101st_tut3sol

101st_tut3sol - MATH1111/2010-11/Tutorial III Solution 1...

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MATH1111/2010-11/Tutorial III Solution 1 Tutorial III Suggested Solution 1. Consider the vectors a = (1 1 ± 1) T ; b = (2 0 1) T ; c = ( ± 1 1 ± 2) T ; d = (1 2 1) T : (a) Reduce the matrix ± a b c d ² in row echelon form. (b) Could one decide by part (a) whether a ;b ;c ;d form a spanning set for R 3 ? How? What else can you tell from the augmented matrix? (c) Find a basis from a ;b ;c ;d for R 3 . (d) Do a ;b form a basis for R 3 ? If no, could you add suitable vector(s) other than c or d to get a basis? Ans . (a) The matrix is 0 @ 1 2 ± 1 1 1 0 1 2 ± 1 1 ± 2 1 1 A . It can be reduced to the row echelon form 0 @ 1 2 ± 1 1 0 1 ± 1 2 = 3 0 0 0 1 1 A : Remark. The reduced row echelon form is 0 @ 1 2 ± 1 0 0 1 ± 1 0 0 0 0 1 1 A : (b) Yes. a ;b ;c ;d form a spanning set for R 3 . Below are two ways to explain: Explanation 1 . Let y = 0 @ y 1 y 2 y 3 1 A 2 R 3 . Consider c 1 a + c 2 b + c 3 c + c 4 d = y : In matrix form, the augmented matrix is 0 @ 1 2 ± 1 1 1 0 1 2 ± 1 1 ± 2 1 y 1 y 2 y 3 1 A ; which can be reduced to 0 @ 1 2 ± 1 1 0 1 ± 1 2 = 3 0 0 0 1 ² ² ² 1 A : ( ² denotes an entry whose value does not matter.) From the row echelon form, we see that the system is always solvable for c 1 ;c 2 ;c 3 ;c 4 . i.e. R 3 = Span( a ;b ;c ;d ).
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MATH1111/2010-11/Tutorial III Solution 2 Explanation 2 . From the row echelon form we see that a ;b ;d are linearly independent, as they are the vectors corresponding to the columns with leading 1’s. As dim R 3 = 3, any 3 linearly independent vectors form a basis. (See Theorem 3.4.3 (I).) In particular,
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This note was uploaded on 05/04/2011 for the course MATH 1111 taught by Professor Dr,li during the Spring '10 term at HKU.

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101st_tut3sol - MATH1111/2010-11/Tutorial III Solution 1...

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