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Scilla [17]
3 years ago
11

50 POINTS!!!

Mathematics
2 answers:
Usimov [2.4K]3 years ago
5 0

Answer:

it's b

Step-by-step explanation:

just because it's long it is very right

Step2247 [10]3 years ago
3 0

Answer:

well by looking at it , it would be for a.

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Suppose T is a transformation from ℝ2 to ℝ2. Find the matrix A that induces T if T is reflection over the line y=1/2x
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Answer:

A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]

Step-by-step explanation:

We have to see how the canonical vectors are transformed throught T. Lets first define T in any basis.

Since T is a reflection, then any element of the line y = x/2 if fixed by T. Therefore T(2,1) = (2,1).

On the other hand, any vector perpendicular to the line direction should be sent to its opposite value. We can take, for example, (-1,2) (note that the scalar product (2,1) * (-1,2) = -2+2 = 0). As a consecuence T(-1,2) = (1,-2). We have

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By summing the first vector with the double of the second one we get, using linearity

T(0,5) = T( (2,1) + 2(-1,2)) = T(2,1) + 2T(-1,2) = (2,1) + 2(1,-2) = (4,-3)

Hence, T(0,1) = (4/5,-3/5)

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Therefore, T(1,0) = (1,4/3)

The matrix A induced by  T has in its first column T(1,0) and in its second column T(0,1). We conclude that

A = \left[\begin{array}{cc}1&\frac{4}{5}\\\frac{4}{3}&-\frac{3}{5}\end{array}\right]

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