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# VjekoFS - Mathematics 33A, Section 3, Final Exam Solutions...

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Unformatted text preview: Mathematics 33A, Section 3, Final Exam Solutions 1. (a) We write the augmented coefficient matrix and then perform Gauss-Jordan elimi- nation (i.e. row operations): 1 2 2 | - 1 1 1- 1 | 1 2 1- 5 | 4 subtract 3 times row I from row II, and subtract 2 times row I from row III 1 2 2 | - 1- 1- 3 | 2- 3- 9 | 6 divide row II by- 1, and divide row III by- 3 1 2 2 | - 1 0 1 3 | - 2 0 1 3 | - 2 subtract 2 times row II from row I, and subtract row II from row III 1 0- 4 | 3 0 1 3 | - 2 0 0 | From the reduced row echelon form we see that x 3 is arbitrarily parametrized, while x 1 and x 2 are expressed in terms of that parameter. We successively write: x 3 = t, x 2 =- 3 t- 2 , x 1 = 4 t + 3 , for arbitrary real parameters t . We can also write the solution in the form x 1 x 2 x 3 = 4 t + 3- 3 t- 2 t . (b) The coefficient matrix is A = 1 2 2 1 1- 1 2 1- 5 and from the computation in part (a) we see its reduced row echelon form: RREF( A ) = 1 0- 4 0 1 3 0 0 As before, we simply read off that the kernel is the set of all: x 1 x 2 x 3 = 4 t- 3 t t = t 4- 3 1 1 Therefore kernel( A ) = span( 4- 3 1 ) and thus one basis for kernel( A ) is: 4- 3 1 . 2. (a) If R is the rotation by 20 ◦ , then R 6 is a composition of 6 rotations, each by 20 ◦ . We conclude that R 6 is the rotation by 2 · 20 ◦ = 120 ◦ , and thus its matrix is R = • cos120 ◦- sin120 ◦ sin120 ◦ cos120 ◦ ‚ = "- 1 2- √ 3 2 √ 3 2- 1 2 # . (b) Answer: ∠ ( R 6 ~v,~v ) = 120 ◦ . R 6 ~v · ~v =- 25 2 . The angle between R 6 ~v and ~v is 120 ◦ , because R 6 ~v is obtained by rotating ~v by 120 ◦ . Also notice that k R 6 ~v k = k ~v k = p 3 2 + (- 4) 2 = 5. Therefore we get R 6 ~v · ~v = k R 6 ~v kk ~v k cos120 ◦ = 5 · 5 ·- 1 2 =- 25 2 . 3. (a) Answer: det A = a . For a 6 = 0. Since A is upper-triangular, we can easily compute its determinant as the product of diagonal entries: det A = 1 · 1 · a = a . We see that A is invertible precisely when a 6 = 0. (b) We use the usual algorithm for finding the inverse: write the identity matrix to the right and then perform Gauss-Jordan elimination until we obtain the identity matrix on the left side. 1 a a | 1 0 0 0 1 a | 0 1 0 0 0 a | 0 0 1 subtract a times row II from row I 1 0 a- a 2 | 1- a 0 1 a | 1 0 0 a | 1 divide row III by a , which we can since a 6 = 0 1 0 a- a 2 | 1- a 0 1 a | 1 0 0 1 | 1 a subtract ( a- a 2 ) times row III from row I, and subtract a times row III from row II 1 0 0 | 1- a a- 1 0 1 0 | 1- 1 0 0 1 | 1 a From the last augmented matrix we simply read off: A- 1 = 1- a a- 1 1- 1 1 a ....
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## This note was uploaded on 02/06/2011 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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VjekoFS - Mathematics 33A, Section 3, Final Exam Solutions...

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