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lab9_sol_f_08

# lab9_sol_f_08 - Math 115 Lab 9 Fall 2008 Topics Rank Column...

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Math 115 - Lab 9 - Fall 2008. Topics: Rank Column Space and Row Space Null Space and Image Space Dimensions of these Spaces You are to provide full solutions to the following problems. You are allowed, and encouraged to collaborate with your classmates, use your notes and textbook and ask the TA for guidance. Direct copying of solutions is not encouraged, nor is it allowed or ethical. Last name: First name: Student number:

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Math 115 - Lab 9 - Fall 2008. Student number: 1. Let A = - 1 0 1 0 2 0 1 0 2 0 2 0 - 1 0 - 2 0 2 0 4 0 3 0 - 1 0 - 2 . (a) Calculate the rank of A . Solution: A can be reduced to the row-echelon form (REF) R = - 1 0 1 0 2 0 1 0 2 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 rank A = 3. (b) Is A invertible? Why? Solution: Since A is 5 × 5 and rank A 6 = 5, A is not invertible. (c) Determine the basis of row A (the row space of A ) and dim(row A ). Solution: A basis of row A is the non-zero rows of R : a basis of row A = { [ - 1 0 1 0 2 ] , [ 0 1 0 2 0 ] , [ 0 0 1 0 2 ] } . dim(row A ) = rank A = 3. (d) Determine the basis of col A (the column space of A ) and dim(col A ). Solution: A basis of col A consist of the columns in A corresponding to the columns with pivot in R . The pivots are in columns 1 , 2 , 3 of R . columns 1 , 2 , 3 of A forms a basis of col A . a basis of col A = - 1 0 2 0 3 , 0 1 0 2 0 , 1 0 - 1 0 - 1 .
(e) Determine the basis of null A (the null space of A ) and dim(null A ). Solution: To determine a basis of null A , we solve the homogeneous system A~x = ~ 0 which is equivalent to (by row reducing A to R ): - 1 0 1 0 2 0 1 0 2 0 0 0 1 0 2 0 0 0 0 0 0 0 0 0 0 x 1 x 2 x 3 x 4 x 5 = 0 0 0 0 0 x 4 , x 5 are the free variables.

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lab9_sol_f_08 - Math 115 Lab 9 Fall 2008 Topics Rank Column...

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