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A7_soln

# A7_soln - Math 235 Assignment 7 Solutions 1 For each...

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Unformatted text preview: Math 235 Assignment 7 Solutions 1. For each quadratic form Q ( vectorx ), determine the corresponding symmetric matrix A . By diagonalizing A , express Q ( vectorx ) in diagonal form and give an orthogonal matrix that diago- nalizes A . Classify each quadratic form. a) Q ( x, y ) = 7 x 2 + 12 xy + 12 y 2 . Solution: We have A = bracketleftbigg 7 6 6 12 bracketrightbigg so A − λI = bracketleftbigg 7 − λ 6 6 12 − λ bracketrightbigg . The characteristic equation is 0 = det( A − λI ) = λ 2 − 19 λ + 48 = ( λ − 3)( λ − 16) . The roots are 3 and 16, so these are the eigenvectors of A . For λ = 3 we get A − 3 I = bracketleftbigg 4 6 6 9 bracketrightbigg ∼ bracketleftbigg 2 3 0 0 bracketrightbigg ⇒ vector z 1 = bracketleftbigg − 3 2 bracketrightbigg . For λ = 16 we get A − 16 I = bracketleftbigg − 9 6 6 − 4 bracketrightbigg ∼ bracketleftbigg − 3 2 bracketrightbigg ⇒ vector z 2 = bracketleftbigg 2 3 bracketrightbigg . Normalizing the vectors, we get that the orthogonal matrix which diagonalizes A is P = bracketleftbigg − 3 / √ 13 2 / √ 13 2 / √ 13 3 / √ 13 bracketrightbigg , and Q = 3 x 2 1 + 16 y 2 1 , where bracketleftbigg x 1 y 1 bracketrightbigg = P T bracketleftbigg x y bracketrightbigg . Since all the eigenvalues of A are positive, it follows that Q ( x, y ) is positive definite. b) Q ( x, y ) = x 2 + 6 xy − 7 y 2 Solution: We have A = bracketleftbigg 1 3 3 − 7 bracketrightbigg so A − λI = bracketleftbigg 1 − λ 3 3 − 7 − λ bracketrightbigg . The characteristic equation is 0 = det( A − λI ) = λ 2 + 6 λ − 16 = ( λ − 2)( λ + 8) . The roots are 2 and − 8, so these are the eigenvectors of A . For λ = 2 we get A − 2 I = bracketleftbigg − 1 3 3 − 9 bracketrightbigg ∼ bracketleftbigg 1 − 3 bracketrightbigg ⇒ vector z 1 = bracketleftbigg 3 1 bracketrightbigg . For λ = − 8 we get A + 8 I = bracketleftbigg 9 3 3 1 bracketrightbigg ∼ bracketleftbigg 3 1 0 0 bracketrightbigg ⇒ vector z 2 = bracketleftbigg − 1 3 bracketrightbigg . Normalizing the vectors, we get that the orthogonal matrix which diagonalizes A is P = bracketleftbigg 3 / √ 10 − 1 / √ 10 − 1 / √ 10 3 / √ 10 bracketrightbigg , and Q = 2 x 2 1 − 8 y 2 1 , where bracketleftbigg x 1 y 1 bracketrightbigg = P T bracketleftbigg x y bracketrightbigg ....
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A7_soln - Math 235 Assignment 7 Solutions 1 For each...

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