Applied Mathematics and Mechanics (English Edition, VOW.8,No.10, Oct.1987)

Published by SUT, Shanghai, China

A MATRIX METHOD OF D I S P L A C E M E N T A N A L Y S I S OF THE

G E N E R A L S P A T I A L 7R M E C H A N I S M Chen Wei-rong ( I ~ , ' ~ ) (Shanghai Second Polytechnical University, "Shanghai)

(Received June 28, 1986; Communicated by Guo Zhong-heng) Abstract An input-output equation o f the general spatial 7R mechanism is derived in this paper by using the method in ['lJ and applying the rotation matrices. The result is the same as [2], but the process o f derivation is simpler. Applying the character o f rotation matrices, it is not difficult to obtain the recurrenceformulas o f direction cosines o f Cartesian unit vectors, to calculate the scalar products and triple products o f these unit vectors, and to derive the 6th constraint equation. Moreover, an algorithm, which consists o f successive applications o f row transformation and expansion based on Laplace "s Theorem, is given to evaluate the 16 x 16 determinant according to its characteristic, so that the evaluation is much simplified.

I.

Introduction

It is well-known that the derivation of the 32nd degree polynomial displacement equation for the general 7R mechanism is the most difficult task in the analysis of single-loop spatial mechanism. It is 'The Mount Everest of Kinematic Problems'-- Freudensteint41. This equation, which is expressed in the form of a 16 • 16 determinant equated to zero, was derived in [2] by applying the theory of spherical and spatial polygonstSl, but its derivation and the evaluation of the determinant are rather complex. In this paper, a displacement analysis of the. general spatial 7R mechanism is given by using the method in [1] and applying the rotation matrices. Imagining a R R R link set is disassembled from the mechanism, its five constraint equations constitute a set of displacement equations of the mechanism. In order, to eliminate the unwanted angular displacements conveniently and obtain the 32nd degree polynomial input-output equation, it is necessary to~erive an equation which is not linearly dependent on these equations, and which is linear in the sinesand cosines oftlae angular displacements. Applying the character of rotation matricest3], it.is not difficult to obtain the recurrence formulas of direction cosines of Cartesian unit vectors, to calculate the scalar products and triple products of these unit vectors. Therefore, the abovementioned five constraint equations can be expressed in the simplest form, and the 6th constraint equation can be obtained conveniently, thus, the derivation is much simpler than .[2]. Using the method in [2], an input-output equation of degree 32 in the tan-half-angle of the 6utput angular displacement is obtained by eliminating the unwanted variables from these equations. The result is the same as [2]. Moreover, an algorithm, which consists of succesive applications of row transformation and expansion based on Laplace's Theorem, is given to evaluate the 16 x 16 determinant according to its characteristic, so that the evaluation is much simplified. Figure 1 illustrates the general spatial 7R mechanism. The frame is link 4, the construction 957

958

Chen Wei-rong

parameters are ( 1.., h.., a , . . ), and the angular displacements are 0.. (m,n = 1.2 ..... 7). The coordinate systems S,,.(x,., Ym' Z,.) and S,.. (x,... Ym.' Z,..) are illustrated in this figure to(>.

-:::_%

o ~

.

:~ ~,.

A aTt

,4

~r

Fig. 1 In this paper we use the symbols for basic rotation matrices and vector expressions in [3] as follows.

,=

0

cos0--sin0

0

sin0

, K0=

cos0

sin0

cos0

0

0

0

1

The expression or column matrix for vector V in coordinate system S= is indicated asV(">, thus

V<'~>=Vx.,i+Vumj+Vz=k,

o r V ( ~ > = ( V x ~ , Vu= V.-~)~

where V x . , VIi = and Vz., are the components o f vector V in coordinate system S,..

Some characteristics of rotation matrices are listed as follows [~1.

I. I;l=l-e,

K;'-----K_8

2. ( M V ~ ' > ) ( M W z ''>)

3 . (MV~=~)• . . . .

(1.1)

_uc=)u(,.> ---r -z

~ .' . . . . ' . - I

(1.9.) •

V(,''>)

(1.3)

where M is the product of some basic rotation matrices. In order to w,-ite the equation in a condensed form, V(0 is abbreviated to V, and the following shorthand notations are used. s,~.---sinam., c m . = c o s a , ~ . ; x . - - - - t g ( g . / 2 ) sm~sia0m, r Cm,

Mm= Ke,,,[ am(,n+l) ---(

Sm 0

~ Sm,Cm.cm,+ 1)

Srtm,Slm,(m+ 1)

Cm.Cm(m+ 1)

~ Cm,Sm.(m,+ 1)

Sire(m§ t)

Cm(m§

\

(1.4)

Displacement Analysis of the General Spatial 7R Mechanism

959

Obviously, we have

II.

