Mathematical Methods for Physics and Engineering - Matematica.NET

8.18 SIMULTANEOUS LINEAR EQUATIONSthan just positive semi-definite) in order to perform the Cholesky decomposition(8.128). In fact, in this case, the inability to find a matrix L that satisfies (8.128)implies that A cannot be positive definite.The Cholesky decomposition can be applied in an analogous way to the LUdecomposition discussed above, but we shall not explore it further.Cramer’s ruleAn alternative method of solution is to use Cramer’s rule, which also providessome insight into the nature of the solutions in the various cases. To illustratethis method let us consider a set of three equations in three unknowns,A 11 x 1 + A 12 x 2 + A 13 x 3 = b 1 ,A 21 x 1 + A 22 x 2 + A 23 x 3 = b 2 , (8.129)A 31 x 1 + A 32 x 2 + A 33 x 3 = b 3 ,which may be represented by the matrix equation Ax = b. We wish either to findthe solution(s) x to these equations or to establish that there are no solutions.From result (vi) of subsection 8.9.1, the determinant |A| is unchanged by addingto its first column the combinationx 2× (second column of |A|)+ x 3× (third column of |A|).x 1 x 1We thus obtain∣ A 11 A 12 A 13 ∣∣∣∣∣ |A| =A 21 A 22 A 23 =∣ A 31 A 32 A ∣ 33∣A 11 +(x 2 /x 1 )A 12 +(x 3 /x 1 )A 13 A 12 A 13 ∣∣∣∣∣A 21 +(x 2 /x 1 )A 22 +(x 3 /x 1 )A 23 A 22 A 23 ,A 31 +(x 2 /x 1 )A 32 +(x 3 /x 1 )A 33 A 32 A 33which, on substituting b i /x 1 for the ith entry in the first column, yields|A| = 1 x 1∣ ∣∣∣∣∣ b 1 A 12 A 13b 2 A 22 A 23b 3 A 32 A 33∣ ∣∣∣∣∣= 1 x 1∆ 1 .The determinant ∆ 1 is known as a Cramer determinant. Similar manipulations ofthe second and third columns of |A| yield x 2 and x 3 , and so the full set of resultsreadsx 1 = ∆ 1|A| , x 2 = ∆ 2|A| , x 3 = ∆ 3|A| , (8.130)where∣ ∣ b 1 A 12 A 13 ∣∣∣∣∣ A 11 b 1 A 13 ∣∣∣∣∣ ∆ 1 =b 2 A 22 A 23 , ∆ 2 =A 21 b 2 A 23 , ∆ 3 =∣ b 3 A 32 A ∣33 A 31 b 3 A ∣33A 11 A 12 b 1A 21 A 22 b 2A 31 A 32 b 3∣ ∣∣∣∣∣.It can be seen that each Cramer determinant ∆ i is simply |A| but with column ireplaced by the RHS of the original set of equations. If |A| ≠ 0 then (8.130) gives299