has the desired form. On the other hand, its product with z is 1. So the inverse x


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1 First homework ssignment p. 5 Exercise. Verify tht the set of complex numers of the form x + y 2, where x nd y re rtionl, is sufield of the field of complex numers. Solution: Evidently, this set contins 0 nd. It is lso esy to check tht it is closed under the ddition nd multipliction. Then most of the xioms of the field follows directly from the corresponding properties of complex numers. The nontrivil prt of this exercise is to prove tht for ny element z = x + y 2, where x nd y re rtionl, its inverse lso hs this form. Consider n element z = x y 2. Then the product zz = x 2 2y 2 z is rtionl. Hence, the numer x 2 2y 2 hs the desired form. On the other hnd, its product with z is. So the inverse x is given y the formul y x 2 2y 2 x 2 2y 2. 2 Exercise 3. Are the following two systems of liner equtions equivlent? If so, express ech eqution in ech system s liner comintion of the equtions in the other system. x + x 2 + 4x 3 = 0 x + 3x 2 + 8x 3 = x 2 + 5x 2 3 = 0 x x 3 = 0 x 2 +3x 3 = 0 Solution: For instnce, let us express the first eqution of the first system s the liner comintion of the equtions of the second system. Use the method of undetermined coefficients: x + x 2 + 4x 3 = (x x 3 ) + (x 2 + x 3 ), where, re the numers tht we wnt to find. Compring coefficients with x in oth sides, one gets =, nd compring coefficients with x 2, one gets =. Then it is esy to check tht the coefficients with x 3 in oth sides coincide. Hence, the first eqution of the first system is the liner comintion of the equtions of the second system with coefficients nd. To complete the solution one just need to pply the sme method nd express the other equtions of the first system vi the equtions of the second one, nd vice vers. Exercise 5 Let F e set which contins exctly two elements, 0 nd. Define n ddition nd multipliction y the tles:
2 Verify tht the set F, together with these two opertions, is field. Solution The solution ws explined in the first recittion. Exercise Prove tht ech sufield of the complex numers contins every rtionl numer. Solution: Let F e sufield of C. Since F is field it must contin distinguished element 0 which plys the role of the dditive identity. But this element is unique (regrded s complex numer) therefore 0 = 0, where 0 is the usul 0 of the complex numers. Therefore 0 F. Anlogously we my conclude tht F. But if F then we must hve tht + = 2 F (notice tht + = 2 since the ddition inside of F is the sme ddition s the usul ddition of complex numers). Using induction we conclude tht N F (if n F then since ddition is closed n + F ). But F must contin ll the dditive inverses of ll its elements, in prticulr, since N F we must hve Z F. Also, since F is field, it must contin ll multiplictive inverses of its elements, therefore Z := { n Z} F. Now let with, Z, 0 e ny rtionl numer, we n know Z F nd F therefore = F. Tht is, F contins every Z rtionl numer. pp. 0 Exercise 2. If 3 2 A = find ll solutions of AX = 0 y rowreducing A. Solution: Denote with S(A, B) the opertion of swpping rows A nd B. And denote with c A + B the opertion of multiplying row A times c nd dding it to row B S(I,III) 2 2 I+II 0 3 I+III II II+III III 6 0 0
3 Therefore the only solution of AX = 0 is X = 0. Exercise 3. If A = find ll solutions of AX = 2X nd ll solutions of AX = 3X. Solution Notice tht the eqution AX = 2X is equivlent to the eqution AX 2X = 0 which is equivlent to AX 2IX = 0 where I is the 3 3 identity mtrix. This lst eqution cn e fctored s (A 2I)X = 0. Writing explicitly A 2I we get A 2I = which is row equivlent to thus the solutions re the elements in the set{(x, x 2, x 3 ) x = x 2 = x 3 } Anlogously for the system AX = 3X we get tht the solutions re the elements in the set {(x, x 2, x 3 ) x = x 2 = 0}. Exercise 8. Consider the system of equtions AX = 0 where ( ) A = c d is 2 2 mtrix over the field F. Prove the following. () If every entry of A is 0, then every pir (x, x 2 ) is solution of AX = 0. Solution: 0 x + 0 x 2 = 0. () If d c 0, then the system AX = 0 hs only the trivil solution x = x 2 = 0. Solution I: Define A in the following wy: A = d c ( d ) c Notice tht A A = I therefore, if AX = 0 then AA X = 0, i.e. IX = X = 0 Solution II: Since t this point we re not supposed to know how to multiply mtrices let us work horrily complicted solution. goes y cses: The proof 3
