Hindawi Publishing Corporation Mathematical Problems in Engineering Volume 2013, Article ID 321032, 9 pages http://dx.doi.org/10.1155/2013/321032 Research Article Inversion of General Cyclic Heptadiagonal Matrices A. A. Karawia Mathematics Department, Faculty of Science, Mansoura University, Mansoura 35516, Egypt Correspondence should be addressed to A. A. Karawia; [email protected] Received 23 December 2012; Revised 26 February 2013; Accepted 27 February 2013 AcademicEditor:JoaoB.R.DoVal Copyright © 2013 A. A. Karawia. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We describe a reliable symbolic computational algorithm for inverting general cyclic heptadiagonal matrices by using parallel 2 computing along with recursion. The computational cost of it is 21 − 48 − 88 operations. The algorithm is implementable to the Computer Algebra System (CAS) such as MAPLE, MATLAB, and MATHEMATICA. Two examples are presented for the sake of illustration. 1. Introduction heptadiagonal matrices of the form (1) and for solving linear systems of the form: The ×general cyclic heptadiagonal matrices take the form: =, (2) 1 1 1 1 0 0 ⋅⋅⋅ 0 1 1 where =(1,2, ..., ) and =(1,2, ..., ) . To the best of our knowledge, the inversion of a general 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ( 2 2 2 2 2 2 ) ( ) cyclic heptadiagonal matrix of the form (1)hasnotbeen ( ) ( 0 ⋅⋅⋅ ⋅⋅⋅ 0 ) considered. Very recently in [5], the inversion of a general ( 3 3 3 3 3 3 ) ( ) ( ) cyclic pentadiagonal matrix using recursion is studied with- ( ) ( . ) out imposing any restrictive conditions on the elements of ( 0 ... ) ( 4 4 4 4 4 4 4 . ) ( . ) thematrix.Also,inthispaperwearegoingtocompute ( . ) ( 0 dddddddd . ) the inverse of a general cyclic heptadiagonal matrix of the =( ) , ( . ) ( . ) form (1) without imposing any restrictive conditions on the ( . dddddddd . ) ( 0 ⋅⋅⋅ 0 ) elements of the matrix in (1). Our approach is mainly ( −3 −3 −3 −3 −3 −3 −3 ) ( ) ( ) basedongettingtheelementsofthelastfivecolumnsof ( 0 ⋅⋅⋅ ⋅⋅⋅ 0 ) −1 ( −2 −2 −2 −2 −2 −2) ( ) in suitable forms via the Doolittle LU factorization [10] ( ) ( 0 ⋅⋅⋅ ⋅⋅⋅ 0 ) along with parallel computation [7]. Then the elements of −1 −1 −1 −1 −1 −1 −1 the remaining ( − 5) columns of may be obtained 0 ⋅⋅⋅ ⋅⋅⋅ 0 using relevant recursive relations. The inversion algorithm ( ) (1) of this paper is a natural generalization of the algorithm presented in [5]. The development of a symbolic algorithm is considered in order to remove all cases where the numerical algorithm fails. Many algorithms for solving banded linear where ≥7. systems need to pivoting, for example Gaussian elimination The inverses of cyclic heptadiagonal matrices are usually algorithm [10–12]. Overall, pivoting adds more operations required in science and engineering applications, for more to the computational cost of an algorithm. These additional details, see special cases, [1–9]. The