8.18 SIMULTANEOUS LINEAR EQUATIONS
than just positive semi-definite) in order to perform the Cholesky decomposition
(8.128). In fact, in this case, the inability to find a matrixLthat satisfies (8.128)
implies thatAcannot be positive definite.
The Cholesky decomposition can be applied in an analogous way to theLU
decomposition discussed above, but we shall not explore it further.
Cramer’s rule
An alternative method of solution is to useCramer’s rule, which also provides
some insight into the nature of the solutions in the various cases. To illustrate
this 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 equationAx=b. We wish either to find
the solution(s)xto these equations or to establish that there are no solutions.
From result (vi) of subsection 8.9.1, the determinant|A|is unchanged by adding
to its first column the combination
x 2
x 1
×(second column of|A|)+
x 3
x 1
×(third column of|A|).
We thus obtain
|A|=
∣
∣
∣
∣
∣
∣
A 11 A 12 A 13
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 33
∣
∣
∣
∣
∣
∣
,
which, on substitutingbi/x 1 for theith entry in the first column, yields
|A|=
1
x 1
∣
∣
∣
∣
∣
∣
b 1 A 12 A 13
b 2 A 22 A 23
b 3 A 32 A 33
∣
∣
∣
∣
∣
∣
=
1
x 1
∆ 1.
The determinant ∆ 1 is known as aCramer determinant. Similar manipulations of
the second and third columns of|A|yieldx 2 andx 3 , and so the full set of results
reads
x 1 =
∆ 1
|A|
,x 2 =
∆ 2
|A|
,x 3 =
∆ 3
|A|
, (8.130)
where
∆ 1 =
∣
∣
∣
∣
∣
∣
b 1 A 12 A 13
b 2 A 22 A 23
b 3 A 32 A 33
∣
∣
∣
∣
∣
∣
, ∆ 2 =
∣
∣
∣
∣
∣
∣
A 11 b 1 A 13
A 21 b 2 A 23
A 31 b 3 A 33
∣
∣
∣
∣
∣
∣
, ∆ 3 =
∣
∣
∣
∣
∣
∣
A 11 A 12 b 1
A 21 A 22 b 2
A 31 A 32 b 3
∣
∣
∣
∣
∣
∣
.
It can be seen that each Cramer determinant ∆iis simply|A|but with columni
replaced by the RHS of the original set of equations. If|A|= 0 then (8.130) gives