MATH221 April 2010
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• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q4 • Q5 • Q6 • Q7 (a) • Q7 (b) • Q7 (c) • Q8 (a) • Q8 (b) • Q9 • Q10 • Q11 • Q12 (a) • Q12 (b) •
Question 01 (b)
Then the matrix is an echelon form for (you may assume this, you don't have to do the row reduction again.)
(b) Find a basis for the column space of .
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The pivot columns of are the same as the pivot columns of the echelon form .
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To find a basis for the column space of we need to find linearly independent columns of . From part (a) we know that the dimension of the column space is 3.
Since and share the same pivot columns, we know that the columns 1, 2 and 4 are linearly independent. Therefore, a basis for the column space of is given by
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