From the library of Prof. George A. Miller 951 Sorat oe oo 223 MGlid RA Tare wa “ MATHERATICS IAT HERAT] ‘ Return this book on or before the Latest Date stamped below. | University of Illinois Library 3 r-& m, 59 a? os - ~ — “S oy ~ ¥ ie My, ¢. Sys T a\ AT) } vy cS ia 27F ie ? 4969 NOV 3 1962 | 2 AY 1964) woy @ 3 MU |GET 2 § sons i 1987 at Verfo¥ L161—H41 fend Heal dla ye te ee mp ale-b)! ie--y/f Sep Mn J 4a(* “all -(e- ). '-3(0y bd ie “CIKITKIN T= 1077 nr ah | negok hen meas hee aulgesulp #(-d KF PF K-J= — 646 ", x-¥X/ [Pa ~ 45 fur frelt Brig oy ~ | Drang jg” ang EMER IX = 19 57 IXEK IK A > 4d | Huw Ax THEO, ae Fey, : “s a DETERMINANTS. AN INTRODUCTION TO THE STUDY OF, WITH EXAMPLES AND APPLICATIONS. BY G. A. MILLER,:A.M., PH.D., PROFESSOR OF MATHEMATICS, EUREKA COLLEGE. NEW YORK: D. VAN NOSTRAND COMPANY, 23 MURBAY AND 27 WARREN STREETS. 1892. ee a ile Tain 5 -Ll/ OO C offen, ihe By D. VAN ‘NOSTRAND Co. CO NTN ES: HISTORY OF DETERMIN ANTS NATURE OF DETERMINANTS INVERSIONS AND PERMUTATIONS EXAMPLES OF DETERMINANTS MEANING OF NOTATION . DIFFERENT METHODS OF NOTATION DETERMINANTS OF THE SECOND ORDER DETERMINANTS OF THE THIRD ORDER INCREASING THE ORDER BY BORDERS COMPLEMENTARY MINORS APPLICATION TO LINEAR EQUATIONS CONSISTENCE OF EQUATIONS FACTORS OF A DETERMINANT . MULTIPLICATION OF DETERMINANTS ° SYMMETRICAL DETERMINANTS. ELIMIN ATION lil PAGE DETERMINANTS. History. — The earliest known record of determinants is found in the writings of Leibnitz, who lived 1646-1716. He com- municated his discoveries to L’ Hospital in a letter dated April 28, 1693, and in a later letter he expressed his belief that their study would lead to many valuable discov- eries. We have no evidence that he or his correspondent pursued the study beyond a few of the most elementary principles. No trace of determinants can be found in the writings of the mathematicians who succeeded Leibnitz until 1750. In this year Cramer rediscovered them while working on the analysis of curves. From 1750 to 1826 the subject was studied by only a few eminent mathematicians, whose writings were suited only for advanced students. Among these writers were Bé- zout, Laplace, Lagrange, Gauss, Cauchy, ete. 1. 2 In 1826 Jacobi commenced a series of papers on this subject in “Crelle’s Jour- nal.” These papers continued for nearly twenty years, and by them the subject was made available for ordinary students. Many new and important theorems were also added. In late years the subject has been enriched by the writingsof many mathematicians, pre-eminent among whom are Sylvester and Cayley. Nature. — Given the equations aye + by =m ' age + boyy = Mz Multiplying the first of these equations by b,, and the second by 6,, we have a,box + Diboy = bem, f Al aeb,x + bboy = byme Subtracting the second squaden from the first, the result will be (Ab, — a2b,) & = bem, — bymg. B. By multiplying the first of equations A by Gz, and the second by a, and subtracting the first of the resulting equations from the second, we obtain (1b, aes gb;) Y = AyMs at Ag™My,. C. 3 From equations B and C the values of x and y are found to be b,m, — byme AyMz — A,M, We [stati UST Sa) 9, — ae ae nS Ab, — Aeh, yb, — Ab, If the notation Fs ii be employed to de- note 2 x 4—3 x 1;10e., the difference of the products obtained by multiplying di- agonally, we may write my, 0b, a 7 By eela ETE) PR ee ede cide a, 6, ay b, de dg He be Ook (ips UF e b; 3 4I, Ms be |, A, be}, etc., are determinants. Observations. — The denominator is the same for both unknowns, and is a determi- nant formed by writing the coefficients of the unknowns in order. This determinant is called the determinant of the system. The numerator of the value of x is ob- tained from the determinant of the system by substituting the second members of the equations (A) in order for the coefficients of x. 4 The numerator of the value of y is ob- tained by substituting the same quantities for the coefficients of y. The product of the elements.in the diagonal ending in the upper left-hand corner is positive. This diagonal is called the principal diagonal. 'The product of the elements in the diagonal ending in the upper right-hand corner is negative. This diagonal is called the secondary diagonal. Direction. — Employ the determinant notation in the solution of the following problems. ~ de+2y=16. e+td3y=14. Solution: —" | go 2 4 16 __ |i4e BL 1 14 Tai ANP T % 8 DB ens 4x 3—1x 2 =10, the denominator. 16x 35—14«* 2=20, the numerator of the value of a. 4x14—1x16=40, the numerator of the value of y. Hence e=20+10=2,¥y=—40 +10=—4. 9 Sx+t2y=7. 3 can as agen 5. 4y—x“=7. C= oe 5 Suggestions. — Write the equations in the form of equations A before writing the determinants. ‘The direct solution of problem 3 is shorter than this general solution. It has been given to explain a method which will be frequently employed in equations of more than two unknown quantities. Examples 2 and 3 lead to the following determinants : — on (ae peed 7 7 A ee gee TAG —1 4 —1 4 2—5 —5 5 7 —21 — 21 0 Jit | Soa en Ase eae, (eo 4(3¢e+y= ae gf2xt+6y=—1h. 2e+3y =18. 4e—9y=4., 5 We Sy x+5y = 20. 3x2 + y = 13. d“2+4y=15. fiemark.— It is generally unnecessary to write the values of the unknowns in the way indicated, since the multiplication may be performed without changing the equa- | 6 tions when they are written in the ordinary form ; e.g., in problem 4 we can easily see that the value of the denominator is 7, (3 X 3— 2X 1), the numerator of the value of x is 14, (9 X 3—13 1), and the nume- rator of ithe value, of y is, 21; (8x13: — 2X9). The work should generally be performed mentally. Direction. — Solve the first six of the following problems mentally, and verify the results : 2 90) ete (ax + by =e. yey ana 12. YEN B, dx + ey =f. : Ae ud es 10 (4a+y=5. 13, $mety d. 2 Yj =": (x +ny =k. [ x 414% —¢, Suggestion.—The de- 14. a 0b nominator is: % y 1 1 ee Wwe ts a \d e I a b (t 42 m0. Lind | 2 D 15. | ( de | dé =f. Suggestion. — Find the Let det Y values of a and aN then a y T invert ; the results will be the values of x and y. oy aes em toe an 99 16. | 32 2y y—5_ lla +152 6 4. 12 E _sy¥4+1 ee Suggestion. — Combine the coefficients of x and y. f iauek AgX + boy a Co = VL) age + bsy + C3% = Mz Multiply these three equations respect- ively by the factors : Given ae + by + 2 =m M. bo 10s Becy b, bs C3], bs Cz by Cy we obtain ) ? fy 39 $558 Sy 8 8D = : CD = : 'p 1%) 7) Tg Tg 69 di) SAG ee on § == 2 4) Say + Lat ou — ere! Tu T T 7 Say \'e@ "9 “9 %9 ‘ JO OnT[VA oy} VOUSFT ‘SOlezZ 04 voNper x pue A Fo squo1loy om 7909 ef, “Ioyyos04 suotenbes ve1yy esoyy PPV By 8 Bq 8 Bq S Sq B e es TTA =z gy Veo + Al 2 q 89+ x 2 q &y as) - 1%) tg Ig T@ Tg tg i 4 ou, — = 2 pee cone a “@— @ ee oy) — oo 0 ™ 19 19 Tg 19 Ig fy 8 Sus Sp 8 8p 8 \s : a =2|, : tot Al” “Wt4+a]° Vu 2.89 AG) £9 8g bank) That the coefti- cient of z is zero is seen in a similar way. We shall show by expanding, that the coefficient of y is zero. by — by = —_— bi bees a bobsC1. Us "Cs by, cy bs = b1b3€e ee O.05C): by Ce Since the sum of the second members is zero, the sum of the first members must also be zero. It should be observed that the multi- plier of an equation is the determinant formed by writing the coefficients of y and z in the other equations in order, e.g. : Let us eliminate y and z from the fol- lowing equations by the method just em- ployed. xe+3sy—z2z = 10. Eb ey eats,” 2e—y+5z2=18. The multi- 38—1|._|—2 1| |-—2 1 pliers are : . i —1 5 —1 5 o—1 The values of the determinants are 14, 9, —1. Multiplying we have: AQx — 2y + 142 = — 28. 92 + 2iy.— 92 = 90. 10 Adding the three equations we have: 49 « = 49. Direction. — Eliminate y and z from the following equations, then find the value of x. (3e+2y—2 = 27. 1 x—y+d2==9, " |ja—2y—42= — 36. Result, « = 4. oor ee ners chage am 2x+y—sz2=a—d5e 2. atZi2z2=6. a Result, «2 = 2. fet 2y=40. Sy —z#= 26. Result, « = 20, Suggestion. — The multipliers are : ere bie a Oe ane 8 Beer | | Semel aloha When the three equations are written in the ordinary way, the operations should performed mentally, e.g. : 11 Let the value of x in the following equations be required : n+ y—3f = 22 Zz xe—2y—z=—1 da2+y—22=33 The required multipliers are readily found to be 5, —1,—7. Multiplying the coefficients of x in order, by these factors, and adding the products, we obtain — 12 a. Treating the absolute terms similarly, we obtain —120. The value of « must be — 120 + —12 = 10. Solve the following equations for 2, in this way, and verify the results : s3a+2y—z2=6. 