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# P6-Desperate-Matrix algebra watch

1. i hate the transformations involved in this. i have attached my questions. please anyone..would be grateful to you. thanks you.
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2. Matrix algebra.pdf (36.0 KB, 126 views)
3. (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.
4. (69c)

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)
5. (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
6. (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.
7. (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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