Integrating Total Physical Response Strategy in a Level I Spanish ...

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• cours - matière : spanish

Integrating Total Physical Response Strategy in a Level I Spanish Class David E. Wolfe, Ph.D. Temple University • Gwendolyn Jones Bishop McDevitt High School • Wyncote, PA Reprinted from Foreign Language Annals, 14, N. 4, 1982 David E. Wolfe (Ph.D., The Ohio State University) is Professor of Foreign Language Education, Temple University, Philadelphia, PA; Gwendolyn Jones (M.Ed.

• spanish course
• explicit instruction for the rest of the meeting
• tpr-world
• moderate impact on instruction with tpr
• método tpr español
• modern language journal
• test

Sujets

• Spanish
• Learning
• Bishop McDevitt High School (Harrisburg, Pennsylvania)
• Super School
• Ohio State University
• Georgetown University
• Université du Texas
• The University of Western Ontario

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• Publié par: vien0
• Nombre de lectures: 26
• Langue: English

Extrait

MATH 203, Section 001 Fall, 2011 Exam 4

Name

Solutions

1. LetP 1 denote the vector space of polynomials of degree at most 1.Let   a   3 T:P 1 →Rbe given byT(ax+b) =ab. DetermineifTis a linear a transformation. Giveclearreasons for your answer.    10 2    T(5(2x=+ 3))T(10x+ 15)= 150, but5T(2x6 == 5+ 3) 10 2   10   306=T(5(2x+ 3)), soTis not a linear transformation. 10 (Of course, there are many other examples that work.)

Tofmrarsn.ntaoiiseartalin

× T.onorsftimaraennartonsiilat

2 2. SupposethatBis the basis forRsuch that the change of coordinate   4 1 matrixP B fromBto the standard basis is given byP B =.If 3 0   2 [v~] B find= ,~v. −3     4 12 5 v~=P B [~v] B = =. 3 0−3 6 Answer:   5 6

2 3. SupposethatBis the basis forRsuch that the change of coordinate   4 1 matrixP B fromBto the standard basis is given byP B =. 3 0 Find the change of basis matrix from the standard basis toB.

−1 The change of basis matrix from the standard basis toBisP= B     1 0−1 1 0−1 0 1 3 ( )= () = −4 . det(P B )−3 −3 4−3 41 3

Answer:   1 0 3 −4 1 3

   1 1 2 4. Letthe basesBandCforRbe given byB={,}and 1−1    2 1 P C={,}. Findthe change of basis matrix C←B . 1 0    1 1 P P The matrix C←B is given by C←B =, so we must 1−1       C C 1 21 12 solve the equations=x+yand=x+ 1 10−1 1   1 y. Wecan do both of these simultaneously if we row-reduce 0 the matrix whose first two columns are the elements ofCand whose second two columns are the elements ofB. [Ofcourse, it is fine to do these systems separately.]     2 1 11 10 1−0 11 1−1 ∼ ∼. 1 0 1−1 21 01 11−1 3   1−1 P Therefore, C←B =. −1 3

Answer:   1−1 −1 3

    1 1 1     3 5. Suppose thatBis the basis forRgiven by{ 1 , 1 , 0 } 1 0 0 3P andCis a basis forRsuch that the change of basis matrix C←B is    1 0 23    1 1−Find 2 . 1 . 3−2 11 C         3 11 13 1         Since + 1 + 02 = 1 , 12 = . Therefore, 1 10 01 1 B       3 10 21 3       2 =1 1−1 1= 1 . 1 3−1 22 1 C

Answer:   3   1 2   1 01   6. Oneeigenvalue of the matrixA = 1−1 0 Find an eigen-is 2. 1 01 vector corresponding to the eigenvalue 2.   x   We want a non-zero vector~v= y such thatA~v= 2v~, that z is, we want a non-trivial solution of the homogeneous system ~ Av~−2~v= 0find such a solution, we row-reduce the coefficient. To     −1 0 1−1 01 10−1 1     matrix. 1−3 0 ∼ 0−3 1 ∼ 0 1− .Therefore, 3 1 0−0 01 000 0 z zis free, withx=zandy=one eigenvector is. Hence, 3   3   1 . 3

Answer:   3   1 3

  0−2 0   7. Findthe characteristic polynomialp(λ) of the matrixA = 23−1 . 0 1 1   λ2 0   p(λ) =det(λI−A) =det( −2λ− ) = 3 1 λ[(λ−3)(λ−1) + 0−1λ−1 3 2 1] + 2[2(λ−1)] =λ−4λ+ 8λ−4.

3 2 p(λ) =:λ−4λ+ 8λ−4

  3−2 8. LetA= .Find all eigenvalues ofA. −1 4

The characteristic polynomial ofAisp(λ) =det(λI−A) =   λ−2 2 2 det=λ+ 7λ= (+ 10λ−2)(λ−5). Therefore, 1λ−4 the eigenvalues are2and5.

Answer: 2and5

  3 5 01 2−11 2   9. LetA= .Then the characteristic polynomial ofAis   1 2−1 0 6 3 02 4 32 54 3 2 p(λ) =λ−3λ−26λ−15λ+7. FindA−3A−26A−15A+A. [Hint: The easiest way to do this is to use the Cayley-Hamilton Theorem.]

4 32 By the Cayley-Hamilton Theorem,p(A) =A−3A−26A−15A+ 7I= 0. 5 43 2 Long division gives thatλ−3λ−26λ−15λ+λ=λp(λ)−6λ, 5 43 2 soA−3A−26A−15A+A=AP(A)−6A=A∙0−6A=   −18−30 0−6 −12 6−12−6  .   −6−12 6 0 −36−18 0−12

Answer:   −18−30 0−6 −12 6−12−6    −6−12 6 0 −36−18 0−12   3 4 10. LetAThen the eigenvalues of= .Aare−1 and 5, with 2 1    1 2 corresponding eigenvectorsand .Find a matrixPsuch −1 1 −1 thatP APis a diagonal matrixDand findD.

We takePto beP B whereBis a basis consisting of eigenvectors    1 2 ofAlinearly independent eigenvectors,. Sinceand are −1 1   1 2 we can takeP=matrix. TheDhas the corresponding −1 1   −1 0 eigenvalues on the main diagonal, soD=. 0 5

  1 2 P=: −1 1

  −1 0 D=: 0 5