2xe¢+3y—3z2=1. x—y+4z2=9. ety+eze=15. 4eu—y—2z= 10. 2xe+3y—2= 19. Resuming equations M (page 7) :— aye + by + cz = m, ane + bey + 2 =m, > M. Ase + bsy + C32 = Ms 3 a, Cy Qe Ce ) 12 ay, Cy a3 C3 5) Ag Ce Ag C3 Multiplying the three equations respect- ly by the factors we obtain 1Veé Go & 8&4 & 89 § ao tg + 9 ®*p ig 9 8p 19 nee aD boaep fh 0 *D| oy at 89 &p By 89 &p yp Ponta Tp een VUIF{ “S9019Z 0} GONPeI x pue x Jo squeTO “Wooo oY, *1044050} suotyenbe ve1y4 eseyy PPV Gy & Gy & by & Gy & 0 &n 2D oD a %p g peers € € ? € 0h es Sheth 9th a, 9+ Tit, D 89 8p $929 89\8y 89 8p La vee tq — 3 — Tg) aloe a eee h\y, tp| 4 — *l15 ty | O87) 89 8p 20220 89 8p Lith oth to+x 77) &9-4)) ©9-2) By | 4 69 &p This result could have been obtained directly from the value of x on page 8 by 13 interchanging the coefficients of a and y. This is evident, since the value of an un- known in a group of equations depends, not on its name, but on its coefficients ; e.g., the value of x in equations M is the same as the value of y in equations N, or the value of z in equations P. aye + by + 42 Met + boy + Coz asx + b3sy + C3 I We ll —y\yw a bye + ayy + G42 boa + aay + Cae bsx + asy + Cs ll zs —\~ A ea + by + az Coe + boy + aoe C3@ + bsy + A3% = Ms I Il ll = 3 Sy ~ Observe, too, that the value of an un- known in @ group of equations is not affected by interchanging all of the coefii- cients of any or all other unknowns. We have already seen (pages 7 and 8) how the value of x may be found when three equations involving three unknowns 14 are given. Finding the value of « in this manner from equations N, we obtain the value of y in M (page 18), and from « in equations P, we find that in M Deis. a a 2 Ae 1 UY 1 m —m m i lt hs ele UA tg le Sie C1 bg As “aves by, ay gts 1% Data bs ae bak Ce The forms of the values of the three un- knowns in equations M lead to the follow- _ ing practical principle, by which simple, simultaneous equations involving three unknowns may be solved. The value of any unknown is a fraction whose numerator is obtained by multiplying — each absolute term (changing the signs of alternate terms) by the determinant formed by writing the coefficients of the other unknowns in the remaining equations in order. The denominator is obtained by multi- plying the coefficients of the required un- known (changing the signs of alternate terms) by the same determinants. The form of the value of z obtained in 15 this way will differ from that just given in having a and 6 interchanged. Such in- terchanges are allowable (page 13). The learner should make himself very familiar with the above rule, since it ap- plies to all simultaneous equations, as will be proved later. Direction. — Find the value of the un- known underscored in the following equa- tions. Oe te Uh Solution. —The multiphers are Ee Sh Ze 2 — oF pics Ore UR 1 si. or, — 23, — 5, 17. oa ~9X 238-9 X5 48X17 ; CLS EH Pee ONS Since only the ratio of the multipliers is essential, when they have a common factor it should be rejected; and when some involve fractions all should be mul- _ tiphed by the L. C. D. of the fractrons. a , < No i, \ a 1 2° 0 Ai A a k f HAD A OR Speen F ; a ee ne Nae ave. 242+ y—2=3. ED BG) IR ete Result, 2 = 3 4Aexe+t9y—22=15 3.<~7xex+38y4+82=71 e+tiliy—d5z2=—6 Result, y = 1 e+3y—z=30. 4.4 2y+2=12. xe—2z2=4 22+ a4 = 7h. z+ty—se=14. hi Result, Y — Ar G la 1 Ye We shall once more resume equations M. a,x + by + 2 =m M. 5ae+t+y—22=0. 5. Agt + boy + C22 = Me age + bsy + C32 = Mz The value of x obtained from these equations is (page 8) aS Vaife may by Cy Tee by Cy ee bs Cs bs C5 by Ce g by Co| _ ‘ een ca by Gy bs C3 bs C3 be Ce ve The denominator, written in the ordinary way, is AyDoC3 — Aydslg — Ab1C3 + abst, + O3b1Co ae Og0oCy- The numerator may be obtained from this expression by replacing the a’s by corre- sponding m’s. The denominator and nu- merator are also written as follows : — a bb Mm Ob, Ga Os" sea Wink Op ule Gs Os- C35 Wise Ogee yh. These are determinants of the third order. The determinants previously employed are of the second order. ‘The order depends on the number of elements in one side of the square. One method of developing a determi- nant of the third order may be inferred from the expressions from which the above determinants were obtained. The follow- ing method is probably more generally used. Connect the elements of the determinant as indicated in figure (p. 18). Multiply the united elements together. Prefix the posi- tive sign to the products of the elements D 18 in the lines parallel to the principal di- agonal, and the negative to those in the lines parallel to the secondary diagonal. The learner may easily verify this rule by comparing the results obtained in this way with the denominator writ- ten in the common 5 way. Determinants is a method of notation by which large expressions may be written in small forms, exhibiting, in a remarkable manner, the laws which govern the ex- pressions ; e.g., the expressions Ayb, — gb, 3 A1b203 — Ayb3Cg — An) 1C5 ha ADs + 3b4Cg — Agbg¢,3 and AybeC3d, — AgbyCgdy — Aybslody + gb, Cody + Aghseyd, — Agboeyd, — Aybetyds + Agbyeyds + AydyCgdg — Agby Cod, — AgbsCyds + AybeC:d3 + Ayb3cyd, — Ash, Cqdy — AybyCgdy + gb, Csdq + Agbslide, — Agbgcydy — Agbgeydy + Agbetyd, + AghsCsd, — AghoCsd, — AghyCod, + aybscody, are written a, b 4 dy a, OF Cy Ua, Os teat Cos a, bg Gin bg Cy az bg Cs ds A, Oe], a, 63 C3|, a, bs Cy dy], respectively. The larger the expression the more will be gained by writing it in the form of a determinant. While the third of these expressions is so long that it is difficult to get its picture in its en- tirety well defined in our minds, the equivalent determinant will present no such difficulties. Itis this brevity which makes determinants such a strong instru- ment of investigation. By the perusal of the preceding intro- ductory pages, the learner should have formed some idea of the nature and uses of determinants, which needs only to be extended. We might have pursued the study in this way, but for simplicity and logical arrangement we shall now change the method, and study the subject from definitions, The preceding pages, while not necessary to the scientific development of the subject, have been given to make the treatment more intelligible, and to 20 encourage the learner, while investigating many apparently useless principles, by the knowledge that the subject is important. The learner cannot expect to see the value of the subject fully until the elements of it have been so thoroughly mastered that he can make the application readily and intelligently. The words of Professor Sylvester may aid in forming a correct view of Determi- nants. He says: “It is an algebra upon an algebra; a calculus which enables us to combine and foretell the results of alge- braical operations in the same way as algebra itself enables us to dispense with the performance of the special operations of arithmetic.” INVERSIONS AND PERMUTATIONS. 1. Definitions. —A change from the natural order is called an inversion. The different orders in which several things can be put are their permutations. 2. When a group of different consecu- tive integers are written in their natural order, e.g., 1, 2, 3, 4, 5, there are no inver- sions. 21 3. When these integers are written in any other order, e.g., 2, 1, 3, 5, 4, the num- ber of inversions is determined by the number of times a larger-pumber precedes a smaller. Illustrations. —In 2, 1, 3, 5, 4 there are two inversions —2 before 1 and 5 before 4; in 2, 3,4, 5, 1 there are four inversions — 2 before 1, 3 before 1, 4 before 1, and 5 before 1. 4, Determine the number of inversions in the following groups of integers. Doh Sects OS Oa eet Ae De ond Ans. 3, 4,5. Boul 2, Onaes 2: Ons, Weary cee ANS 03/85) 0: od, 27 S91, 3,222, Feb, 3, lot, 2 Ans. +0,.1, 1; 2, 2: 4, 1, 2, 3, 4,5, 6: 6, 5, 4, 3,2, 1: 2, 3, 5, 4,6, 1 ATS Om 15.6) 5. If two adjacent integers in the series 1, 2, 3, 4 be interchanged, what is the effect on the number of inversions ? 6. Can the number of inversions be made odd by interchanging two adjacent integers in the group 2, 3, 4,1, 5? 