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    i hate :mad: the transformations involved in this. i have attached my questions. please anyone..would be grateful to you. thanks you.
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    (16)
    det(M - lambda I) = 0
    (4 - lambda)(3 - lambda)(1 - lambda) - 3(-2 - (1 - lambda)) = 0
    (4 - lambda)(3 - lambda)(1 - lambda) - 3(lambda - 3) = 0
    (3 - lambda)[(4 - lambda)(1 - lambda) + 3] = 0
    (lambda - 3)(lambda^2 - 5lambda + 7) = 0

    The discriminant of (lambda^2 - 5lambda + 7) is 25 - 28 < 0.

    So there is exactly one real eigenvalue: lambda = 3.

    --

    (27)
    The points on the target line (the line after the transformation has been applied) can be written as t(1, 4, 3), where t is a scalar.

    So we want to find the vector v such that Mv = (1, 4, 3). Row reduction gives v = (10/7, 1/7, -3/7).

    The points on the original line can be written as s(10/7, 1/7, -3/7), where s is a scalar.

    Answer: x/10 = y = -z/3.

    --

    (44)
    The transformation

    (i) stretches horizontally by a factor 3, giving y = (2/3)x,

    then

    (ii) stretches vertically by a factor 4, giving y = (8/3)x,

    then

    (iii) translates by (1, 1), giving (y - 1) = (8/3)(x - 1), ie, y = (8/3)x - 5/3.
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    As B(C+I) = BC + B = (B+I)^2 then

    BC = B^2 + B + I

    c^(-1) B^(-1) (p,q,r)^T = (1,1,-3)^T

    So (p,q,r)^T = BC(1,1,-3)^T = (B^2+B+I)(1,1,-3)^T

    Now

    B(1,1,-3)^T = (-4,-5,-2)^T

    B^2(1,1,-3)^T = B(-4,-5,-2)^T = (3,-7,-6)^T

    Hence (p,q,r) = (3,-7,-6) + (-4,-5,-2) + (1,1,-3) = (0,-11,-11)
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    (110)

    A(2,3,-1)^T = (-2,-3,1)^T = -1(2,3,-1)^T - eigenvalue = -1

    A(2,-1,1)^T = (6,-3,3)^T = 3(2,-1,1)^T - eigenvalue = 3
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    (Original post by RichE)
    (110)

    A(2,3,-1)^T = (-2,-3,1)^T = -1(2,3,-1)^T - eigenvalue = -1

    A(2,-1,1)^T = (6,-3,3)^T = 3(2,-1,1)^T - eigenvalue = 3

    thank you once again for all your help. can see the silver line in the gray now.
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    (Original post by Jonny W)
    (16)
    det(M - lambda I) = 0
    (4 - lambda)(3 - lambda)(1 - lambda) - 3(-2 - (1 - lambda)) = 0
    (4 - lambda)(3 - lambda)(1 - lambda) - 3(lambda - 3) = 0
    (3 - lambda)[(4 - lambda)(1 - lambda) + 3] = 0
    (lambda - 3)(lambda^2 - 5lambda + 7) = 0

    The discriminant of (lambda^2 - 5lambda + 7) is 25 - 28 < 0.

    So there is exactly one real eigenvalue: lambda = 3.

    --

    (27)
    The points on the target line (the line after the transformation has been applied) can be written as t(1, 4, 3), where t is a scalar.

    So we want to find the vector v such that Mv = (1, 4, 3). Row reduction gives v = (10/7, 1/7, -3/7).

    The points on the original line can be written as s(10/7, 1/7, -3/7), where s is a scalar.

    Answer: x/10 = y = -z/3.

    --

    (44)
    The transformation

    (i) stretches horizontally by a factor 3, giving y = (2/3)x,

    then

    (ii) stretches vertically by a factor 4, giving y = (8/3)x,

    then

    (iii) translates by (1, 1), giving (y - 1) = (8/3)(x - 1), ie, y = (8/3)x - 5/3.
    thank you thank you. u have been a huge help.
 
 
 
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