The Recurrence

2x= 1 - - x ~. s,~= l + x . 2 , e,.-- l+x-------~.

(1 5)

x m s . , + c = = 1, x = c = - - s m = - - x m

(1.6)

Formulas

of Direction

Cosines

Let m, and n be positive integers, l ~ m ~ n ~ 8 (Link 1 may be considered as link 8). The expressions in system S mfor unit vectors in the x + ~,y.+ t, z+ t directions (i.e.i., ,, ], + ~, k, + ~ ) are as follows.

..+,,+-,

.....

,.+,i~,~'= (sl'~. =Ms. ) j - - E , . . . . i + F , , . . . . j

-""'"

+ G.,+,... k .

j

(2 .1)

=(E".

where(U=...., V,~...., W = . L . . . . ) , ( E , , . . . . , F=...., G,~., .... ) and (Xm ..... Y . . . . . . Z,..~ ....) represent three components (direction cosines) in system S,, of unit vectors i + ~,j, + and k . . ~, respectively. The subscripts m... n (a sequence consists of all integers between m and n) denote that the component is linear in the sines and cosines of the angular displacements 0=, 0=+~, "",On. Particularly, if n > m ~ - I , we have .-t

n-I

(2.2) Obviously, the matrix

/ U .... .

E,.... .

X,.

(2.3) W

..

G,,+t .... Z.,+,....

is the coordinate transformation fnatrix from system S,+, to S,,. If re=n, then from(2.1) we have ir=+l----l~.~----'f r ,,i

+ V=j + W(,.+l,.,k

and from (1.4) we obtain U,~----c,~, Y=----s=, W~=+l>,~=O Similarly,

]

E m----'--sMcm(=+l~, F=----~=e=(=.~, G(,,.~.,=s=~.,+~) I X"l~S.lSm(m+l)p Ym~--'=--emsm(m+l), Z ( m + l ) r ~ C n t ( m + l )

(2.4)

It is to be noted that the subscript (m+ 1)m which indicates the component is a constant. Suppose m < n 9 Let V~.+, ... . , F:+,... ., Y"-+t.... be the direction cosines of unit vector

960

Chen Wei-rong

]~m+l)-lrin system an+ I. Obviously, we have

"-+, .... --JI'o-,-+,,( ~.+, .... =j[,o.,..,,

tl

n .,),1

J*m+l

(2.5)

( fl ,, )J] J-m+l

~-., .=(,o.,.+,, ( f l . ,

)~]

J-m+l

Since

"

J-m+l

substituting (2.1) in the above equation and using (2.5), after transformations we obtain the following left-hand recurrence formulas. ra...

n~craUm+l

...

n~smVrn+ l

...

n

V~'... n=C(ra_t)mP'm~.., n--S(m-l)mWra+l .:.. V,, .... --s,,,Um+, .... + c~V,. +, .... W,~ . . . . = s<,~_,),~V,...., + c(,._ o , . W , . +, ....

}

(2.6)

It is to be noted that th.e above 2nd equation is also true when m = n. T h e other two sets o f left-hand rei~urrence formulas are obtained by cyclic exchange o f c o m p o n e n t s in (2.6) according to the form of

~"

U

-'~, ~

X~-E

V

"~ and

r

Y~F

Z~G

E , , ... , = c , ~ E,~+, ... , - s , , , F ~ + , ..., F [ .... ----c<,,,_,,~F,,, ... , - s ( , , _ , ) , , G , , , +, ... ,,

(2.7) F,~ ... , = s,,E,~+l .... +c,~ F~*+I ..., G., .... = s(,,,_,;,,,F . . . . .+e(,,,_t~.,G.,+~ .... X,~ ... ,----c,~X,~+, ... , - s , ~ Y~+I ...,

] /

Y . . . . ,-----s,~X,~+l .... + e ~ Y~+, ..., Z ~ .... = s(.,- I . . Y . . . . . + c ( ~ _ , , , Z , ~ +, ...,

(2.8)

[

)

F r o m ( 2 . 6 ) - ( 2 . 8 ) and (2.4), we have

V~--~c'(~_l)~s~ F~

m c ( m _ 1) mere(Ira,+ l ) C m - - s ( m _ l ) m ~ m ( m + 1.)