22 5. The permutations of any given group of numbers are divided into two classes. The first class embraces all the permuta- tions in which there is an even number of inversions, and the second class all those in which there is an odd number of inver- sions, ne., 4,302,410: 2,3, eed; 1) 2,%are permutations of the first-class, and 1, 3, 2: 2,1, 3: 2, 3, 4,1, 5, are permutations of the second class. 6. TLheorem.— Any interchange of two numbers in a group of different numbers alters the class of permutation of the group. First Part of the Demonstration. When a number and its neighbor inter- change places and all the others remain undisturbed. daebhe group: 2) 1p e acy 0, Ope cates 7, 11, interchange any two neighboring numbers, as 6 and 9, and write the two resulting groups as follows : Ono ASS ibe a ah ree 36 iA Sd AB ie! observe, — (a) The number of inversions which the 23 integers outside of the brackets have, with respect to each other, is the same in both groups. (b) The number of inversions which the integers outside of the brackets (preced- ing) have, with respect to those within the brackets, is the same. (c) The number of inversions which the integers within the brackets have with those outside of the brackets (following) is the same. (d) The number of inversions which the integers within the brackets have, with respect to each other, is changed from 0 to 1, or vice versa. Hence, in this case, the class of the permutation is changed. Second Part of the Demonstration. When any two numbers’in a group interchange places. In the above group let 5 and 11 inter- change places, there being » intermediate numbers. 11 may be brought to the place which 5 now occupies, by n successive left neighbor interchanges, After this is done, 5 may be brought to the place which 11 . moo 24 occupies in the original group by » — 1 right neighbor interchanges. Each of these neighbor interchanges has altered the num- ber of inversions by unity. Then +n — 1 interchanges must have altered the number of inversions by an odd number. If the number of inversions in the original per- mutation was even, it will now be odd; and vice versa. Hence, the class of permutation of the group has been altered. Q. E. D. This important principle is sometimes stated thus: If, in a series of numbers which are all different, any two are inter- changed, the others remaining undisturbed, the number of inversions is increased or decreased by an odd number. EXAMPLES, Interchange the numbers underscored in the following groups, and compare the re- sults with the theorem just proved: 1. 2,3, 4,1: 4, 4 5, 2,3: '2, 1: 3, 1, 2 2 1,4, 5, 3,2: 2, 3,1: 3,1, 6, 4, 5, 2. *7. This theorem is also proved as follows : * Articles marked with an _agterisk may be fe Si eee pale lpn 25 If all the remainders obtained by sub- tracting each number from all the preced- ing numbers be multiplied together, the product will be positive or negative as the permutation of the group is of the first or second class. Let 7 and s be any two numbers of the group except the two that are to be inter- changed, and 7 and k& be the two to be in- terchanged. The class of permutation of a given group is determined by (—bUG%—)(r-—BH Mrs), IT (7 — s) denoting the “ product of all such factors.” The sign of IJ (r — 7) (r — k), IT (r — s) is not affected by interchanging 7and k, while the sign of (¢ — k) will be reversed. The class of permutation of the group will, therefore, be altered by the interchange oft and k&. Q. E. D. 8. Cor.—If a number is transferred to another place, all the others maintaining their relative positions, the resulting per- ae the same mutation is of the | a, different number transferred has moved over an ae number of places. class, if the Se 26 Illustration. —In the permutation 4, 3, 5, 1, 2, transfer 1 to the place of 3; the re- sult will be (4, 1, 3, 5,2) a permutation of the same class, since 1 has been trans- ferred over an odd number of places (1). Again, transfer 1 to the place of 4; the re- sult will be (1, 4, 3, 5, 2) a permutation ofa different class, since 1 has been transferred over an even number of places @» This transferrence can be accomphshed by successive neighbor interchanges. EXAMPLES. Move the numbers underscored to the places marked with carets, and determine the resulting permutation by the theorem (neighbor interchanges) and the corollary : 1B Poa igs Pease hae ately ol alert me) GER ave 4 9. Theorem. — The number of permuta- tions of the group 4, 3, 2, .... 7, 5, com- posed of 7 elements is n/. Demonstration. — The first place of this group may be filled in m ways [any one of the given elements may occupy it]. After the first place has been filled, the second 27 may be filled in » — 1 ways [any one of the remaining elements may occupy it]. After the first two places have been filled, the third may be filled in » — 2 ways [any one of the remaining elements may occupy it], etc. Therefore the number of permu- tations of the group (the different orders in which 2 thing?can be put) is n / “aw (n—1), (m—2) ...2K Ln. Lilustration. — The following are 4! permutations of 1, 2, 3, 4: 244 3 144 3 1. 344 2 344 : afes [afta eg 1432 3, 244 i 4 243 : 445 i 345 i Ss the 28 Writing them in the common way, we have 1, 23,4 noe Aas reo ye Oe A ye OAtoA 8 9, 3, 1,4 Beto A cnet 1. Ane 3,2,1,4 Be outa to 4p 948 1s, 4) 2 qe os Ledeen Dae eda Ore 153 Oot 8941 aie arb BFA oes Neca a LB IND £o5 D4 10. Allthe permutations may be formed from any one permutation by the succes- sive interchange of adjacent numbers (see illus.). EXERCISE. 1. Verify that the following groups have 2!, 3!,and 5! permutations respec- tively. Lee Ae el yO Mela Las ee De 2. How many different words of ten letters each could be formed from an alphabet containing ten letters, no letter being used twice in the same word ? 3. How many different words, no letter being used twice in the same word ? 29 4, Prove that the following groups have 2!, 3!, and 4!1 combinations respectively, permutations of the letters and of the sub- scripts being allowed. Ay, bz + Gy, be, Cg : Gy, dg, Cs, dy. 11. The class of apermutation is changed by each interchange of two numbers (6). The number of permutation of any group is even, since n/ is even when n exceeds unity. Hence the number of permutations in which there is an even number of inver- sions is equal to the number of permuta- tions in which there is an odd number of inversions (10). 12. This important principle may also be proved as follows : — A group of n elements has n/ permuta- tions. Let « and y represent the number of permutations having an even and an odd number of inversions, respectively. Thena+y=n/. Inall of the permuta- tions interchange two given elements. All the resulting Pao gts are different. 1 The different groups that can be made of n things, without regard to order, are their combina- tions. 30 The x permutations have become the y¥ - permutations, and vice versa, without an Incredseun tris... Ci Yue Gt a * 13. When eachelementof a given series is replaced by the following element, and the last by the first, the elements are said to be cyclically interchanged. It is evident that a cyclical interchange of a given num- ber of elements (7) may always be effected by n —1 interchanges of two adjacent ele- ments. If the elements Gib; Cage a were written on the circumference of a circle, and the circumference cut between a and /, the elements would be in their natural order, while the cutting between a and 6 would cause the elements to be cyclically interchanged. The term cycli- cal interchange is extended to include the . case where each Pato is replaced by another element of a group, and the last by the first. Rok xe * 14. Every ant. tthe as be obtained from a given Lata ee cyclical interchanges. Ea 31 Given SiGe oaths Areas to obtain Ze oH bnOs Ay OME Solution. — Replace 5 by 2, 2 by 7, 7 by 1, 1 by 6, 6 by 3,3 by 5. This constitutes the first cycheal interchange. 4 is said to form a cycle by itself. It is evident that this method can be employed in all cases. * 15. Two permutations belong to , Le eee if the difference between the number of elements and the number of groups by whose cyclical interchanges one is obtained from the other is | ead Demonstration. —Let n be the number of elements, and » the number of cyclical interchanges of a1, a2, dz, . . . @, elements respectively. Then must the number of single interchanges be (aj —1) + (@—1)+(@-1)+... Atta ek) = hsp. For a4, + d,+a,+ ... +4a,=n by hypothesis. — In the given example n = 7 and p = 2, 32 ‘. 