motivation of the current operations are sometimes necessary for the algorithm to work paper is to establish efficient algorithms for inverting cyclic at all. 2 Mathematical Problems in Engineering { −1 The paper is organized as follows. In Section 2,new { , if =1, { 1 symbolic computational algorithm, that will not break, is { { 1 1 constructed. In Section 3, two illustrative examples are given. { , if =2, { 2 Conclusions of the work are given in Section 4. { + { 2 2 1 1 { , if =3, { 3 { + + { −1 −1 −2 −2 −3 −3 2. Main Results { , if =4(1) −5, = { { In this section we will focus on the construction of new { −1 −−5−5 −−6−6 −−7−7 { , =−4, { if symbolic computational algorithms for computing the deter- { −4 { minant and the inverse of general cyclic heptadiagonal { − − − { −1 −4 −4 −5 −5 −6 −6 , =−3, matrices. The solution of cyclic heptadiagonal linear systems { if { −3 of the form (2) will be taken into account. Firstly we begin { { − − − with computing the factorization of the matrix .Itisas { −1 −3 −3 −4 −4 −5 −5 , =−2, if in the following: { −2 { , =1, { if { 1 =, (3) { (ℎ − ) { 1 1 {− , if =2, { 2 { (ℎ +ℎ ) where { 2 2 1 1 {− , if =3, { { 3 1 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 { (ℎ−1−1 +ℎ−2−2 +ℎ−3−3) {− , if =4(1) −4, 2 1 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 ℎ ={ 3 3 1 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 { ( ) { −ℎ−4−4 −ℎ−5−5 −ℎ−6−6 ( 4 1 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0) { , if =−3, ( 4 4 ) { ( 1 ) { −3 ( ) { ( . .) { −ℎ−3−3 −ℎ−4−4 −ℎ−5−5 ( . dd d d d d d d .) { , =−2, ( . .) { if ( ) { −2 =( . .) , { ( . dd d d d d d d .) { −∑−2 ℎ ( ) { =1 ( 0⋅⋅⋅0 −3 1 dd0) { , if =−1, ( −3 −3 ) { ( −6 ) −1 ( ) ( 0 ⋅⋅⋅ ⋅⋅⋅ 0 −2 1 d 0) , =1, ( −2 −2 ) { 1 if −5 {− V + , =2, ⋅⋅⋅ ⋅⋅⋅ 10 { 2 1 2 if 1 2 3 −2 −1 −2 { ℎ ℎ ℎ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ℎ ℎ ℎ 1 {−3V2 −3V1, if =3, 1 2 3 −3 −2 −1 { { ( ) {−V−1 −V−2 − V−3, if =4(1) −4, { (4) { −3 { −3 V = −−3V−4 −−3V−5 − V−6 +−3, if =−3, 00⋅⋅⋅0 V { 1 1 1 1 1 1 { −6 02 2 2 2 0⋅⋅⋅0 2 V2 { { −2 003 3 3 3 ⋅⋅⋅ 0 3 V3 {− V − V − V + , =−2, ( 000 d 0 V ) { −2 −3 −2 −4 −5 −2 if ( 4 4 4 4 4 ) { −5 ( ) { ( . ) { −2 ( . ddddd d d d . ) { ( ) {−1 − ∑V, if =−1, =( . ) . (5) ( . ddddd d d d . ) { =1 ( . ) ( 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 −3 −3 −3 V−3) , =1, ( 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 V ) { 1 if −2 −2 −2 {− , =2, 0 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 V { 2 1 if −1 −1 {− − , =3, 0 0 0 ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ 0 { 3 2 3 1 if { ( ) {− − − , =4(1) −5, { −1 −2 −3 if { −3 { = − − − −4 + , =−4, The elements in the matrices L and U in (4)and(5)satisfy { −4 −5 −4 −6 −7 −4 if { −7 { { {− − − −3 + , =−3, { −3 −4 −3 −5 −6 −3 if , =1, { −6 { 1 if { { − , =2, { −2 { 2 2 1 if − − − + , =−2, { − − , =3, { −2 −3 −2 −4 −5 −2 if { 3 3 2 3 1 if −5 { − { − − − , =4(1) −2, 2 { −1 −2 −3 if { , if =2, = −3 { { −2 { 1 { { {−1 − ∑, if =−1, { ( − ) { 3 3 1 , =3, { =1 = { if { −1 { 2 { { { − ∑V ℎ , =, { ( − −(/ ) ) if { −2 −3 −3 { =1 , if =4(1) −2, { −1 Mathematical Problems in Engineering 3 3 { , =3, Set 3 =−(1 ∗1 +2 ∗2)/3 { if = 1 { ( −(/ ) ) ℎ =−(ℎ ∗ +ℎ ∗ )/ { −3 −3 Set 3 1 1 2 2 3 , if =4(1) −2, { −2 Set V3 =−3 ∗ V1 −3 ∗ V2 Set 3 =−3 ∗2 −3 ∗1 {1, if =1, = {2 −21, if =2, { −−1 −−2, if =3(1) −3, Step 2. Compute and simplify. 