33a_mid1_sol

33a_mid1_sol - Math 33A, Midterm 1 solutions 1. We first...

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Unformatted text preview: Math 33A, Midterm 1 solutions 1. We first write the augmented coefficient matrix and then perform Gauss-Jordan elimi- nations (row operations): 1 2- 1 2 | 3 3 6- 1 0 | 5 subtract 3 times row I from row II 1 2- 1 2 | 3 0 0 2- 6 | - 4 divide row II by 2 1 2- 1 2 | 3 0 0 1- 3 | - 2 add row II to row I 1 2 0- 1 | 1 0 0 1- 3 | - 2 From the RREF we see that variables y and w are going to be arbitrary parameters, while x and z are going to be expressed in terms of these parameters. We successively write: w = s, z = 3 s- 2 , y = t, x =- 2 t + s + 1 , for arbitrary real parameters s and t . We can also write the solution in the form x y z w = - 2 t + s + 1 t 3 s- 2 s . 2. The result is: - 1- 2 1 2 3 2 5 8 3. First observe that A = 1 2- 3 2 3 2 1 2 ! = 1 2 1- 3 3 1 although this is not crucial and we could have left A in the trigonometric form. Now we compute both products: BA = 1 2 a b c d 1- 3 3 1 = 1 2 a + b 3- a 3 + b c + d 3- c 3 + d AB = 1 2 1- 3 3 1 a b c d = 1 2 a- c 3 b- d 3 a 3 + c b 3 + d 1 Comparing corresponding entries in the first column, we obtain a + b 3 = a- c 3 and c + d 3 = a 3 + c , which gives b =- c and d = a . In that case entries in the second column are automatically equal. We conclude that B has the form B = a- c c a for arbitrary numbers a and c ....
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33a_mid1_sol - Math 33A, Midterm 1 solutions 1. We first...

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