2 — p =an odd number, and the two permutations belong to different classes. Given Zt Sp1L, os 0105, ke to obtain 1, 2, 4, 3, 6, 8; 7, 5. Here » =8 and p=2. Hence the two permutations are of the same class. 16. Observe that numerals are not es- sential to the preceding demonstrations. Any elements — numeral, literal, mono- mial, polynomial, etc. — may be regarded in their natural order if arranged in one way. ‘The inversions, when they are ar- ranged differently, are found the same way as if they were numerals, and their natural order were the natural order of these nu- merals; e.g., if a + b, d, e, ¢ be considered the natural order, there are four inversions inega+b,e, d. EXAMPLES OF DETERMINANTS. ¢, dy C, ds a, oy A, by ve ae ee Ne ae gh cata ] are determinants of the second order. dean Cree as Z yi 4 —6 rome nC s Ol 3—2 6 4 Leen ee Catala 8 10—4 8 Gtebpe Opa dts 2 GB =e ts Oy Fey park Ue | D Bea pad. | Ga On Cg dy ex ef 6 0 1 Gs WeUerees g h- t | ghek 5 | are determinants of the fourth order. CAD aye, n is a general determinant; i.e., a determi- nant of the m™ order. MEANING OF THIS NOTATION. I. 18.1 Number of Terms.—The square form of elements between two vertical 1 These definitions should be regarded arbitrary for the present. The reasons will appear later. 34 lines represents all the terms which can be formed containing one element and only one of each row (horizontal line) and col- umn (vertical line) as a factor. iOe 19. Natural Order.— The elements in the diagonal beginning in the upper left- hand corner are said to be in their natural order. This diagonal is called the Princi- pal Diagonal; the other diagonal is called the Secondary Diagonal. sh EE 20. Subscripts. —In order that inver- sions may be readily seen, the subscript will generally indicate the row, and the letter the column, to which an element belongs. This notation does not imply any relation or dependence of the ele- ments. It will, however, prove very help- ful in studying the laws governing deter- minants, and will generally be employed. 35 IV. 21. Inversions. — The number of inver- sions in a term is determined by the num- subscripts, ber of inversions in the ne when r) ( letters the { subscripts are in their natural order, or by the number of inversions in both when neither are in the natural order. V. 22. Signs. —Terms in which the num- ber of inversions 1s even are positive, the others are negative. VE 23. Arrangement. — Since each term must contain all the letters and all the subscripts of the determinants (18), all the terms may be written in the natural order of the letters or in the natural order of the subscripts. The former is the more common method, and should be employed by the beginner before determining the sign of the terms (21, 22). When elements 36 have subscripts (a2 03 ¢,; dy), and the num- ber of mversions of both subscripts a letters are counted, the interchange of ax elementsgwill alter the number of inver- sions by an even number, since the sum of two odd numbers must be even (6, page oe -48). These considerations lead to the im- portant principle: the sign of a term is not affected by commuting its elements (21, 22). This principle could have been inferred from the fact that determinants result from algebraic elimination; hence the commutative law must hold. | Vil 24. Number of Terms. — A determinant of the m™ order consists of »/ terms. Demonstration. — All the terms may be formed by keeping the letters in their natu- ral order (23). The first place may be filled in 2 ways, since there are different a’s; the second in m—1 ways. Any one of the nm different 0’s may be chosen except the one whose subscript is the same as the subscript of the a which has been chosen oT to fill the first place (18). The third place may be filled in » — 2 ways, for the same reason, etc. nm (n—1) (n—2)...2XK1=2! VILLI: 25. Development by Permuting the Sub- scripts. — All the terms of the determinant may be obtained from a given term by keeping the letters in a given order, and permuting the subscripts (24, 18). IX. 26. Other Methods of Notation. — Many different methods of writing a determinant are employed. Some will be explained later. The following abbreviated methods will be employed in this work. Gli Ope ails ean Ge ay, by @2 by ly a, b, Oy Ua En aay Az be|, dg b3 C3}, On On ln will be written respectively (a be), (dy be €3)y (G1 bg . . « U,), OF > (Qy be), St (a4 by C3), = (a De Aas Ln)5 38 i.e., only the Principal Diagonal will be given. All the other terms can easily be obtained from it (25). We shall also fre- quently employ the Greek Jetter A to de- note a determinant. The learner should make himself quite familiar with the preceding definitions and. deductions, since they involve almost the entire theory of elementary determinants. We shall now proceed to the study of sepa- rate determinants, beginning with the simplest form. 27. Determinants of the Second Order.— | CaO 1 a, bz = Ay bg — Ae 0}. QUERIES. 1. How are the terms identical with (a; 62) obtained? (18, 25, 24.) 2. What determines the signs? (21, 22.) 3. Do all determinants of the second order represent two terms? (18, 25.) a= , oe odgee 39 4. If one of the elements is 0, what is the value of one of the terms ? 5. Explain when two elements become “0’s. (Two cases.) 6. When three elements become 0’s. 7. Does the order of the factors of a term affect the value of the term? (23.) 8. When is the value of a determinant -of the second order negative ? 9. When will the two terms have the ‘same sign ? 10. If the two rows or the two columns of (a, 6.) are identical, what is its value ? 11. When the rows in order are made the columns in order, is the value of the determinant affected ? 12. Is the value of the determinant altered by interchanging the two rows? the two columns ? 13. What effect does the multiplication of a line of (a, 6.) have on the value of the -determinant ? 40 28. Direction. —Find the value of the following numerical determinants. 3.4 ib ie! 4 —1 TESS Sr N29 ie Col As ts | Ans. 2,38, 14 pi id —4 —7 4-5.| —2 —3]|, |—38 —5}. Ans. 14, —1. neo abe —d a rid 6-8.|c d|]|, |d bys 1-0 10 1G Ans. a’d — 6’c, ab’c + d’, 0. 4 0 “|1-1 9-10.|0 8}, Berigat Ans. 32, 4. ee ate 11-12. |3 1], de ai i 2 4 a 13-14. |.3 _5\, E igi ahs 15-16.|4 0], aH RE eee =| 17-18.|—3 «x ka 7 ) Al 29. Determinants of the Third Order. ar bh Gy a bg Cy We Caw Cas A, by Cs — Ay bg Cy + Ae b3 — Ag b, C3 + sg 0, Cg — Ag bg ¢. (QUERIES. 1. How may the terms identical with (a, 6, cs) be obtained ? (18, 25, 24.) 2. Give the reasons for the signs of the terms (21, 22). 3. Determine the signs of the following terms in three ways (21, 22, 23), and com- pare the results. Cb3M, AgbyC3, Aglaby, 020341, 0 g14Co. 4. If h and 7 represent the number of inversions of letters and subscripts re- spectively, (—1)*** will determine the sign of a term, and is generally called the sign factor of the term. Explain. 30. An easy method for developing a determinant of the third order is explained on page 18. The learner should become very familiar with this method, since it is the one most commonly employed. 42 31. Instead of connecting the terms as indicated on page 18, two of the columns may be repeated thus, % bh & & ho MUM sO, aU. a3; 03; Cs a3 bg The products of elements in the Princi- pal Diagonal and in the two lines parallel to it which contain three elements consti- tute the positive terms. ‘The negative terms are found similarly with respect to the Secondary Diagonal. 32. The development may be effected directly from definitions (18, 22) as fol- lows: Explanation. — Since no two —e,;| elements of aterm belong to bs "¢3| the same row or column (18), the Lae elements that may be combined with a, are 0, ¢,, 63, ts. Hence the only two terms containing a, are a, 6, cs and d,b3¢,. We find the terms which contain a, and those which contain a; in a similar manner. The sum of these terms with the proper signs constitutes the develop- ment of A. 43 QUERIES. 1. What are the terms which contain a, ? 2. What terms contain a; ? 