133S7 - Muhammad Ulhaque due 11/14/07 at 11:00 PM....

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Unformatted text preview: Muhammad Ulhaque due 11/14/07 at 11:00 PM. Assignment 7 MATH133, Fall 2007 A Solution: A has 2 eigenvalues. One of them is eigenspace is and its corresponding 1 ¢ A ¢ 13 3 61 10 3 The determinant of A is . )0 ¢ ¢ &( ¢ 1 ¢ 4.(1 pt) Let   Hint: Use geometric reasoning! Solution: 4 3 T xy x y 2 x y 25 25 4 3 7 24 24 7 25 x 25 y 25 x 25 y T34 34 T4 3 43 since T fixes any point on the line and sends a normal vector to minus itself. Correct Answers:       ¢  ¢ ¢    $ ¢ ¢ ¢     #  !  %        "  ¢ ¢ ¦¤ ¥ span   The other eigenvalue of A is eigenspace is and its corresponding   ¦¤ ¡ ¥ span 2 1 2      © )0 &( ¢ &( )0 &( &( )0 ¢ ¢ ¢  ¢ ¢ &( 2.(1 pt) Let T be the reflection about the line 4x the euclidean plane. The standard matrix A of T is 3y 0 in x 2y z x y z   ¢       © )0 &( )0 x y z x y z x y z x y z © )0 )0 &( Solution: 1 30 0 2 3 39 2 1 30 0 2 3 39 2 1 30 2 3 0 39 2 Correct Answers: 0 0 -1 1 0 1 3 1 0 )0 )0 ¢ ¢ ¢ ¢ ¢ ¢ )0 )0 ¢ ¢ ¢ &( ¢ &( &( &( &( λ 24 λ2 4λ 3 0 2 9λ which implies λ 1 3 are the eigenvalues. The eigenvectors with eigenvalue 1 are the solutions of 6x 24y 0 and the eigenvectors with eigenvalue 3 are the solutions of 8x 24y 0. Correct Answers: 3 3 1 1 4 1 det 5 )0 ¢ ¤ ¡ ¡  " ¢         © x,y 5 24 29 x y λ x has a non-zero solution for y is an eigenvector to the eigenvalue 2 is an eigenvector to the eigenvalue -1 is an eigenvector to the eigenvalue -2 ¢ ' &(  ¢  ¢   ¤ ¢ ¢  ¢ ¢ ¡ ¢ ¤ ¤¤ ¢ ¢ ¨¡ ¦¤ ¡ ¥ ¡ span 1 30 0 20 392 Show that 2, -1 and -2 are eigenvalues of A by providing corresponding eigenvectors: xyz )0 ¢ ¢ ¢ The other eigenvalue of A is eigenspace is and its corresponding 3.(1 pt) Let   ¦ §¤ ¡ ¥ span      ¤ ¡ ¢ A  5 24 29 Compute the eigenvalues and eigenspaces of A: One eigenvalue of A is and the corresponding eigenspace is  ¢ £¡ 1.(1 pt) Let  -0.28 0.96 0.96 0.28 -1 4 -3 1 3 4 c00 1 xyz c101 xyz c310 is . Correct Answers: -2 2 C 7.(1 pt) Assume that A and B are 6 5 and det B 9. Then det 5A 6 matrices with det A ¢ A AA 0 ¢ ¢ ¢ ¢ 9 4 2 3 5 6 5 0 0 1 1 3 0 0 0 9 6 0 1 9 0 1 0 0 0 BAA) ¢ ¢( @ ¢ &@@@ ¢   ¢ -3 1 2 -8 0 6 -9 -1 1 5 1 -2 -8 0 5.(1 pt) Use cofactor expansion along any row or column that seems convenient to compute the determinants of the matrices (i) 28 63 30 0 0 A 0 5 30 2 3 4 2 det A= . (ii) 01 0 1 1 1 1 1 B 0 10 1 11 1 1 det B= . Correct Answers: -18 4 6.(1 pt) The determinant of the matrix BAA) 0 A )A 0 ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ¢ ( &@@ ( &@@         Solution: 1 1 6 1 M11 1 M12 2 M13 3 2 10 2 61 8 10 3 3 2 13 2 M21 0 M22 6 M23 3 2 10 2 13 3 9 10 3 3 2 13 2 M31 1 M32 1 M33 1 1 6 1 13 3 5 10 3 We have Ci j 1 i j Mi j . The adjoint of A is the transpose of the cofactor matrix C Ci j . T 1 2 8 1 0 1 06 9 26 1 adj A CT 1 15 8 95 det A a11C11 a21C21 a31C31 3 1 0 1 1 26 1 A 1 det1 A adj A 3 8 95 Correct Answers:  A  )0 ¢ 3  333 ¢ ¢ 3  333 ¢ ¢  33 33 ¢ ¢ ¢ ¢ ¢ &( )0 ¢ ¢   ¢¢  ¢ &( )0 ¢ 7 ¢ ¢ 33 ¢ 33 ¢ ¢ 33 ¢ 5 33 ¢ 33 ¢ 5 33 "  9  ¢ ¢ &( ¢ ¢ 62 $  ¢  3  ¢ 333 ¢ ¢  3 3 33 ¢ ¢  33 33 ¢ ¢  &( " 1   )0 and  adj A            )0 The minors Mi j of A are M11 , M12 , M13 , M21 , M22 , M23 , M31 , M32 , M33 . The cofactors Ci j of A are C11 , C12 , C13 , C21 , C22 , C23 , C31 , C32 , C33 . Finally, the adjoint of A is  6 9 -1 -1 5 1 0 -1 -2 6 -1 -8 9 5 -0.333333333333333 0 0.333333333333333 0.666666666666667 -2 0.333333333333333 2.66666666666667 -3 -1.66666666666667  &( 2  33 33 33 33 3 33 ¢ 3 3 33 34 3 33 ¢ 33 3 33 4 8 "                   ¢ ¢ 33 ¢ 33 ¢ ¢ 33 ¢ 33 33 33 A Prepared by the WeBWorK group, Dept. of Mathematics, University of Rochester, c UR D  ¢ ¢  ¢ &( k k k 1 5 4 2k2 4 1 5 k1 For k ,k and k , the matrix A is ? . For all other values of k, A is ? . Solution: k1 5 4 k2k2 4 det A k 4k 2k 3 k1 5 k1 so that A is not invertible if k = 4, 2, 3 and invertible otherwise. Correct Answers: 4 2 3 NOT INVERTIBLE INVERTIBLE 3 3       333  ¢ ¢ ¢ ¢ 33 ¢  333  3 ¢ 33       )0 det A5 det B 4 det A 1 BA det AT A det A 1 T BT 1 Correct Answers: 78125 3125 0.000152415790275873 -9 25 -0.0222222222222222 8.(1 pt) Let         " "  " "       ...
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This note was uploaded on 12/29/2010 for the course COMP 202 taught by Professor Verbrugge during the Fall '07 term at McGill.

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