# Suppose p is a plane through and l is a line through

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Chapter 7 / Exercise 13
Trigonometry
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19.SupposePis a plane through(0,0,0)andLis a line through(0,0,0). The smallestvector space containing bothPandLis eitheror.20.True or false forM=all 3 by 3 matrices (check addition using an example)? Problems 21–30 are about column spacesC(A)and the equationAx=b.21.Describe the column spaces (lines or planes) of these particular matrices:A=120000andB=100200andC=102000. 22.For which right-hand sides (find a condition onb1,b2,b3) are these systemssolvable? 142284142x1x2x3=b1b2b3.142914x1x2=b1b2b3.23.Adding row 1 ofAto row 2 producesB. Adding column 1 to column 2 producesC. A combination of the columns ofis also a combination of the columnsofA. Which two matrices have the same column?A=1224andB=1236andC=1326. 24.For which vectors(b1,b2,b3)do these systems have a solution?111011001x1x2x3=b1b2b3and111011000x1x2x3=b1b2b3.25.(Recommended) If we add an extra columnbto a matrixA, then the column spacegets larger unless. Give an example in which the column space gets largerand an example in which it doesn’t. Why isAx=bsolvable exactly when the columnspacedoesn’tget larger by includingb?26.The columns ofABare combinations of the columns ofA. This means:The columnspace of AB is contained in(possibly equal to)the column space of A. Give anexample where the column spaces ofAandABare not equal.27.IfAis any 8 by 8 invertible matrix, then its column space is. Why?28.True or false (with a counterexample if false)? A.(d) The column space ofAIequals the column space of A .
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Chapter 7 / Exercise 13
Trigonometry
McKeague/Turner Expert Verified
NOT FOR SALE Strang-5060bookMay 5, 200513:52772.2SolvingAx=0 andAx=b7729.Construct a 3 by 3 matrix whose column space contains(1,1,0)and(1,0,1)butnot(1,1,1). Construct a 3 by 3 matrix whose column space is only a line.30.If the 9 by 12 systemAx=bis solvable for everyb, thenC(A)=.31.Why isn’tR2a subspace ofR3?2.2SOLVINGAx=0 ANDAx=bChapter 1 concentrated on square invertible matrices. There was one solution toAx=b,and it wasx=A1b. That solution was found by elimination (not by computingA1).A rectangular matrix brings new possibilities—Umay not have a full set of pivots. Thissection goes onward fromUto a reduced formRthe simplest matrix that eliminationcan give.Rreveals all solutions immediately.
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