3. Why is the sum of the terms con- taining 1, 2, a3 the development of A ? EXAMPLES. 33. Direction. — Find the values of the following determinants of the third order. Use the method explained on page 18. De A eric Taio 1 2, 3 0 5 Solution. — 2 SLES is ee Os 5xX0x* —3= 0, 3x —4xX2= — 24, iA 3x1ix—-3=— 39; 5x —4x5 = — 100, 2x0x2= ; LOO! — 14 —(— 109) = 95, the value of A. Meee 37 CL ee Oe eel c Be oar ide 3! oor. 0 2 4| ese | Oe Oued Tk By Owes Babe Ans, — 12, 1, 0.. Ae ITT x —y 3 oyey 6 4 3 —1 5-6.|8 10 14}, — 2x 2 iThO Ans. 0,112 —ay + 24 | 4 —1 3| — 3 Gee? Trish 258 15). Ans. 296. What general principle may be deduced from 4? Prove by expanding a Of & a, A, as Bini On we Ch Cas ee 2—T7 4 Lee 0 | 3 6 1|=|-—7T7 6 2 9. | 0 2—l Ale i sae What principle may be inferred from 8 and 9? (Compare with 11, page 39). Prove by expanding 45 a, b, GY GQ, 0; ox aye 0, | Ay be Co|==| Ay by Cov |S=lage by Cy ll. |as 05 Cs Ga Og: Cot, Gye: Ont Cabs ay bd, Cy ay by Cy Az b, C3 a A, bs Co 12. Ao bg Co 3 bs C3 bh & G = —|b, a, Cy bs 3 Cs aq b H+2h bb GY on) be Co = aR) co 2 bo by Co LS iia 0 Ce dg +26, bs Csl, aq b Y Ag be Ag) = 0. 14.) Gs Ope Ce Notes. — In what follows 4 will be used to repre- sent she eterminant before and 4’ to represent it after fe eveformatiod! 34, Rows and Columns Interchanged. — If the rows in order are made the columns in order, or vice versa, the value of the determinant is not altered; i.e., A = A’. 46 Demonstration. — Denoting by A and A’ respectively the original determinant, and the determinant after the rows in order have been made the columns in order, or the columns in order have been made the rows in order, then will A = A’. Terms of A formed with respect to the columns are identical with terms similarly formed from A’ with respect torows. The signs of these terms are the same; for if a,b,c... represent the numbers of the columns, andl, m,n ... represent the numbers of the rows of A from which the elements of any term have been chosen, then will J, m,n ... represent the num- bers of the columns, and a, b,c... rep- resent the numbers of the rows of A’ from which the elements of the identical term have been chosen. Therefore all the iden- tical terms have the same sign (21), and A= AX (See 33, 8.and' 9.) 35. Rows or Columns Mutually Inter- changed. — The interchange of any two rows or of any two columns changes the sign, but not the absolute value of the determinant. AT Demonstration. — The terms chosen simi- larly from A and A’ differ only with re- spect to two subscripts or two letters, which must be interchanged in all of the terms to make them identical. ‘Therefore all of the terms similarly chosen have dif- ferent signs, and A=—A’. (6, 21, 22), (33, 12). 36. Identical Lines. — When two rows or two columns are identical, the determi- nant equals 0. Demonstration. — Interchange the identi- eal lines. This will not alter the determi- nant. 9). 9a =A Dy=(oo), Aiz=—— AY Hence A’ = —A’, and-A——= ALS When a change of sign does not affect the value of a quantity, it must be 0. (83, 14) ot. Multiplication of Lines. — Multiply- ing or dividing a row or a column of a determinant multiplies or divides the de- terminant by the factor. Demonstration. — Each term of A’ con- tains the given factor, or is divided by it (18). Terms similarly formed from A and A’ are made identical by the removal of 48 this factor. .:. A’ =A or + by the factor. (33, 10, 11.) Corollary. —If the elements of a line are equal multiples of the elements of a parallel line, the value of the determinant | is 0. Suggestion. — Divide by a factor which will make the elements identical. Then consult 36. 38. Transposition of an Klement. — Any element of a determinant may be trans- posed to any desired place. Demonstration. — The element may be brought to the required row by interchan- ging two rows, then to the required column by interchanging two columns (35). When these two interchangings are required, the sign of the determinant will not be affected, since the sign has been changed twice. It is more common to effect this trans- formation by transposing the lines to the required places without affecting the rela- tive positions of the other lines. This can 49 obviously be effected by successive inter- changes of adjacent parallel lines. The sign of the resulting determinant is deter- mined by (—1)"**’, where 1 represents the number of ; Bd ns over which the line containing the given element has been transposed (8). EXERCISE. 39. Direction. —Consult first the articles to which reference is made. Then prove the equality by solving the numerical de- terminants. 2 DE cic eemeed 4 0 0} =|3 0 —5 Be Ie Be 61 eT 0 6 aor Ll La | 0 5 —-3)/=-|0 -3 5 85.;2 —2 3], |2 °3 -2), 0 5-38 Shines i Raa igt ha 2-28 2 2 FPO) 5| = 0. 36. 7—3 3 —3 DaroreD PESTA Y pied UD 37. -Cor.15 10 20 ide BG | }2 4 1 Pa ts | SToauGe LL. 2s OF 1 2416) Gal? 1800 1 ete orl | Sa sno The element 7 in Ais SNS to the PN é- upper left-hand corner in A’, and to the upper right-hand corner in A”. Explain the signs and prove the equality. 1. Explain the difference by illustra- tions of interchanging elements or lines, 51 and transposing (transferring, moving) an element or a line over a given number of elements or lines. (6, 8, 34, 38). 2. Find the values of the following by . multiplying rows or columns to reduce to a simpler form, before evaluating the determinants. Crean 2 4—2 1 (a).|\4 —6 8 (6).|2 4 8 E 3 Al, + 5 Of, Geib) O (c).|ac bd ce a ab 0 Solution. — 6.9 3h 2) Spe 4—6 8|=2.3.2;/2-—2 4|= 2058 1 athe tae 2 3 tek 12.2;1—1 2 Livngle <2 SOA ee 64h) FIDE OG | — 240. 3. Do all the principles which apply to rows apply to columns? (34.) LIBRARY UNIVERSITY OF ILLINOtS. 52 4. Write the terms of (a, 0, ¢; d,) which contain both a; and ¢. (26, 25, 18, 32.) 5. Write the terms of (a bo. cs dy és) which contain a, 6; ¢, Also those which contain ¢, bs. 6. Of how many terms are the two pre- ceeding determinants composed? (24.) 7. Show that in a determinant of the n” order only two terms have n — 2 ele- ments in common and that these have opposite signs. (18, 22.) 8. In the determinant /a, db, c,... 1 n) (page 37) a/ terms have n — a elements in common, and (n— a)/ have a elements incommon. Explain. 9. How is the value of a determinant affected by changing the signs of all the elements in 2 rows? (87.) Prove that Ais ee Wik kg Bede A ae Tee Sb De aaa ea (100... | eens Pee ree abd betoeee 10, (+ @& d? - abe.) des ae. 03 Suggestion. — Multiply the first column by abed, then divide the rows by different factors. | A+ 4 d, YM Y d,| | Dy cq dy y+ by Cy dy] =}dq Cg dy|+| bg Cy dy Ti | as + dg Cz ds As C2 d. bs Cz ds ? 40. If all of the elements of a line of a determinant are binomials, A = A’ + A”, where A’ = A with the first terms of the binomial as elements in place of the binomial elements, A” = A with the second terms of the binomials as elements in place of the binomial elements. Demonstration. — Each term of A has a binomial factor (18). Decompose each term into two terms with monomial factors ; e.g., example 11 is equal to (a, + 0) & d; — (a3 + b3) ¢, dy + (a, + by) ¢ dy — etc. = M1 Co dz — Az Co dy + A, Cz, d, — etc., which are terms of A’ plus 0, ¢, ds — bg ¢y dy + b, ¢; dy—ete., which are terms of A”, Separate all the resulting terms into two groups — the first containing the first terms = 54 of the binomials, and the second, the second terms of the binomials. Then will the first group equal A’ and the second A”. The reasoning will become clear by com- pleting the solution of this example. Scholium. — The same reasoning may be employed when a line contains polyno- mials of any given form or of different forms. Also when the elements in several or all of the parallel lines are polynomials. 1. Prove that BAO eD qi Sab Ba W Sa Rie St ed er age at abating es 75 2 Bedard i oma a | dn oer sae 2 24+3+5 1 —1 2-—6 2 —4 442-3 3 aa ° 2. Resolve A’ and A” each into two de- terminants, and prove the sum of the four = A. 41. A determinant is unaltered by add- ing to the elements of a line equal multi- ples of the corresponding elements of a parallel line. 55 Demonstration. — A’ = A + A” (40). A” = 0 (387 Cor.). This principle is very important. We shall therefore apply it to a general determinant, to enable the learner to understand the demonstration better. To prove as UW ey) aie se eee L Ao bo siete k, aoe le a deere st). ok a a UL Deen aLOR iL avant at ay b, eee k, + ml, eee bi Gai Us te ales eee es eee ae ate as ees yiaty EB Gee by oh tale dl m represents any number, and ky, ky, ..., Hepes aks ABU? Gite se, De eam are the elements of any two columns. Any number of columns may precede, be aa mediate, or follow these two columns. 