The type of the conic section for the equation 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 = 10 then transform it to principal axes. The equation 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 = 10 represents hyperbola _ . The transformation form is x = 1 5 [ 2 − 1 1 2 ] y _ . The transformation equation is x 1 = 2 5 y 1 − 1 5 y 2 and x 2 = 1 5 y 1 + 2 5 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 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 = 10 . Let the function be Q = 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 . It can be written as, Q = 3 x 1 2 + 4 x 1 x 2 + 4 x 1 x 2 − 3 x 2 2 = 10 . From the above function, it is observed that the matrices A = [ 3 4 4 − 3 ] and x = [ x 1 x 2 ] . The characteristic equation of A = [ 3 4 4 − 3 ] is | A − λ I | = 0 . | [ 3 4 4 − 3 ] − λ [ 1 0 0 1 ] | = 0 | [ 3 4 4 − 3 ] − [ λ 0 0 λ ] | = 0 | 3 − λ 4 4 − 3 − λ | = 0 Simplify further and obtain the values of λ as follows. ( 3 − λ ) ( − 3 − λ ) − 16 = 0 λ 2 − 25 = 0 ( λ − 5 ) ( λ + 5 ) = 0 λ = 5 or λ = − 5 Thus, the values of λ are 5 and − 5 . 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 = 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 becomes, Q = 5 y 1 2 − 5 y 2 2 . From the given equation Q = 10 , then 5 y 1 2 − 5 y 2 2 = 10 . 5 y 1 2 − 5 y 2 2 = 10 5 y 1 2 − 5 y 2 2 10 = 1 1 2 y 1 2 − 1 2 y 2 2 = 1 From the equation 1 2 y 1 2 − 1 2 y 2 2 = 1 , it is observed that the equation represents hyperbola. Thus, the equation 3 x 1 2 + 8 x 1 x 2 − 3 x 2 2 = 10 represents hyperbola _ . Use eigen values to find the normalized eigen vectors and the matrix X as follows. Now obtain eigen vector for λ = 5 as follows. A − λ I = [ 3 4 4 − 3 ] − 5 [ 1 0 0 1 ] = [ 3 4 4 − 3 ] − [ 5 0 0 5 ] = [ − 2 4 4 − 8 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ − 2 4 4 − 8 ] ∼ [ 4 − 8 − 2 4 ] ( R 2 ↔ R 1 ) ∼ [ 4 − 8 0 0 ] ( R 2 → R 1 + 2 R 2 ) ∼ [ 1 − 2 0 0 ] ( R 1 → 1 4 R 1 ) From above matrix, it is observed that the system of equation is { x 1 − 2 x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = 2 x 2 . Let x 2 = 1 , then the value of x 1 = 2 . Thus, the eigen vector is [ 2 1 ] . It is known that the normalized eigenvector for X is X ^ = X | X | . So, the normalized eigen vector for [ 2 1 ] is 1 5 [ 2 1 ] . Now obtain eigen vector for λ = − 5 as follows. A − λ I = [ 3 4 4 − 3 ] + 5 [ 1 0 0 1 ] = [ 3 4 4 − 3 ] + [ 5 0 0 5 ] = [ 8 4 4 2 ] Use row reduction to obtain the eigen vectors as follows. A − λ I = [ 8 4 4 2 ] ∼ [ 8 4 0 0 ] ( R 2 → R 1 − 2 R 2 ) ∼ [ 1 1 2 0 0 ] ( R 1 → 1 8 R 1 ) From above matrix, it is observed that the system of equation is { x 1 + 1 2 x 2 = 0 0 x 1 + 0 x 2 = 0 . So, the value of x 1 = − 1 2 x 2 . Let x 2 = 2 , then the value of x 1 = − 1 . Thus, the eigen vector is [ − 1 2 ] . It is known that the normalized eigenvector for X is X ^ = X | X | . So, the normalized eigen vector for [ − 1 2 ] is 1 5 [ − 1 2 ] . Thus, the matrix X is X = 1 5 [ 2 − 1 1 2 ] . Hence, the transformation form is x = 1 5 [ 2 − 1 1 2 ] y _ . The transformation equation is x 1 = 2 5 y 1 − 1 5 y 2 and x 2 = 1 5 y 1 + 2 5 y 2 _ .

Question

Want to see the full answer?

Answer 1

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

7. Show that for R sufficiently large, the polynomial P(z) in Example 3, Sec. 5, satisfies the inequality |P(z)| R. Suggestion: Observe that there is a positive number R such that the modulus of each quotient in inequality (9), Sec. 5, is less than |an|/n when |z| > R.

9. Establish the identity 1- 1+z+z² + 2n+1 ... +z" = 1- z (z1) and then use it to derive Lagrange's trigonometric identity: 1 1+ cos cos 20 +... + cos no = + 2 sin[(2n+1)0/2] 2 sin(0/2) (0 < 0 < 2л). Suggestion: As for the first identity, write S = 1+z+z² +...+z" and consider the difference S - zS. To derive the second identity, write z = eie in the first one.

8. Prove that two nonzero complex numbers z₁ and Z2 have the same moduli if and only if there are complex numbers c₁ and c₂ such that Z₁ = c₁C2 and Z2 = c1c2. Suggestion: Note that (i≤ exp (101+0) exp (01-02) and [see Exercise 2(b)] 2 02 Ꮎ - = = exp(i01) exp(101+0) exp (i 01 - 02 ) = exp(102). i 2 2