V~

~ ~ c ( m _ l) rosin(m+ l ) C m . ~ s ( m _ l ) m C m ( m + 1)

} (2.9)

Displacement Analysis of the General Spatial 7R M e c h a n i s m

961

Wra, ~ - s (m_ l) r,aSm

G,~ ~S(m_l)mCm.(ra+l)Cm-Jl-e(m_l)mSm(m+l) }

(2.10)

Zm~---S(m_t)mSm(m+l)Cm'q-C(ra_l)mCm(m+l) Since tl

( ,..i:/"'),..k,

substituting (2.3) in the above equation, we have the following right-hand recurrence formulas.

X,.,...=s.~.+~;(U .... ._,s,,- E,....._~e.) + e.~.+~X,.....-~

1

Y . . . . . = s,,<.+~ (V,,,... , , _ , s . - - F. . . . . ._Lc.) + c.~,,+ ~>Y,,, . . . . .

(2.11)

Z . . . . n =Sn(~l+l)(Wnl ' . - I S . - - G ~ ,.. n-IOn) JI-C~:(n+l) Zra ....-1 The other two sets of right-hand recurrence formulas are obtained in the same way as stated above.

..., = e , ( , + r

U,. .... = U .... ._~c,,+ E,....._~s. V . . . . . = V , , , . . . ~ , _ I c . + F , . . . . . . ,s.

i .

W,,,....=W,.,..._~e.+G,....._~s.

J

( - - U,~ ... , - i s , + E , , ... , _ l c , ) + s,c, + t)X,~ ... ,_,

F,n ... , = e , c , + l ) ( - - V,~... , _ i s , + F , ~ . . . . _ t e , ) + s , ( , + t ) Y,, ... , - i G,~ ..... =e,,<.+l>

(2.12)

(2.13)

( --IV,,,... , , _ l s . + G . . . . . . le,,) +s.<.+l> Z,,, ...._,

Noting that re(or n) is the left-hand (or right-hand) integer o f subscript sec~uence m...n, variable 0m can be separated from other variables contained in a c o m p o n e n t using the left-hand recurrence formulas ( 2 . 7 ) - ( 2 . 9 ) , and 0, can be separated using the right-hand recurrence formulas (2. l l) - (2.13).

III.

The S c a l a r P r o d u c t s and Triple P r o d u c t s of Cartesian Unit V e c t o r s

Suppose l ~ m ~ n ~ . ~ 8 . Using (1.2), it is easy to obtain the scalar products of two Cartesian unit vectors as followg.

Therefore,

id,t=U,...,_t ] Similarly,

i~,k,=X,+... ,_t k,~i,,----W,~+ 1 ... , - i / k..k,----Z,§

t ... ,-i

Particularly, in the case o f n = m + 1 we have i.,i~,+t = U . , ,

imk.+l=X,,

k~i,.+l=O,

k , ~ k , , + t = c~,~=+l)

(3.1)

962

Chen Wei-rong

Soppose l < m < n follows9

9 The triple products of three Cartesian unit vectors can be obtained. ~s m-l

,,.. ~,, ,.,=((,n.,.,)~, (~, ,, 1~, (n"-'M,1~) From (1.2), ~1.3) and the above equation we have |

n-I

(,.q_ ,;,1~, (n,, 1~)

,~., ~,, ~.,=(~, and from (2.3) fD

Therefore,

(k,,,, kl,

(3.3)

k,,)=Wt+1...,,-iY,,...,-l-Gl§

Similarly. Z

,,. ~, ~.,=(,

~--1

(n M;,)~(n,,

1~ )

= G ~ . l ... ,,-tZ,,+ 1... , - , - - Z z + t :.. , , - Y , , . . . ,-1

(3.41

Jm~

Suppose

l~m(n~6

.. Some shorthand notations are introduced as follows.

