Assignment. 1. For each of the following matrices find

Added on - 22 Sep 2019

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Assignment1. For each of the following matrices find the multiplicative inverse, or explain why itdoesn’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)=1Minors 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 lettersa b c d e f g h i jso 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 = 0And 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, . . .