Linear Algebra Solutions 78

Linear Algebra Solutions 78 - / 5 , 2 / 5] t and [-2 / 5 ,...

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Then X t AX = - 4 x 2 1 + y 2 1 and the original equation 4 xy - 3 y 2 = 8 becomes - 4 x 2 1 + y 2 1 = 8, or the standard form - x 2 1 2 + y 2 1 8 = 1 , which represents an hyperbola. The asymptotes assist in drawing the curve. They are given by the equations - x 2 1 2 + y 2 1 8 = 0 , or y 1 = ± 2 x 1 . Now x 1 y 1 = P t x y = " 1 5 - 2 5 2 5 1 5 # x y , so x 1 = x - 2 y 5 , y 1 = 2 x + y 5 . Hence the asymptotes are 2 x + y 5 = ± 2 µ x - 2 y 5 , which reduces to y = 0 and y = 4 x/ 3. (See Figure 2(a).) 3. 8 x 2 - 4 xy + 5 y 2 = X t AX , where A = 8 - 2 - 2 5 and X = x y . The eigenvalues of A are the roots of λ 2 - 13 λ + 36 = 0, namely λ 1 = 4 and λ 2 = 9. Corresponding unit eigenvectors turn out to be [1
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Unformatted text preview: / 5 , 2 / 5] t and [-2 / 5 , 1 / 5] t . Hence if P = " 1 5-2 5 2 5 1 5 # , then P is an orthogonal matrix. Also as det P = 1, P is a proper orthogonal matrix and the equation x y = P x 1 y 1 represents a rotation to new x 1 , y 1 axes whose positive directions are given by the respective columns of P . Also P t AP = 4 9 . 81...
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