40 at io-equal te & pa art eqpal To ay b, eve ky eee l, Qe Devt et, so). McA Ont ee teaneaee PRANAB AE Pt a GARBER tide MIST GAA Ay aT db, eee ml, eee G Ay Dorit tem gious a3 ifs Aan Sh IR co aay i GREDCD, | Se wambs oe, The first of these determinants is A, and the second (A”) is equal to 0... alten. ), ay 0 D, Cor.—A determinant is unaltered by adding to the elements of one line equal multiples of the corresponding elements of two or more parallel lines. Scholium. — Denoting by 1, la Js, lg, .. « the parallel lines of a determinant, the de- terminant is not altered by writing in- stead, J; + mls, lp + nls + 14, ly — lay lg, 2. but we cannot substitute 2, + 2,, 2, + 1, Js, l,,... because after 7, + /, was substituted for J,, the lines were J, + ly, la, ls, iy...» 57 We have therefore no reason to suppose that the addition of 2, will not alter the determinant. In the case supposed A’ = 0 (36). It is evident that we could have added the transformed line J, + J,, giving Ly + la, 2 by + dy sy Ue. This principle is very useful in the re- duction of numerical determinants. The following examples will show how the principle is applied. To find the value of the following de- terminant we divide the ah t6;5 Ki ard 1 3 BV oo Oe = OIG: quays 10 =—6 6 row by 2, and subtract the quotient from the second and third rows. Direction. — Find the values of the fol- lowing by reducing according to 41. 3 Ae Ga Sta ee A esa Tate aoa i yee Orgeherets Ge ok hs aah themed ec igee > a 42. If all the elements except one of a line are zeros, the determinant is equal to MER i: 58 the determinant formed by striking out the lines which contain the significant element, multiplied by (—1)’** times the significant element; e.g., a,0 0 0G Ss by Cz = A, Ke a Ls bo Co = — ao m1 4 | as bs Cy AC BT ies Ps Demonstration. — Since all the terms of A must contain one element from the line in which there is only one significant ele- ment (18), all the terms which do not reduce to zeros must contain this element. The elements which unite with any given element in a determinant are found by permuting the other subscripts with the other letters; for all the permutations of 1, 2, 3, 4, ... which may be formed by keeping 1 in its place are found by prefix- ing 1 to all the permutations of 2, 3, 4, ... (See 18, 23, 25); but permuting the other subscripts with the other letters will evidently lead to the determinant formed by striking out the row and column con- taining the given element. If the significant element is in the 59 upper left-hand corner the sign is positive. It may always be transposed to that place without altering the value of the de- terminant by prefixing the sign factor (St) § PEO3) ew Aas (Le A, Again ( — 1)"+* determines whether an element causes the sign of a term from which it is removed, tochange. Therefore, when the significant element is removed from all the terms which do not reduce to zeros and placed before a parenthesis en- closing them, it must be preceded by the sign factor (—1)"*+*% When the paren- thesis is written as a determinant this sign factor remains. Prove that eae On Wer () Hoe eed =215 Bt rey 1 0 % Sed 0 De mn BUA Go) AT rad eames Rhy a A’ may easily be found by cancelling the row and the column containing the sig- nificant element. The given determinant 60 and those on page 58 should, for this pur- pose, be written | ; : : =| ) , by ; Cy é 2 be Co 3 . ce ds 53 3 QD bs es]. 2. When a line contains only one sig- nificant element, is the determinant altered ; _ changing. the other, elements in the Pre rpendicular ORE Conteinig. the given element? (1.) 43. The order of a determinant may be increased indefinitely, for the borders b] TNO) O: Teer ay eames OL OIE y 0 0 ( may be placed on any determinant with- out altering the value. Prove that Le AO, a by) |e, by Late Wren, PEN a 61 [4 0—-0 see 3d 2—3 2 Ki 11 =|5 sae yi 12-—1 6 en 0 01 2-5 4 aslor 3|=|0'0% 74 i 3. |2—1/~|02—-1} 100 2-11. q b&b GY | Gy 0-0 No; | a, b, C|/=|0+a,+0 2b | 4 | ag Des i405 RO iets Outage Oait_ Cs pei a Cy ODD ance Oma ey =10 0b, Cy|+|de b, Col tl]Q 2 Ce 0 bs C3 bs Cz as bs C3 b, ¢ Gea ieee = Hl, . ae bs a na, be el (40, Scholium; 42.) shot t as O0Oa—x ak¥—x2 0b—2 oF — x a— x sara Oe Gt Oy, — 0 Lit ea Baas OEE (0 | do ORS Oats ome Ca fis) Be Ob BEL Pa La 62 44, Any determinant may be so reduced that all the elements except one of a given row are Zeros. Demonstration.— If none of the elements of the given row are zeros, reduce them to their L. C. M. by multiplying the col- umns; then subtract one of the columns from all the others. The result will be the required determinant. If some of the elements in the given row are already zeros, treat only the significant elements in this way. Illustration, — 3 4 a 1|~ |12al 12al 12al 12a b ¢ 21|_|4abl Back 241 120) _ a 4)1|~|40°l. 3461 481 12al|> 1nOe Be A i Olan | Oni 12al 0 0 0 Aabl 3acl — 4abl 241—4abdl 12a — 4abl 4a7l 3abl — 4a7l 481— 4a72 12al — 4a7l 4al — 4al 361—4al 24a —4al "| 3acl—4abl 241—4abl 12a—Aabl = 12a | 8abl — 40° 481 — 40° 12al — 40° — Aal v6/— 4dal 24a—A4al |. 63 The reduction would have been much simpler if the last row Sy been chosen for the given row. ; ; 45. The order of a Syrian may ae decreased by the method employed on page 61, problem 4, or according to 44. The judicious use of the principle ex- plained in 41 simplifies these transforma- tions greatly. EXAMPLES. Ge sioa Lig, £4 Triage s 18728738: °-.8 Bt At ls 7 O18 SOR 40 ebay 13 A ees Pate lee RAT eye Any abel pe aae Wet sect hl gE erp riya Sa. a 1 et ums ac O (2 e4e bob}. 12) o2 P71 Tre OL, AG is eid le As, ees io iiak She eee | een {ead Ar i) eis ee eae” Me (oy peony (em) Ty, | 2 tM Fees b Onis Oescek A: \Ts) on (mess Explanation. — A’ is obtained by sub- tracting multiples of the last column from 64 the others. A” is obtained from A’ by subtracting the sum of the first three columns from the last, ete. Direction. — Find the values of LO eae Maes eee 2°00) 0 —1 DO aie eo AA ie OO ee lS Our eel 0 Prot dati AIG Sh Poe a bi, 7 —2 ane os hp eA O—2 5 8 4. Dees S gated ake a Ans. 0. 2 (56+ 2), — 972 46. Prove that a b ki & Gy by Ce ky, ls dz ob, Cg kz (3) = Pe pa a : ip : a, b bh Cj ky, || Aa by be Ce | ky (| 1 a b| | ky Reet ete amend Qn), Det. Knits 65 Hints, — Multiplying the first column by — 6, and the second by a, we add the results; then multiply the second by — ¢ and the third by 0,, ete. Operating on all of the columns (x — 1 times) in this way we obtain. (—1)-? b, Cy eee ky l, i= 0 0 — Ay by + ay by — by + by Ce —,b,+4,6; —bs +, Cy 0 l 1 Pie ice RE Lal ads —hgh+hl; ls a apa oe a we By making the last column the first the sign factor (—1)"—' will disappear (35), since it may be accomplished by ”~? success- ive interchanges of adjacent columns. We now reduce the determinant to one whose order is nm —1 (42). Thend q4...k, L A eb NG OMEAI Aone " ; OO aA eas tan GrneDs 66 Scholium. — The simplest line should be made the first row before this method is employed. Illustrative Solution. — Weanling oe bas EN eae poe eZ 4 Dic ee — 6 wie i Melero ied 14 10 25 —55|- 4-26 16] = al pe 11 5 2 ay | 10 25 4 j ms rie 96| |— 26 16° — 2.5.10} 144 40 A aSOR Ase TG) eee erie 1g | Seated a 100 — 324 510 es tse 00 i00 = — 972. — 324 510 67 These reductions could have been sim- plified by (37). We have not employed many other principles for the purpose of impressing the one under consideration. It should be observed that only the ex- treme elements in the first row may be zeros. For the reduction of numerical determinants containing few or no zero we consider this method superior. ae We have now indicated three general«& Zet methods by which all determinants may Ai, be reduced (43, 4, 44 and 42, 46) as far as © /4. may be desired. Which of these methods should be employed is determined by the nature of the problem. Frequently sev- eral methods are used in the solution of one problem. Determinants of higher orders should generally be reduced to the second order before evaluating. AT. ais ee a » by ey dy 3 b, ie d, tty tt | Hae byey dy, ly Ayeg duly (a). by Cy dy (O)}; | Ogarege dg ay ets | 4 &% d,| ; (°). Cy dy , (@). ds 68 (6), (c), (d) are respectively first, second and third minors of (a, db, ¢3 dy). Dott Gerke bs; C3 ds by ty dy A, by (e) a, and dz Os :(f) Cay and 7 (g) (a, bs ¢4) and dy Cy (e), (f), (g) are respectively complement- ary minors. 