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415a7.3

# 415a7.3 - Section 7.3 Systems of Linear Algebraic Equations...

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Section 7.3: Systems of Linear Algebraic Equations; Linear Indepen- dence, Eigenvalues, Eigenvectors Problem 6. Let x (1) = (1 , 1 , 0) , x (2) = (0 , 1 , 1) , and x (3) = (1 , 0 , 1) . To determine whether these vectors are linearly independent, we check if the following equation c 1 x (1) + c 2 x (2) + c 3 x (3) = 0 (6.1) has only the zero solution, i.e., ( c 1 , c 2 , c 3 ) = 0 . Eq.(6.1) can be expressed as c 1 1 1 0 + c 2 0 1 1 + c 3 1 0 1 = 0 0 0 or, equivalently, 1 0 1 1 1 0 0 1 1 c 1 c 2 c 3 = 0 0 0 . (6.2) In terms of the augmented matrix, one has 1 0 1 | 0 1 1 0 | 0 0 1 1 | 0 . 1. Adding ( - 1) times the first row to the second row: 1 0 1 | 0 0 1 - 1 | 0 0 1 1 | 0 . 2. Adding ( - 1) times the second row to the third row: 1 0 1 | 0 0 1 - 1 | 0 0 0 2 | 0 . Hence one has the system c 1 + c 3 = 0 , c 2 - c 3 = 0 , 2 c 3 = 0 that is equivalent to Eq.(6.1), the solution of which is ( c 1 , c 2 , c 3 ) = 0 . Hence x (1) = (1 , 1 , 0) , x (2) = (0 , 1 , 1) , and x (3) = (1 , 0 , 1) are linearly independent. Alternatively, one has det 1 0 1 1 1 0 0 1 1 = 2 negationslash = 0 . 1
This implies that the only solution of Eq.(6.1) is the zero vector ( c 1 , c 2 , c 3 ) = 0 . Problem 7. Let x (1) = (2 , 1 , 0) , x (2) = (0 , 1 , 0) , and x (3) = ( - 1 , 2 , 0) . To determine whether these vectors are linearly independent, we check if the following equation c 1 x (1) + c 2 x (2) + c 3 x (3) = 0 (7.1) has only the zero solution, i.e., ( c 1 , c 2 , c 3 ) = 0 . Eq.(7.1) can be expressed as c 1 2 1 0 + c 2 0 1 0 + c 3 - 1 2 0 = 0 0 0 or, equivalently, 2 0 - 1 1 1 2 0 0 0 c 1 c 2 c 3 = 0 0 0 . (7.2) In terms of the augmented matrix, one has 2 0 - 1 | 0 1 1 2 | 0 0 0 0 | 0 . 1. Multiplying the first row with 1 / 2 : 1 0 - 1 / 2 | 0 1 1 2 | 0 0 0 0 | 0 .

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415a7.3 - Section 7.3 Systems of Linear Algebraic Equations...

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