The type of the conic section for the equation 7 x 1 2 + 6 x 1 x 2 + 7 x 2 2 = 200 then transform it to principal axes. The equation 7 x 1 2 + 6 x 1 x 2 + 7 x 2 2 = 200 represents ellipse _ . The transformation form is x = 1 2 [ 1 − 1 1 1 ] y _ . The transformation equation is x 1 = 1 2 y 1 − 1 2 y 2 and x 2 = 1 2 y 1 + 1 2 y 2 _ . Theorem used: Principal axes theorem: “The substitution x = X y transforms a quadratic form Q = x T A x = ∑ j = 1 n ∑ k = 1 n a j k x j x k , ( a j k = a k j ) to the principal axes form or canonical form Q = y T D y = λ 1 y 1 2 + λ 2 y 2 2 + ⋯ + λ n y n 2 , where λ 1 , λ 2 , ⋯ , λ n are the eigen values of the matrix A and X is an orthogonal matrix with corresponding eigen vectors x 1 , ⋯ , x n , respectively.” Calculation: Given that, the equation is 7 x 1 2 + 6 x 1 x 2 + 7 x 2 2 = 200 . Let the function be Q = 7 x 1 2 + 6 x 1 x 2 + 7 x 2 2 . It can be written as, Q = 7 x 1 2 + 3 x 1 x 2 + 3 x 1 x 2 + 7 x 2 2 . From the above function, it is observed that the matrices A = [ 7 3 3 7 ] and x = [ x 1 x 2 ] . The characteristic equation of A = [ 7 3 3 7 ] is | A − λ I | = 0 . | [ 7 3 3 7 ] − λ [ 1 0 0 1 ] | = 0 | [ 7 3 3 7 ] − [ λ 0 0 λ ] | = 0 | 7 − λ 3 3 7 − λ | = 0 Simplify further and obtain the values of λ as follows. ( 7 − λ ) ( 7 − λ ) − 9 = 0 λ 2 − 14 λ + 40 = 0 ( λ − 10 ) ( λ − 4 ) = 0 λ = 10 or λ = 4 Thus, the values of λ are 10 and 4 . From this formula Q = y T D y = λ 1 y 1 2 + λ 2 y 2 2 + ⋯ + λ n y n 2 , the function Q = 7 x 1 2 + 3 x 1 x 2 + 3 x 1 x 2 + 7 x 2 2 becomes, Q = 10 y 1 2 + 4 y 2 2 . From the given equation Q = 200 , then 10 y 1 2 + 4 y 2 2 = 200 . 10 y 1 2 + 4 y 2 2 = 200 10 y 1 2 + 4 y 2 2 200 = 1 1 20 y 1 2 + 1 50 y 2 2 = 1 From the equation 1 20 y 1 2 + 1 50 y 2 2 = 1 , it is observed that the equation represents ellipse. Thus, The equation 7 x 1 2 + 6 x 1 x 2 + 7 x 2 2 = 200 represents ellipse _ . Use eigen values to find the normalized eigen vectors and the matrix X as follows. Now obtain eigen vector for λ = 10 as follows. A − λ I = [ 7 3 3 7 ] − 10 [ 1 0 0 1 ] = [ 7 3 3 7 ] − [ 10 0 0 10 ] = [ − 3 3 3 − 3 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 3 3 3 − 3 ] ∼ [ − 3 3 0 0 ] ( R 2 → R 1 + R 2 ) ∼ [ 1 − 1 0 0 ] ( R 1 → − 1 3 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = x 2 . Let x 2 = 1 , then the value of x 1 = 1 . Thus, the eigen vector is [ 1 1 ] . It is known that the normalized eigenvector for X is X ^ = X | X | . So, the normalized eigen vector for [ 1 1 ] is 1 2 [ 1 1 ] . Now obtain eigen vector for λ = 4 as follows. A − λ I = [ 7 3 3 7 ] − 4 [ 1 0 0 1 ] = [ 7 3 3 7 ] − [ 4 0 0 4 ] = [ 3 3 3 3 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ 3 3 3 3 ] ∼ [ 3 3 0 0 ] ( R 2 → R 1 − R 2 ) ∼ [ 1 1 0 0 ] ( R 1 → 1 3 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = − x 2 . Let x 2 = 1 , then the value of x 1 = − 1 . Thus, the eigen vector is [ − 1 1 ] . It is known that the normalized eigenvector for X is X ^ = X | X | . So, the normalized eigen vector for [ − 1 1 ] is 1 2 [ − 1 1 ] . Thus, the matrix X is X = 1 2 [ 1 − 1 1 1 ] . Hence, the transformation form is x = 1 2 [ 1 − 1 1 1 ] y _ . The transformation equation is x 1 = 1 2 y 1 − 1 2 y 2 and x 2 = 1 2 y 1 + 1 2 y 2 _ .

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Answer 2

Expert Solution & Answer

Answer 3

Expert Solution & Answer

Answer 4

Expert Solution & Answer

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Answer 8

Chapter 8 Solutions