1. A3(A3+A2)-1+A2(A2+A)-1+2(A3+A2)-1A4
Question see also:
https://web2.0calc.com/questions/matrix-problem-simplify-as-little-as-possible-each
1.
A3(A3+A2)−1=A3[AA2+IA2]−1|Identity matrix I =A3[(A+I)A2]−1|Formula:(AB)−1=B−1A−1=A3[(A2)−1(A+I)−1]|(A2)−1=(AA)−1=A−1A−1=(A−1)2=A3(A−1)2(A+I)−1=AA2(A−1)2(A+I)−1|A2(A−1)2=I=AI(A+I)−1|AI=A=A(A+I)−1
2.
A2(A2+A)−1=A2(AA+IA)−1|Identity matrix I =A2[(A+I)A]−1|Formula:(AB)−1=B−1A−1=A2[A−1(A+I)−1]=AAA−1(A+I)−1|AA−1=I=AI(A+I)−1|AI=A=A(A+I)−1
3.
2(A3+A2)−1A4=2[A2A+A2I]−1A4|Identity matrix I =2[A2(A+I)]−1A4|Formula:(AB)−1=B−1A−1=2[(A+I)−1(A2)−1]A4|(A2)−1=(AA)−1=A−1A−1=(A−1)2=2(A+I)−1(A−1)2A4=2(A+I)−1(A−1)2A2A2|(A−1)2A2=I=2(A+I)−1IA2|IA2=A2=2(A+I)−1A2
summary
A3(A3+A2)−1+A2(A2+A)−1+2(A3+A2)−1A4=A(A+I)−1+A(A+I)−1+2(A+I)−1A2=2A(A+I)−1+2(A+I)−1A2=2A(A+I)−1I+2(A+I)−1A2|I=A−1A=2A(A+I)−1A−1A+2(A+I)−1A2=[A(A+I)−1A−1+(A+I)−1A] 2A|A(A+I)−1=[(A+I)A−1]−1={[(A+I)A−1]−1A−1+(A+I)−1A} 2A|(A+I)−1A=[A−1(A+I)]−1={[(A+I)A−1]−1A−1+[A−1(A+I)]−1} 2A=[(AA−1+IA−1)−1A−1+(A−1A+A−1I)−1] 2A|AA−1=A−1A=IIA−1=A−1I=A−1=[(I+A−1)−1A−1+(I+A−1)−1] 2A=[(I+A−1)−1(A−1+I)] 2A=[(I+A−1)−1(I+A−1)] 2A|(I+A−1)−1(I+A−1)=I=I2A=2A
2. -2 A-2(A-2+A-3)-1+2A3(A2+A3)-1+ 3/4 A(A-1+I)-1+ 3/4 A3(A+I)-1
−2A−2(A−2+A−3)−1+2A3(A2+A3)−1+34A(A−1+I)−1+34A3(A+I)−1= ?
1.
−2A−2(A−2+A−3)−1+2A3(A2+A3)−1(A−2+A−3)−1=[(A−1+I)A−2]−1(A2+A3)−1=[(A+I)A2]−1=−2A−2[(A−1+I)A−2]−1+2A3[(A+I)A2]−1|A−2=A−1A−1=(AA)−1=(A2)−1=−2A−2[(A−1+I)(A2)−1]−1+2A3[(A+I)A2]−1|[(A−1+I)(A2)−1]−1=[A2(A−1+I)−1]=−2A−2[A2(A−1+I)−1]+2A3[(A+I)A2]−1=−2A−2A2(A−1+I)−1+2A3[(A+I)A2]−1|A−2A2=A−1A−1AA=A−1(A−1A)A=A−1(I)A=A−1A=I=−2I(A−1+I)−1+2A3[(A+I)A2]−1=−2(A−1+I)−1+2A3[(A+I)A2]−1|[(A+I)A2]−1=[(A2)−1(A+I)−1]=−2(A−1+I)−1+2A3[(A2)−1(A+I)−1]|(A2)−1=A−2=−2(A−1+I)−1+2A3A−2(A+I)−1=−2(A−1+I)−1+2AA2A−2(A+I)−1|A2A−2=I=−2(A−1+I)−1+2AI(A+I)−1=−2(A−1+I)−1+2A(A+I)−1|A(A+I)−1=[(A+I)A−1]−1=−2(A−1+I)−1+2[(A+I)A−1]−1=−2(A−1+I)−1+2(AA−1+IA−1)−1=−2(A−1+I)−1+2(I+A−1)−1=−2(A−1+I)−1+2(A−1+I)−1=0−2A−2(A−2+A−3)−1+2A3(A2+A3)−1=0
2.
−2A−2(A−2+A−3)−1+2A3(A2+A3)−1=−2(A−1+I)−1+2A(A+I)−1=0−2(A−1+I)−1+2A(A+I)−1=02(A−1+I)−1=2A(A+I)−1(A−1+I)−1=A(A+I)−1
3.
−2A−2(A−2+A−3)−1+2A3(A2+A3)−1+34A(A−1+I)−1+34A3(A+I)−1=0+34A(A−1+I)−1+34A3(A+I)−1=34A(A−1+I)−1⏟=A(A+I)−1+34A3(A+I)−1=34AA(A+I)−1+34A3(A+I)−1=34A2(A+I)−1+34A3(A+I)−1=34(A2+A3)(A+I)−1=34A2(I+A)(A+I)−1=34A2(I+A)(I+A)−1⏟=I=34A2I=34A2