By Martyn R Dixon; Leonid A Kurdachenko; Igor Ya Subbotin

ISBN-10: 0470496363

ISBN-13: 9780470496367

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**Extra info for Algebra and number theory : an integrated approach**

**Example text**

This provides us with a very simple way of writing systems of linear equations. We can use the algebra of matrices to help us solve such systems also, as we shall see. Now we consider multiplication of a matrix by a number, or scalar. 7. Definition. Let A = [aiJ] be a matrix from the set Mkxn(IR) and let a E JR. The product of the real number a and the matrix A is the matrix aA = [cij] E Mkxn(IR), whose entries are defined by C;J = aa;1, for every pair of indices (i, j), where 1 ::::; i ::::; k, l ::::; j ::::; n.

Theorem. Let a, b, c E Z. Then the following properties hold: (i) if a I b and b I c, then a I c; (ii) if a I b, then a I be; (iii) if a I b, then ac I be; (iv) if c =/= 0 and ac I be, then a I b; (v) if a I band c I d, then ac I bd; (vi) if a I b and a I c, then a I (bk + cl) for every k, l E Z. Proof. (i) We have b = ad and c = bu for some d, u E Z. Then c = bu = (ad)u = a(du), so that a I c, since du E Z. (ii) We have b = ad for some d E Z. Then be = (ad)c = a(dc), so that a I be. (iii) We have again b =ad.

1 = 8s. 1 is a right inverse to f. (vi) Assertion (iii) implies that the mapping f is injective, and by hypothesis, f is surjective, so f is bijective. Let b 1 , b2 be two distinct elements of B. Since f is bijective, there are distinct elements a 1, a2 E A such that b1 = f(al ), b2 = j(a2). Since the mapping h is injective, f2(aJ) =f. f2(a2). In tum, it follows that which shows that f 1 is injective. SETS 25 Furthermore, g2 o h =cA. Since f is a bijective mapping, it has a right inverse g, so fog= cs.