15. Inverse Matrix

Again let U be a (p,q) matrix and E the unit matrix with r row and columns, when obviously

E_nU = U, UE_n = U.

Thus, nit matrices have with respect to matrix products the same property as the number 1 has for the product of numbers, whence arises its name. Besides, in most cases, the linkage or some other relationship will determine the number of rows in the unit matrix, whence its subscript can be omitted and you can simply write E.

If there the equation Mx = E holds for the column x of the variables x₁, ··· , x_n , where M is a square matrix, then necessarily M = E_n. You confirm this by setting for i = 1, ··· , n consecutively x_i = 1, and all other x_i = 0. According to the rule for matrix multiplication, Mx is then exactly the i-the column of M and x is the i-th column of the unit matrix E_n.

We have already encountered (14.4), in fact, when dealing with the rule for multiplication of determinants (cf. (13.2)). This rule can now be expressed as follows: If C = AB, then |C| = |A| |B| or more simply |AB| = |A| |B| or, in words,

Rule 5: The determinant of a product of matrices equals the product of the determinants.

Naturally, you can transfer by induction this rule to the product of an arbitrary, finite number of matrices.

A matrix of linear transformation is regular, if its determinant is non-zero. Matrices or substitutions with vanishing determinants are singular. It follows from (15.1) that: The product of regular matrices is regular. We will demonstrate below the different characters of regular and singular substitutions. Obviously, an important question is whether a linear substitution, say (14.2), can be inverted, that is, whether and how you can express y₁, ··· , y_n in terms of z₁, ··· , z_n. For the case when |A| 0, we have already dealt with this task in Section 12. In fact, you can interpret (14.2) as a system of n linear equations for y₁,···, y_n. According to (12.3), you find the unique solution

y = (A_1kz₁ + ···+ A_nkz_n)/|A|, (k = 1, ··· , n). (15.2)

In any case,

y = Pz, (15.3)

with a uniquely determined matrix P, which follows from (15.2). If you set z=Ay, you find y = PAy. Since the y₁, ··· , y_n are variables, then, as pointed out above, you must have PA = E. Since E with respect to matrix multiplication corresponds to the number 1, you call P the inverse matrix of A and denote it by A^-1. According to (15.2), you have then

The first expansion formula tells you that

AA^-1 = E. (15.5)

If there exist for any square matrix A a matrix P with the property PA = E, then it follows from (15.1) that |P||A| = |E| = 1, whence necessarily |A| 0. Hence the inverse matrix only exists for matrices the determinants of which do not vanish, that is, only for regular matrices.

There exists for every regular matrix A and only for such matrices a uniquely determined inverse matrix A^-1 with the property

A^-1A = AA^-1 = E.

If in (15.5) you change over to determinants, you find |A||A^-1| = 1 and also |A^-1|=| A|^-1.

If you write the system of equations (12.1) in the form Ax = c, you find, after multiplication from the left by A^-1, the fomula

x = Ex = A^-1Ax = A^-1c,

which is, of course, Cramer's Rule, as you discover by comparison of (12.3) with (15.4).

For two regular matrices A and B also AB is regular. Obviously, B^-1A^-1AB=B^-1EB = B^-1B = E. It follows from the uniqueness of the inverse matrix that

(AB)^-1 = B^-1A^-1.

Correspondingly, you prove for the product of k regular matrices

(A₁ A₂ ··· A_k-1A_k)^-1 = A_k^-1A_k-1^-1 ··· A₂^-1A₁^-1 (15.6)

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