Linear Systems of Equations

Back

A linear equation system is a collection of coefficients (a_ij), variables (x_i) and constants (d_i) of the following kind:

a₁₁x₁ + a₁₂x₂ + ..... + a_1nx_n = d₁
a₂₁x₁ + a₂₂x₂ + ..... + a_2nx_n = d₂
...............................................
a_n1x₁ + a_n2x₂ + ..... + a_nnx_n = d_n

This can be rewritten in matrix form:

Ax = d

where:

	é	a11	a12	....	a1n	ù
A =	ê	a21	a22	....	a2n	ú
	ê	....	....	....	....	ú
	ë	an1	an2	....	ann	û

is the matrix of coefficients and:

	é	x1	ù
x =	ê	x2	ú
	ê	...	ú
	ë	xn	û

is the vector of variables or unknowns, while:

	é	d1	ù
d =	ê	d2	ú
	ê	...	ú
	ë	dn	û

is the vector of constants.

Defining the "augmented" matrix [A, d] as an n ´ (n+1) matrix with d in the last column, it can be shown that there exists a solution x to the system Ax = d if and only if rank is such that r(A) = r([A, d]). The system has no solution if r(A) < r[A, d]. This is obvious as d ought to be a linear combination of the elements in A if a solution x to exist.

If r(A) = r([A, d]) = n, then the solution is unique (thus if A is non-singular, then Ax = d has a unique solution, x). If r(A) = r([A, d]) < n, then the solution is not unique - in fact, there will be an infinite number of solutions. Consider the following:

Theorem: If Ax = d has more than one solution, then it has an infinite number of solutions.

Proof: Suppose x* and x** are two solutions to the system Ax = d. Then, for any scalar a ¹ 0, we know that a Ax* = a d and (1-a )Ax** = (1-a )d, so A[a x* + (1-a )x**] = d, thus the constructed vector a x* + (1-a )x** is also a solution. As this is true for all a ¹ 0, then there are an infinite number of such vectors.§

Thus a system, Ax = d has either no solution, a unique solution, or an infinite number of solutions. If A is a square matrix and non-singular (has no linearly dependent rows/columns, i.e. |A| ¹ 0), then we can solve for the unique solution vector x*.

(1) Inverting

The straightforward procedure is by inverting the matrix:

Ax = d Þ x* = A^-1d

where A^-1 is the inverse of A and expressed as: A^-1 = (adj A)/|A|, where adj A is the adjoint matrix of A (defined as the transpose of a matrix whose elements are the cofactors of the corresponding elements in A) while |A| is the determinant of A.

(2) Cramer's Rule

A more practical procedure is to consider the following:

x_i* = |A_i|/|A|

where |A_i| is the determinant of matrix A with the ith column replaced by the solution vector d. So, for x₁:

	ê	d1	a12	....	a1n	ú
|A1| =	ê	d2	a22	....	a2n	ú
	ê	....	....	....	....	ú
	ê	dn	an2	....	ann	ú

We can do this for all x_i, i = 1, ..., n and thus obtain the solution vector x*.

(3) Homogeneous Systems

If it turns out that d = 0 for our earlier system of equations, i.e. Ax = 0, then we have a homogeneous system. This causes a problem in solving for x. In fact, there are only two alternatives: either we have the trivial solution x = 0 or else x is indeterminate.

We can see triviality immediately from Cramer's rule. As x_i = |A_i|/|A|, then because the system is homogeneous, the vector 0 is in the ith column of A_i. Thus, expanding by that column, x_i = 0/|A| = 0. This will be true for all i = 1, ..., n, thus every x_i = 0. The only alternative to the trivial case is if it also happens to be the case that |A| = 0, so then x_i = 0/0 which is indeterminate, i.e. there are infinite number of solutions for x_i.

However, we can still say something about x in the homogeneous case. Namely, if we have indeterminacy (i.e. if |A| = 0), we can still sometimes obtain proportional values x_i/x_j for every i, j = 1, .., n even though we cannot obtain x_i and x_j directly.  But for this we need to analyze eigenvalues.

Back

Top

Selected References

Next