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Monthly Archives: October 2013
Linear Algebra and Its Applications, Review Exercise 2.8
Review exercise 2.8. Do the following: a) Find a matrix whose nullspace contains the vector . b) Find a matrix whose left nullspace contains the vector . c) Find a matrix with column space spanned by and row space spanned … Continue reading
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Tagged column space, left nullspace, nullspace, row space, spanning set
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Linear Algebra and Its Applications, Review Exercise 2.7
Review exercise 2.7. Find the most general solution to the following system of linear equations: Answer: This system corresponds to the system where The general solution to this system is a combination of a particular solution to plus the homogeneous … Continue reading
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Linear Algebra and Its Applications, Review Exercise 2.6
Review exercise 2.6. Given the matrices find bases for each of their four fundamental subspaces. Answer: The second column of is equal to twice the first column so the rank of (and the dimension of the column space of ) … Continue reading
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Linear Algebra and Its Applications, Review Exercise 2.5
Review exercise 2.5. Given the matrices find their ranks and nullspaces. Answer: We can use elimination to reduce to echelon form. We first exchange the first and third rows: and then subtract 1 times the second row from the third … Continue reading
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Linear Algebra and Its Applications, Review Exercise 2.4
Review exercise 2.4. Given the matrix find its echelon form and the dimensions of the column space, nullspace, row space, and left nullspace of . Answer: We perform elimination on to reduce it to echelon form. In the first step … Continue reading
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Linear Algebra and Its Applications, Review Exercise 2.3
Review exercise 2.3. State whether each of the following is true or false. If false, provide a counterexample. i) If a subspace is spanned by a set of vectors through then the dimension of is . ii) If and are … Continue reading
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Linear Algebra and Its Applications, Review Exercise 2.2
Review exercise 2.2. Find a basis for a twodimensional subspace of that does not contain , , or . Answer: One approach is to come up with a linear system that has a twodimensional nullspace that excludes the coordinate vectors. … Continue reading
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