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3.2 Rank of a Matrix - 3.2 3.2 The Rank of a Matrix and...

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3.2. 3.2. The Rank of a Matrix and Matrix Inverse 1. (a) F (Theorem 3.5) (b) F ( ) If A M m × n ( F ) , B M n × n ( F ) and B is invertible, then the rank(AB)=rank(A) (Example) A = B = 0 1 0 0 and rank ( A ) = rank ( B ) = 1 AB = 0 0 0 0 and rank ( AB ) = 0 (c) T (d) T (p.153, Corollary to the Theorem 3.4) (e) F (p.153, Corollary to the Theorem 3.4) (f) T (p.153, Theorem 3.4 and Theorem 3.5) (g) T (p.161) (h) T ( ) A M m × n ( F ) , rank ( A ) = dimR ( L A ) n (i) T 2. (a) A = 1 1 0 0 1 1 0 0 0 , rank ( A )=2 162 PNU-MATH
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3.2. (b) A = 1 1 0 0 - 1 1 0 0 1 , rank ( A )= 3 (c) A = 1 0 2 0 1 2 , rank ( A )= 2 (d) A = 1 2 1 0 0 0 , rank ( A )= 1 (e) A = 1 2 0 1 1 0 0 1 1 - 2 0 0 0 0 1 0 0 0 0 0 , rank ( A )= 3 (g) A = 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 , rank ( A )= 1 3. A M m × n ( F ) , rank ( A ) = 0 iff A is the zero matrix ( ) clear ( ) let A = ( e 1 , 0 , · · · , 0) , e 1 6 = 0 Since rank ( A ) = 0 , e 1 is dependent a F s.t e 1 = a 0 It’s contradict to e 1 6 = 0 4. (a) 1 1 1 2 2 0 - 1 2 1 1 1 2 1 0 - 1 2 1 0 1 3 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 , 2 163 PNU-MATH
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3.2. (b) 2 1 - 1 2 2 1 2 0 0 1 0 0 12 0 0 1 0 0 , 2 5. ( A | I n ) ˆ ( I n | B ) , B = A - 1 (a) 1 2 1 0 1 1 0 1 1 0 -1 2 0 1 1 -1 A - 1 = - 1 2 1 - 1 , rank ( A ) = 2 (b) 1 2 1 0 2 4 0 1 1 2 1 0 0 0 -2 1 rank ( A ) = 1, so @ A - 1 (c) 1 2 1 1 0 0 1 3 4 0 1 0 2 3 -1 0 0 1 1 0 -5 3 -2 0 0 1 3 -1 1 0 0 0 0 -3 1 1 rank ( A ) = 2, so @ A - 1 (d) 0 -2 4 1 0 0 1 1 -1 0 1 0 2 4 -5 0 0 1 1 0 0 1/6 15/9 -1/3 0 1 0 -5/18 -4/9 2/9 0 0 1 1/9 -2/9 1/9 rank ( A ) = 3 , A - 1 = 1 6 15 9 - 1 3 - 5 18 - 4 9 2 9 1 9 - 2 9 1 9 (e) 0 -2 4 1 0 0 1 1 -1 0 1 0 2 4 5 0 0 1 1 0 0 1/6 -1/3 1/2 0 1 0 1/2 0 -1/2 0 0 1 1/6 1/3 1/2 rank ( A ) = 3 , A - 1 = 1 6 - 1 3 1 2 1 2 0 - 1 2 - 1 6 1 3 1 2 (f) 0 2 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 0 1 0 1 0 0 1 0 -1/2 -1/2 0 0 0 0 -1/2
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