/-~?.."~: (i,~ x kD (i,, x k ,,," M('"~ 2-..~-- (i,,, • k,) (k,, x ke)

}

N ~ : ~ = (k,,, x k~) (i,, x k,,), --z...5--T4("")--rut,.,. ^v kl) (k,, X ka)

(3.7)

From which

,...,.~,•

L("~"-[

,---,

,;: ) k ][ ( n~.-~I~, )(,• (r

M~ ) k )1

= Z 2 ... ,,-1 ( F , , ... ,_ :Z,§ ... 6 - Y , , ... , - i Y , ... 6) --G~ ... , , - R G,,§ l ... , - , Z , § ... 6--Z,, § 1... , - t Y , ...e)

(3.8)

Similarly, M $ ~". ~--- Z z ... , , _ A V , , ... ,,-1Y= ... ~ - - F,,, ... , , - I X , , ... B) - - G ~ . . . , , - A W,~+ L... ,~-l Y , , . . . 5 - - G , , + l ... , - i X , , . . . N ~~". ' 6 ~ G ~ ... ~ - I ( E,~... , , - I Z , . L. , B - X , ,

~)

(3.9)

... ,a-l Y , ... 6)

- - W 2 ... , , - A F , , ... = - I Z , + t ... 6 - - Y , , ... , - t Y , .. ,)

(3.10)

H ~ ? . : ~ = G2 ... ,~_,(U,,.. ,,=,]/',,... , - - E , , ... ,,_,X,, ... ,) - - W , .., ,,-l (V,, ... , _ , Y , ... ~ - - F , , ... , - i X , ...,)

(3.11)

Displacement Analysis of the General Spatial 7R Mechanism

963

The subscripts of every term in (3.3) - (3.11) do not repeat, so that these terms are linear in the sine and cosines of every angular displacement 0~ (]= 2,3,4,5). IV.

The Displacement Equations

The displacement equations of the general 7R mechanism shown in Figure 1 are the constrain equations of the RRR link set (link 1 and 7) as followsm.

Ra.a,kl--RalB~ks RA~B,ka-----Ra,s~ke

(4.2)

(RA,es, kl, ke)=(Ra;n~, ks, k~)

(4.3)

k~ks=keks (R..~,)z: (Ra~s,) ~

(4.4)

(4.1)

(4.5)

where 6

6

R,,~n, = 3-~1. i . + S-~ h,.k., $

8

Ra,n, ------ ~ t . i . - - ~ h.,k. ram7

Using (4.6),

(4.8)

(4.7)

film8

(4.7), (3.1)-(3.6), (2.4) and (2.10), (4.1)-(4.5) can be reduced to the followin

equations.

O, =s,Rh~s,,c,--I:O

(4.1) p

O2=ser(his71cT-/tsO

(4.2) p

O s = K,cT-hTs,~s~tsT Zz ... e~- K s - - ssTsTtc~

(4.3) p (4.4) v

Q , = K T c I + Kss~

(4.5) I

where e

6

O,= ~.,l,. w,...,,_, + ~h,,Z~...,,_, + K, 01-8

m-8

8

4

Q~=~ t,,x,,...,+ ~h,,Z,,§ fl~ ~ 2

fll ~ 2

4

5

Q~= v~I,,(G~....,_,Z,.+,...,-Z~...,,_,Y,,...,)+ ~-'h.,(W,....,_y,,...~ lrr - 8

m, ~ $

-- G2 ... ,,,-iX,, ... ,) + IR s l ~ Z , o - c , Y , . . . , t) + Io(e,G,~-Z,,Ys) 6

O

o,= 22 ll~m2

~ II--lW+ 1

0

+

+ leG2 ... 6 - h 2 s , X ~ . . . o - K s 4

6

~,,~.ts...:._~ + I2 ( ~2 t.h.x....._, m-'2

3

2: h.t.w.+~...._, )+ :E fl~ M4.2

~rb*p2

n - l~.+ 1

8

~ n ~ ffl.§ 2

h.h.Z.,, ... ._, + K.

(4.8)

964

Chert Wei-rong Kt =hoee~cTl + hTeTx+ hx + hzcx, Kz=hre68 + h8 + hTea7+ hlesTc~l K~ =/7s6~c~ + ll%7svt K'4=/TeeTsTtq-/tsevev~, K~=eB~c~

'

Ks=

h' --

l~+ ~ 2

=

5 )l ,,- -) /

h=

l?l,-- 7

7

-4- s hmh,.. ,c,,,o,,+t;- ~ h,,,h=,.,e.(.+,)--h,h,%,e,, Irl=2

rtt ~ 8

tC7=l~l~--h~hxs~s~t, Ks=helts~ + l~h~s~ k~(i= 1,2.... ) is constant, andQ~ is linear in s and cj (j = 2,3,4,5). The following derivation indicates that Q,Z~... ~-Q~Qz is linear in sj and cj too, so that the 6th constraint equation, which is not linearly dependent on set (4.1)"-(4.5), can be derived conveniently. Multiplying the left and right sides of (4.4)" by corresponding sides of (4.5)" gives ~

l,~l.(i,,i.) (kxk~)+

l=h,(i,~k.)(k,kr)