48. If we delete one line and one col- umn of a determinant, the remaining elements in their relative positions may be regarded as a determinant. This de- terminant is called the first or the prin- cipal minor. There are n? principal minors in a determinant of the n” order. 49. If we delete the same number of rows and columns, the remaining elements in their relative positions constitute a minor whose order is n — a, » being the order of the determinant and a the num- ber of the rows deleted. [ (c), (d), ]. 50. Definition. — If we cancel the same number of rows and columns, the twice cancelled and uncancelled elements respec- 69 tively constitute complementary minors of the determinant. When their signs are prefixed they are called co-factors. [(e), (f); (9) Ol. The principal minors, which are the complements of a, dy a3, ... by, by, b5,.-. are denoted'by Aj. Al) Ay.) 2) Bi, Ba Dy. as In (a) (47) : bz Cy dy b, ¢ ad, A, = bs C3 de A, = bs C3 ds by C4 dy } b4 C4 ds ; b 4 dy As = bo Co ds PoC Oy Gea eth DinsCuwatGy By = Ag C3 ds B, = Ag C3 ds C, ady|, Gy Cy dy, ay A B, = Ad Co ds It is evident that there are as many principal minors as there are elements, and that any principal minor may be ob- tained by deleting the lines containing the ~ corresponding element. 70 52. From 42 and 43, 4, it may be seen that (a, bg ¢3 a4) = a, A,— a, A, + a3 As — a, Ay. = — 6, B, + & By, — bs Bs + b, By. = ¢,0, —e, Cy + ¢3 Cz — oy Cy. 3 ay FF diahe a taee Di A Mere ICATION a "Co-FACTORS. 53. If the elements of a line are respec- tively multiplied by the co-factors of the corresponding elements of a parallel line the sum of the products is 0. Demonstration. —In the determinant s+ (a 6... l,), let us multiply the col- umn containing the a’s and the column containing the b’s by the co-factors of the first column, the results are Ay A, b, eee l, ot oa Ag 1M bs eee be sig, Abel Dade EL Ay b, Ay eee L, Gag) AHO AS) iby [Mamet Sw VN ae 71 These multiplied columns equal respec- tively (52), A... vot Mendes tiie a Meee fae ash BY aCe a Bane tela a tee GR UN IRN S UER Ob ae ote ee The first is the original determinant (52), and the seéond is 0 (36). It should be observed that the second A is obtained from the first by writing 4, 6,,. . . in place OL Gy¥ Gay neous Illustrative Example. — vs 3 2 QC, = 23, + C, = 5, C; = — 17. 1—7 1 Ohl ean OU Multiplying the first and second columns by these co-factors, we have 2 234+1x«5—3x«17=0. BN DB = TSG na 1 ieee APPLICATION. 54. Given QAethy+toe =m dyt + boy + Co % = My > M. a;0+b,y +Cs% = Ms to find the values of x, y, and 2. 72 The determinant of the system is (a, 0, ¢3). Multiplying by x, we obtain Oy te07! Cy Os a0 Cy Got Os Cele hte © Og ty ee nt .Os 80a Agt ~Os C, aAxrz+thytae bh G Mm, 07 cy 2% + 0,y + c,2 b, Col =| me d Ce adgx+bsy +c3% 63 Cz WM, Os - Cs Hence m ob Me by Cy Ms, bg Cg ay Ae b, Ce ag bs C3 We could evidently obtain the values of y and 2 in a similar way. This method is general. The equations may also be solved in the following way. We multiply the equations respectively by A,, —A,, and As. The resulting equa- tions are a,A,;x+b,Ayy+oA,2 =i — Uy Ag t — by Ay yy — Cy Ag? = — my Ay > M’. a; Agx +6; Asy + cs; Ag 2 = mz Az 73 Since any numbers written in the square form may be regarded as a determinant (18), we may consider the co-efficients of the unknowns in M to be a determinant. The notation employed gives it the form of the general determinant of the third order (a, by ¢3). Assigning the values to A,, — Ay, As, which they have in (a, 6, cs), and adding the equations M’ together, we obtain (53) (a, A, — a, As + 3 As) c= My, A, — Mz, Ay + ms; As. When written in the more common form, this becomes ay b, Cy M4 b, Cy Gs (OK Cola = ma bs nes Gem Oeliels Neg Oe Ce Hence m b GY Me bg Cy Ms bs Cs Ca a bb de bg Cy ne Oe. ola a See pages 7-18. 74 55. To find the value of y we multiply equations M by — B,, B,, and — B,, and add the three equations together. To find the value of z we multiply by C,, — C., and C;, and add the resulting equations. The results will be (53) (— 6, B, +6, B,— 0; Bs) y = — m B, Ms Be, — mz Bz. (¢, Cy — ¢2 Cz + cs Cg) 2 = m, Cy — me C, + mz C3. Dividing by the co-efficients of these un- knowns, and writing the results in the ordinary way, we find the values of the three unknowns to be ly My Wty a2 Me Ce dz Ms Cz ihe RS ters a, b, Cy ds bs Cs|, Lllustrative Solution. — Required, the values of «, y, and 2 in 75 2e+S5y—z2=11 x—2y+42=38 Sy—22=4 Ade ees | Se cone 4. abe Mabsh BONS aR aN abiales re meme She fi Wane Ps Tam he ee § ais ty be Explanation. —'The reduction has been effected according to 46. Since both nu- merator and denominator are multiplied by 3, it is omitted. 56. Since the denominator of the frac- tions representing the values of a, y and # respectively is the same, we need to find its value for but one of the unknowns. We therefore find the values of the other numerators and divide them by the com- mon denominator—13. Thus 211-1 fie ied SBe S47 eee, Pe eager ~ Lae oo FE een pean itr wat eae (2 Sold ees My se Oe EA 6 — 21 j pede ~— — $x 3 ~ 76 It is not difficult to see that the method pursued in articles 54 and 55 is independ- ent of the number of equations. This gives rise to the following practical rule for the solution of simple simultaneous equations. The values of the unknowns are repre- sented by fractions whose common denomi- nator is the determinant formed by writing the coefficients of the unknowns in their relative positions. The numerators are ob- tained from the common denominator by replacing the coefficients of the required un- known by the second members in order. EXAMPLES. Direction. — Find the values of the un- knowns in the following groups of equa- tions : 1 ce + y=d. (20— ly =m. Suggestion. —In 1 ane et ny: al Deel bere yt «— z= 14, e—d2+2y = —26. 4, <2 yt22—34 = 22. Suggestion. — Always consider the un- knowns in the same order in all the equations when writing the determinants. In 4 pat DY ee 7 Balun Th! 9G 2s 14; Os 31. Tih i4ioar 2 WN 9 mee ty. oy oa: ES Bh Sivas aH Ou S ee een nee WE ea oe eee me | 2 13 93 2G Te Mia OM Loo lea ERG bei dices : PRO ace Bue h MUR When one line has a common factor, as — 26, 14, 22 in 4, it may be placed before 78 : afl the determinant | nd the other factors” placed instead when “the determinants are- first written. In this way the operation of reduction may often be greatly abridged. 2e+ y+32=3. bins x—2y=— 3b. y—42= 2. 3x2+9y+ 82 = 41. 6 5a2+4yi'— 22 = 20. lle«+T7yi— 62 = 37. 22 a8 Vir emis 26. rf x—tytBe= 24. Sea+ yte= 46. e+ yt e2e=a+b+e. cx + ay + bz = bx + cy +az =ca+ay + bz. Ans. b+e—a,a+c—b,a+b—e. 2e=utyt+z. g Jdy=ututez. "\)\4e=>u+at+y. usa — 14. 4e+t4yt2z2= 388. Se+2y—3t=4. 10.¢ w—y+2t=14. d82—2t+u=9. 2¢t+y=21 (ec Hints. —'Transposing the unknowns hh. 9 to the first members and arranging in the order x, y, 2, u, we find that (ige ted: (eae “Teta (ak PL 4 Gay: F430 =: OT re RR a Me ee ae ily 7 rad Rie a PeyO in Core Grae Oak Rea ipa 7 Wa raneca| Oso aSon Liew 280 Sehawas DOG are eee lint Oia dwee en] eA The reduction has been effected accord- ing to 41,42. When the determinant of the third order contains a number of zeros it is unnecessary to reduce it to a lower order, since it may easily be evaluated. In transforming determinants care should be taken to transform them in such a way as to lead to the smallest possible elements. 80 2e+y+2=17. a4 xetytu=i1i. [ ytetu=9. ax—2by+cze=H7—204+ ax —by — cz = a? — b? — c+. NAfiuda see Lf CONSISTENCE OF EQUATIONS. 57. Definition. — When the number of independent equations is equal to the number of unknowns, the equations are said to be consistent. 2ax + by—cz = 20°+ Cc. 12. 58. Definition. — When there is one more equation than unknowns, the equa- tions are inconsistent unless one of the equations is dependent. 