By Alan G. Hamilton
This can be a brief, readable advent to simple linear algebra, as frequently encountered in a primary path. the improvement of the topic is built-in with a good number of labored examples that illustrate the guidelines and strategies. The structure of the booklet, with textual content and suitable examples on dealing with pages implies that the reader can persist with the textual content uninterrupted. the scholar could be in a position to paintings in the course of the e-book and study from it sequentially. rigidity is put on functions of the tools instead of on constructing a logical method of theorems. a number of workouts are supplied.
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Additional info for A first course in linear algebra, with concurrent examples
To see this, we show that B is the inverse of A, where B = [bid] , x P is the diagonal matrix with b,i = 1/aii, for I S i S p. Calculate the product AB. The (i, i)-entry in AB is P Y- aikbki, k=1 which is equal to alibi,, since for k 54 i we have aik = bki = 0. By the choice of bit, then, aiibii= 1 for each i, and so AB = I. Similarly BA = I. 4). 5 Matrix inverses At the end of Chapter 4 we discovered matrices E and F with the property that EF = I and FE = 1, and we said that they were inverses of each other.
A set of p equations in q unknowns involves a p x q matrix multiplied to a q-vector. Now let us abstract the idea. Can we multiply any matrix with any column vector? Not by the above process. To make that work, there must be as many columns in the matrix as there are entries in the column vector. A p x q matrix can be multiplied on the right by a column vector only if it has q entries. The result of the multiplication is then a column vector with p entries. We just reverse the above process. 7.
Here is the trick: (AB)(BA 1)=A(BB-')A =AIA-' 1 (rule (v) on page 27) =AA-'=I, and (B-1A-1)(AB)=B-1(A-'A)B =B-'IB=I. ) This rule extends to products of any number of matrices. The order reverses. Indeed, we have come across an example of this already. , F, respectively, and (E,Er_1 ... E2E1)(F1F2 ... F,-1Fr)=I, and (F1F2 ... Fr-1Fr)(ErEr_ 1 ... E2E1)=1, so the inverse of ErEr -I ... E2E1 is F1 F2 ... Fr-,F,, and vice versa. 10 Find the inverse (if it exists) of the matrix -1 2 1 1 0 1 1 3 2 2 1 1 1 0 0 1 0 -1 0 1 0 1 3 2 0 0 1 2 2 0 0 0 -1 2 1 0 1 0 1 3 2 0 0 1 1 0 0 1 0 0 1 1 1 1 1 2 0 2 2 i -2 0 ?
A first course in linear algebra, with concurrent examples by Alan G. Hamilton