SOLUTIONS TO HOMEWORK #1, MATH 54 SECTION 001, SPRING 2012

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1 SOLUTIONS TO HOMEWOK #, MATH 5 SECTION, SPING JASON FEGUSON. Bewre of typos. In fct, your solutions my e etter thn mine.. Some prolems hve scrtch work nd solutions. The scrtch work sys how I cme up with my solution nd the solution is wht I would ctully turn in.. Ex...6: Consider ech mtrix in Exercises 5 nd 6 s the ugmented mtrix of liner system. Stte in words the next two elementry row opertions tht should e performed in the process of solving the system First dd times ow to ow to turn the in ow into. Tht will chnge ow to [ 5 5 ], so then multiply the new ow y 5 to turn the 5 into. Multiplying ow y to turn the into is lso resonle.. Ex...: In Exercises 7-, the ugmented mtrix of liner system hs een reduced y row opertions to the form shown. In ech cse, continue the pproprite row opertions nd descrie the solution set of the originl system The solution is (x, x, x, x ) = (, 5, 6, ).. Ex...: Solve the systems in Exercises x x = 8 x +x +9x = 7 x +5x = The solution is (x, x, x ) = (5,, ).

2 . Ex...6: Determine if the systems in Exercises 5 nd 6 re consistent. Do not completely solve the systems. x x = x +x = x +x = x +x +x + x = This mtrix corresponds to the system x x = x +x = x +x = A solution cn e found y picking some numericl vlue for x, for exmple x =, nd then using it to find x, x, nd x, so the system is consistent. 5. Ex...: In Exercises 9-, determine the vlue(s) of h such tht the mtrix is the ugmented mtrix of consistent liner system. h 6 This mtrix corresponds to the system: h + h 6 h + x + hx = (h + )x = Then (x, x ) = (, ) is lwys solution, no mtter wht h is, so the corresponding system is consistent for ll numers h. h In. you lerned nother wy to see tht corresponds to consistent system h + no mtter wht h is, the mtrix is in echelon form nd does not hve row of the form [ ] where is nonzero. 6. Ex...8: Suppose,, c, nd d re constnts such tht is not zero nd the system elow is consistent for ll possile vlues f nd g. Wht cn you sy out the numers,, c, nd d? Justify your nswer. x +x =f cx +dx =g

3 Scrtch Work. Since is nonzero, we cn divide y it to row-reduce: [ f f ( c) + f d c g cf (If you re worried tht c might e zero, you don t need to. When you multiply row y constnt, tht constnt hs to e nonzero. But dding constnt multiple of one row to different row works even when tht constnt is zero.) If d c, the system is consistent no mtter wht f nd g re. Even if d c =, the system is still consistent for some numers f nd g. For exmple, if d c = nd f = g =, then the ugmented mtrix ecomes:. ] But the prolem sys tht the system is consistent for ll numers f nd g, not just some numers. For exmple, if d c = nd f = nd g =, the ugmented mtrix ecomes:, which corresponds to n inconsistent system. [ f f ] [ ( c) + f d c g cf ] If d c, the system is consistent for ll numers f nd g. For exmple, we cn divide the eqution ( ) d c x = g cf y d c to find x, nd then use x + x = f to solve for x. If insted d c =, then if f = nd g = the ugmented mtrix ecomes:, which corresponds to n inconsistent system. Therefore, we cn sy d c nd not much else. Becuse, you cn rewrite the finl nswer s d c. The eqution d c is lso the correct nswer even when =, ut you will need completely different steps in your row reduction to show it. In Chpters nd you ll lern techniques tht don t depend on whether is zero nd tht will give you the finl nswer d c with lmost no work. 7. Ex...: In Exercises nd, determine which mtrices re in reduced echelon form nd which re only in echelon form... c. d. The leftmost nonzero entries in the nonzero rows of the mtrices re mrked with oxes:

4 . The zero row is elow the nonzero rows. The oxed entries re ll nd go from top-left to ottom-right, nd ll entries directly ove nd elow the oxed entries re, so the mtrix is in reduced echelon form.. The oxed entries in the nonzero rows go from top-left to ottom-right, nd ll entries directly elow the oxed entries re, so the mtrix is in echelon form. However, there is ove the oxed entry in ow, so the mtrix is only in echelon form. c. The oxed entry in ows is not to the right of the oxed entry in ow, so the mtrix is not in echelon form, nd therefore is lso not in reduced echelon form. d. The zero row is elow the nonzero rows. The oxed entries go from top-left to ottom-right, nd ll entries directly elow the oxed entries re, so the mtrix is in echelon form. However, the oxed entry in ow is nd not, so the mtrix is only in echelon form. 8. Ex...: Find the generl solutions of the systems whose ugmented mtrices re given in Exercises The generl solution is (x, x, x ) = ( u v, u, v), where u nd v re ny numers. 9. Ex...6: Exercises 5 nd 6 use the nottion of Exmple for mtrices in echelon form. [In other words, ech is nonzero numer, not necessrily ll the sme. Ech is ny numer, not necessrily ll the sme.] Suppose ech mtrix represents the ugmented mtrix for system of liner equtions. In ech cse, determine if the system is consistent. If the system is consistent, determine if the solution is unique.... There is no row of the form [ ], so the system is consistent. Ech column except the rightmost column hs in it, so the system hs no free vriles nd therefore hs unique solution.. There is no row of the form [ ], so the system is consistent. The second column hs no in it, so the second vrile is free. Thus, the system does not hve unique solution.. Ex...: Suppose 5 coefficient mtrix for system hs three pivot columns. Is the system consistent? Why or why not? Scrtch Work. So fr, the only wy we ve lerned to decide if system is consistent is to row-reduce the ugmented mtrix nd see if there is pivot in the rightmost column. So tht is wht we need to do for this prolem. Tht mens we ll need to tke wht the prolem gives us nd use it to figure out something out wht hppens when we row-reduce the 6 ugmented mtrix. The key thing to relize is, the leftmost 5 columns of the ugmented mtrix re the coefficient mtrix. Tht mens if you tke row opertions tht will row-reduce the coefficient mtrix nd pply them to the ugmented mtrix, then wht they will do to the rightmost column is unpredictle, ut they will row-reduce the leftmost 5 columns of the ugmented mtrix.

