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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 = ˆ 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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