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Linear Algebra with Applications (3rd Edition)

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Math 417 homework 3 solutions (Given only for problems that are not straightforward computation; contact the instructor if you still have questions about the others.) Problem 12,20,24 Finding things “by inspection” is of course whichever way you can. Systematic way: compute the reduced row echelon form. Column j of the original matrix is redundant if and only if the rref does not have a leading 1 in that same column j . Gather the non-redundant column in a list; that is a basis for the image. For each redundant column, read a kernel vector out of the rref, as shown in class. Problem 28 The first three vectors are clearly independent, because each has a 1 (nonzero) in a row where the previous ones have 0. Looking again at rows 1 to 3, the fourth vector can be redundant only if it is 2 times first plus 3 times second plus 4 times third vector. In that case k = 2 · 2 + 3 · 3 + 4 · 4 = 29. Hence the fourth vector is linearly independent if and only if k 6 = 29.
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hw3sol - Math 417 homework 3 solutions(Given only for...

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