112. Let A be a 3 x 3 matrix.
now A-1 = 1|A| adj(A)
now adj(A-1) = adj (1|A| adj(A))
= 1|A|3-1 adj(adj A)
=1|A|2 adj(adj A) [because adj(kA) = kn-1 adj A where n is the order of the matrix A]
1.If A and B are square matrices of the same order and A is non singular then for a positive integer n, (A^-^1 BA)^n is equal to
(A)A^-^1 B^nA
(B)n(A^-^1 BA)
(C)A^-^n B^nA^n
(D) None of these
2.Prove that adj(A^-^1)= adj(adjA)/\left|a \right| ^2.
3.If A and B matrices commute then
(A)A-1 and B also commute
(B)B-1 and A also commute
(c)A-1 AND B-1 also commute
(D) all of the above
-
UP 0 DOWN 0 0 6
6 Answers
11
23hey 1st ques ans is A ?
and in 2nd one order of matrix shud hav been given as three , then only the ques is correct , or else if the question is for any general square matrix , then it shud hav been |a|n-1 instead of |a|2
23AB=BA
premultiply A-1
B=A-1BA
post multiply A-1
BA-1=A-1B
similarly AB-1 = B-1A
now post multiply A-1
AB-1A-1 = B-1
pre multiply A-1
so B-1 A-1 = A-1 B-1
hence D
1hey u r right but how u solve 1ST.PLEASE GIVE ME EXPLANATION.
23(A-1BA)n = A-1BA . A-1BA . A-1BA . A-1BA...........A-1BA
= A-1B(A A-1)B(A A-1)B(A A-1)B...........BA
=A-1BnA