4 . = 0 in this cse, since d c 0 we know tht c 0 therefore 0 nd c 0. Therefore we cn perform the following reduction: 0 S(I,II) c d ( ) c I d c II d d c c II+I 0 c d Thus AX = 0 nd IX = 0 hve the sme solutions, ut the ltter system only hs the trivil solution In this cse we cn perform the following reduction: ( ) I ci+ii d c II c d c d 0 d c II+I Thus AX = 0 nd IX = 0 hve the sme solutions, ut the ltter system only hs the trivil solution. (c) If d c = 0 nd some entry of A is different from 0, then there is solution (x 0, x 0 2) such tht (x, x 2 ) is solution if nd only if there is sclr y such tht x = yx 0, x 2 = yx 0 2. Solution: Suppose tht is the nonzero entry. Since d c = 0 we know tht d = c. Thus we cn perform the following reduction of A:: ( ) I ci+ii c d c d 0 0 Thus AX = 0 if nd only if x = x 2, letting x 0 = nd x0 2 = we otin wht we wnt. An nlogous rgument cn e given if we ssume ny other of the entries to e different from 0. pp. 56 Exercise 3. Descrie explicitly ll 2 2 rowreduced echelon mtrices. Solution: The first element in the first row of rowreduced echelon mtrix is either or 0. In either cse the first element in the second row is 0. The second entry in the second row is gin either or 0. However, the former is possile only if the first entry in the first row is not 0. So there re 3 possiilities to fill out ll entries except the upper right: ( ) x x 0 x,, In the first cse, x is necessrily 0, in the third cse x is either or 0, while in the second cse x cn e ritrry. Thus n infinite fmily of 2 2 rowreduced echelon mtrices prmeterized y n element c F ( ) c 0 0 4
5 nd three other 2 2 rowreduced echelon mtrices ( ) ,, exhust ll possiilities. Exercise. Find ll solutions of 2x 3x 2 x 3 + 5x 4 + 2x 5 = 2 x 2x 2 4x 3 + 3x 4 + x 5 = 2 2x 4x 3 + 2x 4 + x 5 = 3 x 5x 2 x 3 + 6x 4 + 2x 5 =. Solution: Consider the ugmented mtrix of this system It is esy to check tht the sum of the first nd the second rows is equl to the sum of the third nd the fourth rows. Hence, the fourth row is the liner comintion of the other rows nd cn e eliminted. Now sutrct the second row times 2 from the first nd the second rows. We get Interchnge the first nd the second rows , nd sutrct the second row times 4 from the third one Consider the system corresponding to the ltter mtrix x 2x 2 4x 3 +3x 4 +x 5 = 2 x 2 +x 3 x 4 = x 5 = 5
6 We get tht x 5 =, x 2 = x 3 +x 4 +2, x = 2x 2 +4x 3 3x 4 3. Hence, for ny choice of (x 3, x 4 ) there is unique pir (x, x 2 ) such tht (x, x 2, x 3, x 4, ) stisfy the system. Then the solutions re prmeterized y two elements, F : x = + 2 x 2 = 2 + x 3 = x 4 = x 5 =. Exercise 0. Suppose R nd R re 2 3 rowreduced echelon mtrices nd tht the systems RX = 0 nd R X = 0 hve exctly the sme solutions. Prove tht R = R. Solution: The min ide is to prove tht ech row of ech mtrix is liner comintion of the rows of the other mtrix. First, suppose tht one of the mtrices R, R hs one zero row. Then the solutions of the corresponding system depend on 2 prmeters. Hence, the other mtrix lso hs zero row, nd the second rows of R, R coincide (they re oth zero). Consider 2 3mtrix A whose first row is the first row of R nd whose second row is the first row of R. The corresponding system gin hs the sme 2prmeter fmily of solutions s the systems RX = 0, R X = 0. Thus rowreduced echelon mtrix equivlent to A should hve one zero row. This is possile only if the rows of the mtrix A coincide (since oth rows of A hve leding coefficients ). Now suppose tht ll rows of mtrices R, R re nonzero. Then the solutions of the corresponding system depend on prmeter. Form 3 4mtrix B whose first two rows re the rows of R nd whose lst two rows re the rows of R. Then B should e row equivlent to mtrix with two zero rows (otherwise the system BX = 0 would not hve prmeter fmily of solutions). Hence, the rows of of R cn e represented s liner comintions of the rows of R nd vice vers. This is possile only if the leding coefficients of their first rows occur t the sme plce. Indeed, if for instnce, the first row R of R strts with more zeros thn the first row of R, then so do the second row R 2 of R nd ny liner comintion of R, R 2. Using similr rguments one cn prove tht the leding coefficients of the second rows of R, R occur t the sme plce. Then it is esy to see tht the only wy for row of R to e liner comintion of the rows of R is to coincide with the respective row of R. 6
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