By William Brown

Goals to hide crucial points of the idea of matrices whose entries come from a given commutative ring. crucial evidence approximately commutative earrings are constructed in the course of the publication, and proofs that keep on with from concrete matrix calculations also are supplied.

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Extra resources for Matrices over Commutative Rings

Sample text

They can be used if you are only interested in changing the roots of an equation without finding the individual values. 6 The equation 4x 2 find: (i) - 3x + l = 0 has roots ex and ~. Without evaluating ex and ~' 1 l (iii) - +- (ii) ex+~ ex ~ (vi) ex4 + ~4 (v) Solution Dividing the equation by 4, we get x2 -~x+~=0 "'+A- __ J__J. "" p4-4 Hence ex~=~ We have the answers to (i) ~and (ii) ~ (iii) .!. + .!. = ~ + ex = ~ ~ .!. = 3 ex ~ ex~ 4 · 4 (iv) ex2 + ~ 2 cannot be evaluated immediately. However (ex+ ~) 2 = ex2 + 2ex~ + ~ 2 Rearranging, ex2 + ~2 = (ex+ ~) 2 - 2ex~ = m2-2 X~ =k (v) 3 ex + ~ suggests working out (ex + ~) 3 3 (ex+ ~) = ex 3 + 3ex2 ~ + 3ex~2 + ~ 3 3 = ex3 + ~ 3 + 3ex~(ex + ~) Hence ex3 + ~ 3 =(ex+ ~) 3 - 3ex~(ex + ~) = 3 {i) -3 X~ X~ = -649 (vi) ex4 + ~4 = (ex2)2 + (~2)2 = (ex2 + ~2f _ 2 ex2~2 = (16I)2 - 2 X (1)2 31 4 = -256 It is possible, by a technique of transforming the equation, to find a new equation where the roots are related to the roots of a given equation.

E. 5 9 (4 sig. 5 -1) (4 sig. 6 The second term of a geometric series is 8, and the fifth term is 64. Find the first term and the common ratio. Solution Here (i) ar = 8 and 4 = 64 4 64 ar ar (ii) + (i) (ii) (cancel by a) 8 r3 = 8 ar Hence r=2 Substitution into (i) gives a =4 The first term is 4, the common ratio is 2. 7 How many terms are there in the geometric series 9 324 4+8+ 16 + ... + 326? l-J r -s · 4-2 324 Tn becomes 41 x (3)n-l 2 -_ --u; 2 324 _ _ _ (J)24 (J)n-1 2 - 224 - 2 n - 1 = 24, n = 25 There are twenty-five terms in the series.

T2 is the second term, that is, T2 = 2, and so on. Let us look at another example: 8 ~)2r+ 1) r=3 Hence the rth term is T, = r starts at 3, T3 2r + 1 = 2 x 3+ 1= 7 T4 = 2 X 4+ 1 = 9 8 Hence L)2r + 1) = 7 + 9 + 11 + 13 + 15 + 17 r=3 This notation can be used quite easily to express a series where the terms alternate in signs. Jr+l =0 (-1)' Here the rth term is T, = - r+ 1 r starts at 0, 0 T,0 = (-1) = 1 0+ 1 Hence SERIES 41 The notation can also be used for a series that carries on indefinitely.