6

S

+ ~

h,.I. (k~i,,)(k,ko) ~+ ~

B ~ TI;+2

~

6

~ ] h=h,,(k,,,k,,)(k,ke)

ftl~2

n ~ ITI + 2

+ K,Z2... 6= (KTe7 + KssT) (Ks-s87sTxeT)

(4.9.)

Multiplying (4.1)'by (4.2)'and rearranging gives 6

~ m.-2

~

4

Y~, /,./,,(i,,,k~)(kli,,)+y-],[ 1 1 - rt; + / ,

I1~ m 9

5

~

l,,,h,,(i,,,ko)(k,k,,)

~t = 11'/,+ 1 3

+ _~ h=/,,(k,,,k6)(k~i,,)]+~ II = I B + Z

6

~

h,.h.(k,,,k,)(k,k,,)+Q6

I/1, = 2 B ~ tB'I- 2

= sr~s~l(hes07cT-- ITs~) (hls71eT--lls0

(4.1o)

where

o5= ~--]~ 5IA 1.(l~W2...,,_lX,...B+h,,Z2,..~_lX,...5

)

tit,=8

4

+

h,(l,,,Wz ..... tZ,+x...s+h,,Z2...,,,_IZ.+t...u) ] nmm--I

+KLQ2+Kz(QI-K,) Then subtracting (4.10) from (4.9) and using (3.7) and following Lagrange's Identities (i,,,i,,) (k~ks) -- (i,.kO (kti,,) "-" (i,,, X kD (i,, X kD **,~,o

(4.11)

Displacement Analysis of the General Spatial 7R Mechanism

965

we have (4.12)

Q~ = K s (KTc~ + K~s,) where 6

O.

lrl*~

I1- rtl+ l

$

4

8

5

t/t~2

n=m+l

n - ,-71+ 2

5

a,,,,,~ ~ z ... s + KeZz... ~--Q, + l~ltseTs,,

(4-13)

l r l ~ 9- t l ~ n l + 2

(4.12) is the 6th constraint equation which we want.

V.

The Input-Output E q u a t i o n

The input-output equation of the general spatial 7R mechanism can be obtained by eliminating ( 0~. 0z , 0~ ) from set (4.1)'- (4.5)'and (4.12) using the method in [2]. Solving for c7 arid S 7 from (4.4)" and (4. l ) . we have % = (K~--Z~ ... ~)/(s~s~, ), s~=

[h~(K~--Z~... ~) -. O~]/(l~s~

)

(5.1)

Substituting (5.1) in (4.2). (4.3); (4.5)" and (4.12) sives

Q~=K~o(K~--Z~... ~)+K~Q,

(5.2)

Q ~ = K , z ( K ~ - - Z z . . . ~) + K , : Q ,

(5.3)

O.~=K,~(K~--Zz... ~) --K,~Q, Q,= K~[K,,( K,--Z~... ~)--_K,~Q, ]

(5.4) (5.5)

where

Kg =lls,7/(L.s~l),

Klo=hl--Kgh6 ]

Kll=hlseT/17,

K , a = h t + Kghe

(5.6)

/q,z='/,.sT/s67 + ItcT~/sT~--heKt, K i4=lTl,/(soTs,,)+K~hi

In order to eliminate (/gz, 0~ ) from set (5.2)"---(5.5), we use the recurrence formulas mentioned above to separate ( 02, /9~ ) from other ~r and then use (1.5). thus we have (a,x~ +b~xs+ct,)x~ + (d~x~ + e , x s + f , ) x 2 - r (g,x] +h~x6 +./~) = 0

(5,7)

where i = 1,2,3,4. The coefficients (% b, .... Ji) are quadratic jn the tan-half-angle x 3 and x 4. Multiplying the four equations of set (5.7) throughout sequentially by x 2, x 5 and x2x 5. Following this procedure give a total of 16 equations from which it is possible to eliminate linearly the 15 unknowns( x ~ x l , ..., xs, x2 )- Then we obtain the input-output equation of degree 32 in x 4 as followst2k