4 −2 1, if =1, For from to do ={ −−1, if =2(1) −4. =( − ∗ / )/ (6) −3 −3 −2 =( − ∗−3/−3 − ∗−2)/−1 We also have: −2=−2 −−2 ∗−3 −1 =−1 −−1 ∗−2 −−1 ∗−3 Det () = ∏. (7) =1 = − ∗−3/−3 − ∗−2 − ∗−1 Remark 1. It is not difficult to prove that the LU decomposi- If =0then =end if tion (3) exists only if ≠ 0, = 1(1) −1 (pivoting elements). Moreover the cyclic heptadiagonal matrix of the form (1) End do has an inverse if, in addition, =0̸ .Pivotingcanbeomitted by introducing auxiliary parameter in Algorithm 1 given Step 3. Compute and simplify. later. So no pivoting is included in our algorithm. For from 4 to −5do At this point it is convenient to formulate our first result. =−( ∗ + ∗ + ∗ )/ It is a symbolic algorithm for computing the determinant −3 −3 −2 −2 −1 −1 of a cyclic heptadiagonal matrix of the form (1)and =−( ∗−3/−3 + ∗−2 + ∗−1) can be considered as natural generalization of the symbolic algorithm DETCPENTA in [5]. End do Algorithm 1. To compute Det() for the cyclic heptadiagonal matrix in (1), we may proceed as follows. Step 4. Compute and simplify. Step 1. Set 1 =1 For from 4 to −4do If 1 =0then 1 =(is just a symbolic name) end if ℎ =−(ℎ−3 ∗−3 +ℎ−2 ∗−2 +ℎ−1 ∗−1)/ Set 1 =1, 1 =1 V =−( ∗ V−3/−3 + ∗ V−2 + ∗ V−1) Set 1 =−1/1 Set V1 =1 End do Set 1 =1 Step 5. Compute simplify. Set ℎ1 =/1 Set 1 =1 −4 =(−1−−5∗−5−−6∗−6−−7∗−7)/−4 Set 2 =2/1 −3 =(−1−−4∗−4−−5∗−5−−6∗−6)/−3 Set 3 =3/1 −2 =(−1−−3∗−3−−4∗−4−−5∗−5)/−2 Set 2 =2 −2 ∗1 −4 =−4−−4∗−7/−7−−4∗−6−−4∗−5 If 2 =0then 2 =end if −3 =−3−−3∗−6/−6−−3∗−5−−3∗−4 Set 2 =−1 ∗1/2 −2 =−2−−2∗−5/−5−−2∗−4−−2∗−3 Set V2 =2 −2 ∗ V1 ℎ−3 =(−ℎ−4∗−4−ℎ−5∗−5−ℎ−6∗−6)/−3 Set 2 =−2 ∗1 ℎ−2 =(−ℎ−3∗−3−ℎ−4∗−4−ℎ−5∗−5)/−2 Set ℎ2 =( −ℎ1 ∗1)/2 V−3 =−3 −−3 ∗V−6/−6 −−3 ∗V−5 −−3 ∗V−4 Set 2 =2 −2 ∗1 V = − ∗V / − ∗V − ∗V =( − ∗ )/ −2 −2 −2 −5 −5 −2 −4 −2 −3 Set 3 3 3 1 2 −2 V−1 =−1 −∑=1 V Set 3 =3 −3 ∗1 −3 ∗2 −2 If 3 =0then 3 =end if −1 =−1 −∑=1 4 Mathematical Problems in Engineering 00000 If −1 =0then −1 =end if 00000 ℎ =( − ∑−2 ℎ )/ (. .) −1 =1 −1 (. .) ( ) (. .) (. .) = − ∑−1 V ℎ ( ) =1 = (. .) , (. .) (. .) (00001) If =0then =end if ( ) (00010) 00100 Step 6. Compute Det() =∏ ( )=0. =1 01000 The symbolic Algorithm 1 will be referred to as 10000 DETCHEPTA. The computational cost of this algorithm ( ) is 52 − 195 operations. The new algorithm DETCHEPTA (9) is very useful to check the nonsingularity of the matrix we get when we consider, for example, the solution of the cyclic () (−1) (−2) (−3) (−4) heptadiagonal linear systems of the form (2).