Assignment. 1. For each of the following matrices find
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Assignment1. For each of the following matrices find the multiplicative inverse, or explain why it doesn’t have one.(a)[3523]Solution: A= [3523]|A| = 3*3 – 2*5 = 9-10= -1Minors areM11 = 3, M12=2, M21=5, M22=3Co factors of matrix areC11 = C11C12 = -C12C21 = -C21C22 = C22Cofactors Matrix [3−2−53]A-1 = 1/|A| * (Cofactor)TA-1 = 1/-1 * [3−5−23]A-1 = [−352−3](b)[236324669]Solution: A= [236324669]|A| = 2*(2*9-6*4)– 3*(3*9-6*4) +6*(3*6-6*2) = 2*(-6)- 3*(3) +6*(6) = 15Minors areM11=-6, M12=3, M13=6,M21=-9, M22=-18, M23=-6,M31=0, M32=-10, M33=-2
Cofactors Matrix [−6−369−186010−2]A-1 = 1/|A| * (Cofactor)TA-1 = 1/15 * [−690−3−181066−2]A-1 = [−2/5−3/50−1/5−6/52/32/55/2−2/15](c)[1257013600140001]Solution: A=[1257013600140001]|A| = 1*(1*1-3*0-0*4)– 2*(0*0-3*0+6*0) +5*(0*0-1*0+6*0)-7*(0*0-1*0+3*0) = 1 Minors areM11=1, M12=0, M13=0, M14=0,M21=2, M22=1, M23=0, M24=0,M31=1, M32=3, M33=1, M34=0,M41=-1, M42=6, M43=4, M44=1,Cofactors Matrix [1000−21001−31016−41]A-1 = 1/|A| * (Cofactor)T
A-1 = 1/1 * [1−21101−36001−40001]A-1 = [1−21101−36001−40001]2. How many ways can we rearrange the letters a b c d e f g h i j so that no vowel endsup in the position where it began?Solution: n=10 (a, b, c, d, e, f, g, h, i, j) Vowels a, e, i (3 vowels)No. of Ways= 2419143. Find a closed form for the generating function for each of these sequences.(a) 7, 3, 4, 6, 7, 3, 4, 6, 7, 3, 4, 6, . . .Solution: 1 mod 4 = 12 mod 4 = 23 mod 4 = 34 mod 4 = 05 mod 4 = 16 mod 4 = 27 mod 4 = 38 mod 4 = 0 And so on...F (n) = {ifnmod4=1,7ifnmod4=2,3ifnmod4=3,3ifnmod4=0,6} n E |N(b) .1, 0.01, 0.001, 0.0001, . . . Solution: a2/a1= a3/a2=...... an/an-1= 1/10a1= 1/10f(n) = an/an-1= 1/10; n≥2; n E |N; a1=1/10(c) 2, 5, 8, 11, 14, 17, 20, . . .
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