## 4. Vectors and linear operators in algebra: the linear systems.

A linear system such

can be interpreted as

The problem is to find, if possible, a vector v(x;y) such that the product between the matrix and v(x;y) equates the vector v’(c;c’) or, more syntheticaly, to solve the equation

From the elementary algebra we obtain

The two fractions can exist only if their denominator isn't null. This denominator, formed only by the coefficients of S, is said the determinant of S.

N.B.: In these pages matrices have their entries between parentheses and determinants have their entries between square brackets.

Using Δ to represent the determinant of S, if Δ is not null, we can write

If we call S-1 the operator in the second side, we can say that the solution of the equation

is

The product between S-1 and S is the identity matrix. Therefore S-1 is the reciprocal (or inverse) matrix of S.

S admits a reciprocal matrix only if its determinant isn't null.

In general, given a square matrix S, we can find its inverse matrix in the following way.

• First, we calculate Δ, the determinant of S.
To calculate Δ we can apply the Laplace's rule.

• Then, we write T, the transpose matrix of S, that is the matrix whose columns are the rows of S and vice versa.
• Finally, we write the matrix whose entries are the cofactors of the entries of T divided by Δ.

Example.

Given the matrix

we have Δ=2, so the inverse matrix is

The product between L-1 and L is

Example.

The linear system

can be represented by the matrix equation

which has solution

This procedure can be applied to more complex systems of n equations with n unknowns. Obviously, the more n is big, the more the calculations grow, and we need powerful instruments of automatic computation such Mathematica (©Wolfram) or WolframAlpha.