Linear Algebra Solutions 81

# Linear Algebra Solutions 81 - 2 = 0 and y 2 = 0 or x 1 2 =...

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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 = 9 0 0 4 . Moreover 5 x 2 - 4 xy + 8 y 2 = 9 x 2 1 + 4 y 2 1 . To get the coeFcients of x 1 and y 1 in the transformed form of equation (3), we have to use the rotation equations x = 1 5 ( x 1 + 2 y 1 ) , y = 1 5 ( - 2 x 1 + y 1 ) . Then equation (3) transforms to 9 x 2 1 + 4 y 2 1 + 36 x 1 - 8 y 1 + 4 = 0 , or, on completing the square, 9( x 1 + 2) 2 + 4( y 1 - 1) 2 = 36 , or in standard form x 2 2 4 + y 2 2 9 = 1 , where x 2 = x 1 + 2 and y 2 = y 1 - 1. Thus we have an ellipse, centre ( x 2 , y 2 ) = (0 , 0), or ( x 1 , y 1 ) = ( - 2 , 1), or ( x, y ) = (0 , 5). The axes of symmetry are given by x
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Unformatted text preview: 2 = 0 and y 2 = 0, or x 1 + 2 = 0 and y 1-1 = 0, or 1 √ 5 ( x-2 y ) + 2 = 0 and 1 √ 5 (2 x + y )-1 = 0 , which reduce to x-2 y + 2 √ 5 = 0 and 2 x + y-√ 5 = 0. See ±igure 3(b). 5. (i) Consider the equation 2 x 2 + y 2 + 3 xy-5 x-4 y + 3 = 0 . (4) Δ = f f f f f f 2 3 / 2-5 / 2 3 / 2 1-2-5 / 2-2 3 f f f f f f = 8 f f f f f f 4 3-5 3 2-4-5-4 6 f f f f f f = 8 f f f f f f 1 1-1 3 2-4-2-2 2 f f f f f f = 0 . 84...
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