0004, a16~

0 0

a~ b, d,

e,

0 0 9~ 0 0 0 h, ], ai d~ 0 0 b~ c~ e, 0 c, 0 0 a, b~ 0 0 0 0 f~ j~ d, e, g, a, 0 0 0 b, c , 0 0 d, g, 0 0 e, / , h,

h, ],

(i=1,

9, h,

0 0 c, f ,

2, 3, 4)

f~

=0,

(5.8)

966

Chen Wei-rong

There are more than seven elements which equal zero in the lst-12th columns of the 16 x 16 determinant A~6. According to this characteristic we can use the following method to evaluate this determinant. At first, every element in 1st-4th columns of the determinant, except a4, c4, g4, and Ja, can be reduced to zero by means of row transformations. Then the determinant can be expanded by applying Laplace's Theoremt61 as follows.

(5.9)

Als=a4c494/,Alz where

A,6 A~6 0 0 As1 Ass Ass Au 0 0 A=l A,~2 0 0 0 0 A,~s A,~, A=5 I A.5 A . l Anz 0 0 Arts An4 0 0 0

An=

(k=l,

0

Akl

Akz

2, 3; I = 4 ,

0

0

Aks

A**

A,6 As. A=e A..

A,7 A,8 Al7 Az8 A.~7 A~s A.7 A.s

5, 6; m-----7, 8, 9; n----10, 11, 12)

After that, all elements in lst-4th columns of the 12 x 12 determinant A l~, except those in I st, 2nd, 7th and 8th rows, can be reduced to zero by means of row transformations as follows. Multiplying the 1st row by E~.,the 2nd row by F~, and adding these two rows obtained to the i' throw (i = 3,4,5,6), the elements, which are in the i'th row and 3rd or 4th columns, can be reduced to zero, where

E~=-(A~tA~-A~I-AID/D, D=AllA2z--AtzAzl, F~=--(AIIA~2--AfIAt2)/D (5.10) Similarly, multiplying the 7th row by Ej, the 8th row by Fj, and adding these two rows obtained to the j'th row (] = 9,10, l l, 12), the elements, which are in the j'th row and 1st or 2nd columns, can be reduced to zero, where Ej and Fj can be calculated by using the equations, which are obtained by changing the numbers 1,2,i in (5.10) to 7,8j, respectively. Then the determinant AL~ can be expanded by applying Laplace's Theorem as follows. A,z=

IB~zl

Azl Azz

(5.11)

Asl Asz

where k,l= 1,2..... 8, IB~,I is an 8 x 8 determinant, in which the 1st a n d 2 n d columns can be obtained from 5th and 6th columns in A,z by removing the elements in 1st, 2nd, 7th and 8th rows. The determinant [B,s[ can also beexpandcd after row transformations as above. Thus the evaluation of this determinant is reduced to the evaluation o f a 4 x 4 determinant. It is obvious that the evaluation is much simplified than using the method in [2]. If the value of input angle 0s ( - - z ~ f f s ~ Z ) is given, then the corresponding value of output angle 8, can be obtained by solving (5.8). From the value of 03 and corresponding value of 84~ we can calculate the values of coefficients .(ai, b~. . . . . ji). Thus the values of x 5 and x 2 can be obtained by using the method in [2]. Multiplying the 2nd equation of set (5.1) by x 7, adding the equation obtained to the 1st equation and using (1.6) leads to the following equation for evaluating the value of x r x~'-

l , [ s , , - - (K,--Zz ... 5~/s,, ]

h,(Ks-Z2...D-{31

(5.12)

Displacement Analysis of the General Spatial 7R Mechanism

967

Now we derive the equations for evaluating the values of x 6 and x,. Obviously, 7

]-[ Mj=E ,i-I

where E is the unit matrix. Therefore, 1

i.e. X~Ti + Ya~j + ZTk = W z ... 5i + G~ ... 5J + Z2 ... Bk Equating corresponding coefficients of i a n d / i n the left and right sides of the above equation, and using recurrence formulas (2.8), we obtain

X7%--Y,se--W2...~

(5.13)

XTsa + Y,~% = 672 ... B

(5.14)

Multiplying (5.14) by x 6, adding the equation obtained to (5.13) and using (1.6), we have x ~ = (X~--W~... D/(Y'~ + G2... u)

(5.15)

xl---- (X2 ... 5-- W~) / ( Y ~ ... s + GT)

(5.16)

Similarly, from

we have

References

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