MATRICES AND VECTOR SPACES
Show that the set of simultaneous equations
2 x 1 +4x 2 +3x 3 =4,
x 1 − 2 x 2 − 2 x 3 =0, (8.123)
− 3 x 1 +3x 2 +2x 3 =− 7 ,
has a unique solution, and find that solution.The simultaneous equations can be represented by the matrix equationAx=b,i.e.
243
1 − 2 − 2
−33 2
x 1
x 2
x 3
=
4
0
− 7
.
As we have already shown thatA−^1 exists and have calculated it, see (8.59), it follows that
x=A−^1 bor, more explicitly, that
x 1
x 2
x 3
=^1
11
21 − 2
4137
− 3 − 18 − 8
4
0
− 7
=
2
− 3
4
. (8.124)
Thus the unique solution isx 1 =2,x 2 =−3,x 3 =4.
LU decompositionAlthough conceptually simple, finding the solution by calculatingA−^1 can be
computationally demanding, especially whenNis large. In fact, as we shall now
show, it is not necessary to perform the full inversion ofAin order to solve the
simultaneous equationsAx=b. Rather, we can perform adecompositionof the
matrix into the product of a squarelower triangularmatrixLand a squareupper
triangularmatrixU, which are such that
A=LU, (8.125)and then use the fact that triangular systems of equations can be solved very
simply.
We must begin, therefore, by finding the matricesLandUsuch that (8.125)is satisfied. This may be achieved straightforwardly by writing out (8.125) in
component form. For illustration, let us consider the 3×3 case. It is, in fact,
always possible, and convenient, to take the diagonal elements ofLas unity, so
we have
A=100
L 21 10
L 31 L 32 1U 11 U 12 U 13
0 U 22 U 23
00 U 33=U 11 U 12 U 13
L 21 U 11 L 21 U 12 +U 22 L 21 U 13 +U 23
L 31 U 11 L 31 U 12 +L 32 U 22 L 31 U 13 +L 32 U 23 +U 33 (8.126)The nine unknown elements ofLandUcan now be determined by equating