5 Now, once we ve row-reduced the leftmost 5 columns of the 6 ugmented mtrix, we re lmost completely done with row-reducing the 6 mtrix. In fct, we will e done row-reducing if we hve lredy found pivot in every row mong the leftmost five columns, nd in tht cse there will e no pivots in the rightmost column. Here we cn finlly use wht the prolem gives us. We re given tht the 5 coefficient mtrix hs three pivot columns. Tht mens tht once you row-reduce the coefficient mtrix, there will e three different pivots, nd they hve to e in three different rows, so y the previous prgrph, we re finished. Think out the 6 ugmented mtrix. If you tke row opertions tht reduce the coefficient mtrix nd pply them to the ugmented mtrix, you will get 6 mtrix tht I will cll A, whose leftmost five columns re n echelon form of the coefficient mtrix. Ech pivot column of the coefficient mtrix hs only one pivot in it, so the coefficient mtrix hs three pivots. Ech of these pivots hve to e in different row, so ech of the three rows of coefficient mtrix hve pivot. Tht mens tht ll three rows of A re nonzero, nd tht the leftmost nonzero entry of ech row of A is in one of the leftmost five columns. Tht mens A is in echelon form nd hs no pivots in the rightmost column, so the system is consistent.. Ex...5: Suppose the coefficient mtrix of system of liner equtions hs pivot in every row. Explin why the system is consistent. Scrtch work. If you look t the solution to the previous prolem, the thing we relly needed to know out the coefficient mtrix ws tht there ws pivot in every row, not tht the mtrix ws 5. Tht mens tht the sme solution will work here. Suppose the coefficient mtrix is m n, nd think out the m (n+) ugmented mtrix. If you tke row opertions tht reduce the coefficient mtrix nd pply them to the ugmented mtrix, you will get n m (n + ) mtrix tht I will cll A, whose leftmost n columns re n echelon form of the coefficient mtrix. Becuse the coefficient mtrix hs pivot in every row, every row of A is nonzero, nd the leftmost nonzero entry of ech row of A is in one of the leftmost n columns. Tht mens A is in echelon form nd hs no pivots in the rightmost column, so the system is consistent.. Ex...6: In Exercises 5 nd 6, write system of equtions tht is equivlent to the given vector eqution. [ 8 x + x + x 5 =. 6 ] Do the sclr multiplictions nd dditions nd then compre entries to get: x +8x + x = x +5x 6x =. Ex...: In Exercises 9 nd, write vector eqution tht is equivlent to the given system of equtions. x + x +x = 9 x 7x x = 8x +6x 5x =5 Do the sclr multiplictions nd dditions nd then compre entries to get: x + x 7 + x =

6 . Ex...: In Exercises nd, determine if is liner comintion of,, nd. =, =, = 5 6, = 8 6 Scrtch Work. Sying is liner comintion of,, nd is the sme s sying there re numers x, x, nd x for which x + x + x =. If you rephrse this question s system, the prolem is sking if: is consistent. So just try to solve this system. x +5x = x + x 6x = x +8x = This mens tht system of equtions corresponding to the vector eqution x + x + x = is consistent, so is liner comintion of,, nd. 5. Ex...: In Exercises nd, determine if is liner comintion of the vectors formed y the columns of A. 6 A = 7, = Scrtch work. ephrse this question in terms of consistency of system of liner equtions, just like in the previous prolem So if,, nd re the three columns of A, the system of equtions corresponding to the vector eqution x +x +x = is consistent, so is liner comintion of the columns of A. 6. Ex...8: Let v =, v =, nd y = h 5. For wht vlue(s) of h is y in the 8 plne generted y v nd v? Scrtch work. The plne generted y v nd v is the set of ll points in tht cn e written s x v + x v for some numers x nd x. So the question is sking to find ll numers h for which y is liner comintion of v nd v. h 5 + h 5 8 h + h 5 h + 7 So if h = 7, the system corresponding to x v + x v = y is consistent, nd otherwise it is not. Therefore, h = 7 is the only numer for which y is in the plne generted y v nd v. 6

7 7. Ex...6: Let A = 6 8 5, let =, nd let W e the set of ll liner comintions of the columns of A.. Is in W?. Show tht the third column of A is in W. Scrtch work. According to the prolem, W just mens everything in tht cn e written s liner comintion of the columns of A. So sking if is in W is roundout wy of sking if cn e written s liner comintion of the columns of A So if,, nd re the three columns of A, the system of equtions corresponding to the vector eqution x + x + x = is consistent, so is liner comintion of the columns of A. This mens is in W.. Let the three columns of A e,, nd. Becuse = + +, is liner comintion of the columns of A, so is in W. 7