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Unformatted text preview: MATH 152 L1 & L2 Applied Linear Algebra and Differential Equations Spring 200405 ¥ Week 10 Worksheet Solutions : Euclidean Vector Q1. (Linear combination of vectors) Consider the vectors v 1 = (1 , 3 , 0), v 2 = ( 1 , 1 , 1), v 3 = (3 , 1 , 1). Is b = (3 , 5 , 1) a linear combination of v 1 , v 2 , v 3 ? Solution c 1 v 1 + c 2 v 2 + c 3 v 3 = b for some c 1 ,c 2 ,c 3 ⇐⇒ c 1 1 3 + c 2  1 1 1 + c 3 3 1 1 = 3 5 1 ⇐⇒ c 1 c 2 +3 c 3 = 3 , 3 c 1 + c 2 + c 3 = 5 , c 2 c 3 = 1 has solutions . Therefore we need to check whether the above system of linear equations has solutions for ( c 1 ,c 2 ,c 3 ) or not. By doing this, we simplify the augmented matrix as follows 1 1 3 3 3 1 1 5 1 1 1  3 R 1 + R 2→ 1 1 3 3 4 8 4 1 1 1 R 3 + R 1→ 4 R 3 + R 2 1 2 2 4 1 1 1  1 4 R 2→ R 2 ↔ R 3 1 2 2 1 1 1 1  2 R 3 + R 1→ R 3 + R 2 / £ ¡ ¢ 1 2 / £ ¡ ¢ 1 1 / £ ¡ ¢ 1 . Since the bcolumn is nonpivot, the system has solutions for ( c 1 ,c 2 ,c 3 ). In fact, the reduced row echelon form implies that we have b = (2) · v 1 + ( 1) · v 2 + (0) · v 3 = 2 v 1 v 2 . Q2. (General solution in vector form) Consider A = 2 2 1 1 1 1 1 2 1 , b = 2 1 2 . (a) Find a special solution x = ( x 1 ,x 2 ,x 3 ,x 4 ) of Ax = b satisfying x 1 = x 2 = 0. Solution Since the vector b is identical to the third column of A , we find a special solution x p = ( x 1 ,x 2 ,x 3 ,x 4 ) = (0 , , 1 , 0) satisfying that Ax p = b . (b) Solve the homogeneous equation Ax = . Solution We simplify the coefficient matrix A as follows A = 2 2 1 1 1 1 1 2 1  2 R 2 + R 1→ R 1 ↔ R 2 1 1 1 2 4 1 1 2 1  R 3 + R 1→ 2 R 3 + R 2 1 1 1 3 1 2 1 1 3 R 2→ R 2 ↔ R 3 1 1 1 1 2 1 1 R 3 + R 1→ R 3 + R 2 / £ ¡ ¢ 1 1 / £ ¡ ¢ 1 2 / £ ¡ ¢ 1 . The general solution of Ax = is given by x h = x 1 x 2 x 3 x 4 =  x 3 2 x 3 x 3 = x 3  1 2 1 , where x 3 is the free variable....
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This note was uploaded on 09/30/2010 for the course MATH MATH152 taught by Professor Kcc during the Spring '10 term at HKUST.
 Spring '10
 KCC
 Math, Linear Algebra, Algebra, Equations, Vectors

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