59. If a group of nm simple equations involving »—1 unknowns is consistent the determinant formed by writing the coefficients of the unknowns and the abso- lute terms in order equals 0. Demonstration. — Writing the absolute terms in the first members and denoting 81 — 1 by e, the general group in question becomes a,¢x¢+bhy+...+h2+me=0. ag C+b, y+... +4,2+ me = 0. Ke a, «+b, y+...+1,2 +m, e=09. Since all of these equations are satis- fied when e= —'1’ by h¥pothesis, théy must evidently be satisfiable when e is re- garded as unknown. Considering e un- known and finding the values of the other unknowns by 56, we find that the deter- minants of the numerators reduce to 0’s, since one of the columns (the absolute terms) consists of zeros: therefore the common denominator must equal zero, otherwise the unknowns could have no value except zero. Q. E. D. 60. If the value of a determinant van- ishes, at least one of its rows is dependent." Demonstration. — To a certain arbitrary multiple of the elements of the second column add arbitrary multiples of the cor- responding elements ce all the other col- umns. he-fac eli sd stiall-allbe 1 See Note A at the end. 82 finite,-greater-or~less-than zero, and be 80 chosen.that none of the-sums of the-mul tiples be-zerox The determinants may be written as follows : — ay b, ee ly a 0 A = As by arn be ay Sy Fi => 0) INS ie ls S$; 8... represent the sums of the mul- tiples. Hence A’=3,8,—s,S.+. e .+s,5, = 0. Therefore s, is expressible in terms of s,, Ss. . ., which proves the theorem. The above proof may seem to fail when all the principal minors, 8, S,... are zero. In this case the minors are treated in the same way as the original determi- nant; and by repeating this process we shall see that either all of the elements in a column are zeros (which would prove 83 the theorem), or that one line is expres- sible in terms of the others, which would again be a proof of the theorem. From the foregoing it follows directly that if the determinant formed by writing the coefficients and absolute terms of equations containing 2 — 1 unknowns in order vanishes, the equations must be con- sistent. Illustrative Solution. — PE ape er ptatis 64 Ah CI ea Pgh | 62+8y—22=6:; on 4 2 10 Dye WGy te 4—-1—1 8| °° 6 48 — 25 6 For if, in the given determinant, the third row be added to the first, and the second to the fourth, the determinant will have two identical lines (41, 36). If we find the values of x, y, and 2 in any three equations, and substitute in the fourth, the equation will be satisfied (59). E.g., 84 LOM Me a5 3 10 2 ee alae s § -1 Bis wee GG cine BAAD a a. San N eB alie | Paola ead Aare Gece DE Combining the first and second, and the third and fourth, we obtain (40) 19 tbe B eT On PSST VS ai eg ag hoy Bae ah, GV Sieea a Ushi ee aa 6|-19 25 Fd as Chl i EP ae Teta 61. Direction. — Find which of the fol- lowing groups of equations are consistent. Verify your results by substituting the values of the unknowns. 2¢+3y = 10: 2 oa 3y = —12. eit Oe 1 0, 2ea—y+2z2=6. Sbxa+2y4+7Tz2=12. 85 etyte=9. 5 xety—z=5. ‘ x—-y+2z=5. —ex+y+2=1. 62. If m simple simultaneous equations involving 7 unknowns are not independent, the value of each unknown takes the form O ° . . ° ~}; 1e., is indeterminate. For the determinant numerators and de- nominators will contain identical lines, or may be transformed into determinants containing identical lines. Illustrative Solution. — ax + by + cz =m. ba + cy t+az=n. (a+ b)x+(6+oy+ateoeze=m+n. a b c b c a | =0. a+b b+e cta m b C n C a m+n b+ece c +a 0 a m c b n a 86 This shows that any one of the un- knowns in the given set of equations may have an indefinite number of values, If we assign a fixed value to one of them, the others will also become fixed. The values of the unknowns are just as inde- terminate as if only two equations were given. HomMocGENEOuS EQUATIONS OF THE FIRST DEGREE. 63. Definition. — A homogeneous equa- tion of the first degree is an equation all of whose terms contain an unknown factor of the first degree; e.g., 2% +3y—z2=0 anda—y=0. 2x%+3y—z2= 51s not homogeneous. 64. ithok the number of homogeneous equations is equal to the number of un- knowns, all the unknowns equal zero, unless one equation is dependent. Demonstration. — Given ax + by = 0. cx + dy = 0. 87 Aliisnin =e pe if w and y are not C equal to zero, Te Since the ratio of Canc the coefficients is the same in the two equa- tions, one may be obtained by multiplying the other by the ratio between the coeffi- cients of the same unknown; le., one equation is dependent. When there are more than two equations we may reduce them to two by elimination; .°. the proof is general. Illustrative Solution. — Eliminating z, we obtain 38a—4y=0. 38a2—y=9. which are satisfied only whenz =y = 0. 65. Any group of simple equations is satisfied by making all the unknowns equal to + infinity, since the ratios between in- finities are indeterminate. Any group of simple homogeneous equations is satisfied by making all the unknowns equal to 88 0 orto In general work we are not concerned with the infinite values, but seek only the finite quantities, which satisfy the equation or set of equations. In homogeneous equations we generally seek for values differing from zero. 66. When the number of homogeneous equations is one less than the number of unknowns, or when the number of equa- tions is equal to the number of unknowns, but one of the equations is dependent, etc., the values of the unknowns may generally be found in terms of one of the unknowns; or, what is the same, the ratio which exists between the unknowns may be determined. Consider the three homogenous equa- tions a,x + boy + Coz = 0. AgX -+. bsy +- C3e = If z is not 0, we obtain the following: i + by +aqz2= 0. a +3,2 = —¢, z dg~ + by % =. — e, z z 89 Since we have three equations and only two unknowns, one of the equations is de- pendent if they are consistent (59). The condition that these equations are consist- ent is ay 6b Qe by Co = 0. ag 63 Cg From this we deduce the important prin- ciple. 67. In order that n homogeneous equa- tions, involving n unknowns, may be simul- taneous, it is necessary and sufficient that the determinant formed of the coefficients of the unknowns in order (the determinant of the system) equals 0. EXAMPLES. 1. Which of these sets consists of simul- taneous equations ? 2e+2y—32=0. a 8a—y—z=0. 34a—2y—Tz=0. xe+t2y—sz2=0. b <2x—y+2y=0. x—3y—z = 0. 90 Factors OF A DETERMINANT. 68. When a determinant is equal to zero after x is substituted for y, y — x is one of its factors. Demonstration. — Let A and A’ represent the determinant before and after substitut- ing. Since A has some terms which contain © y” as a factor by hypothesis, we have A=Sy +81 + Soy? + Ssy¥F +... Where §, 8,, S., . .. are the coefficients of the different powers of y and independ- ent of y, therefore they are unaltered by the substitution of x for y. A’ = S2° + 8,2 + 8.274 S.78? +... Subtracting, remembering that A’ = 0, we obtain A=81 (y¥ — 2) + 82.(y* — 2) + Ss(y—-y)+... Hence A is divisible by y— 2. The ele- ments of A are supposed to contain only positive integral powers of y. 91 Illustrative Solution. — Giver jit ok cpu ee Cinae t Explanation. — When we substitute y for x, two lines become identical. There- fore x —y isa factor. For the same rea- son «—a, y—a are factors. It is not difficult to see that these are all the fac- tors, and that the determinant is equal to (@ — y) (y—4) (w—a). | The given determinant could have been factored by subtracting the rows sepa- rately; e.g., subtracting the second row from the first, we find that « — y is a fac- tor; subtracting the third row from the first, and we see that « — a is a factor; by subtracting the third row from the second, we find the remaining factor y —a. These successive transformations, which leave the determinant unaltered, are often very useful to discover the factors. We shall give one more example where the principle may be employed with advantage. Prove that|a 6b ¢ d be a Ca 6d aa PO) ry Aa aN ORS 3 (a+b+e+d)(a+b—c—d) x (a—b+e—d) (a—b—ec+d). EXERCISE. 1. Eliminate the unknowns from axtbhy+oaz+t+d,=0. age + byy + ez + d, = 0. ase + bsy + 6,2 + dz = 0. age + by + ge +d, = 0. Suggestion. — Find the values of the unknowns from the first three equations and substitute these values for the un- knowns in the fourth equation. /t 2. Prove that 4 h otf ye