0 1 0 ] ⋯ ⋯ ( 5 ) . Now obtain eigen vector for λ = 10 as follows. A − λ I = [ − 5 0 15 3 4 − 9 − 5 0 15 ] − 10 [ 1 0 0 0 1 0 0 0 1 ] = [ − 5 0 15 3 4 − 9 − 5 0 15 ] − [ 10 0 0 0 10 0 0 0 10 ] = [ − 15 0 15 3 − 6 − 9 − 5 0 5 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 15 0 15 3 − 6 − 9 − 5 0 5 ] ∼ [ − 15 0 15 3 − 6 − 9 0 0 0 ] ( R 3 → R 3 − 1 3 R 1 ) ∼ [ − 15 0 15 0 − 6 − 6 0 0 0 ] ( R 2 → R 2 + 1 5 R 1 ) ∼ [ − 15 0 15 0 1 1 0 0 0 ] ( R 2 → − 1 6 R 2 ) Simplify further, A − λ I ∼ [ − 15 0 15 0 1 1 0 0 0 ] ∼ [ 1 0 − 1 0 1 1 0 0 0 ] ( R 1 → − 1 15 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + 0 x 2 − x 3 = 0 0 x 1 + x 2 + x 3 = 0 0 x 1 + 0 x 2 + 0 x 3 = 0 . So, the values of x 1 = x 3 and x 2 = − x 3 . Let x 3 = 1 , then the values of x 1 = 1 and x 2 = − 1 . Thus, the eigen vector is x = [ 1 − 1 1 ] ⋯ ⋯ ( 6 ) . From the equations (1) and (4), it is verified that x = P y , from the equations (2) and (5), it is verified that x = P y and from the equations (3) and (6), it is verified that x = P y .

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18.9. Let denote the boundary of the rectangle whose vertices are -2-2i, 2-21,2+i and -2+i in the positive direction. Evaluate each of the following integrals: L₁ = 2- (a). dz, (b). (d). ₁ = 22+2 [ dz, (e). √, z COS 2 dz dz, (c). L (2z+1)2dz, z(z+1)' (1). [e² si 1 sin z+ dz. (22+3)2