ಮಾತೃಕೆಗಳು, A ,B ,(A+B)\n ಏಕವಚನವಲ್ಲದಿದ್ದರೆ, [A(A+B)^(-1)B]^(-1)=B^(-1)+A ಎಂದು ಸಾಬೀತುಪಡಿಸಿ ^(-1...

If the matrices, A, B and (A+B) are non-singular, then prove that [A(A+B)^(-1) B]^(-1) =B^(-1)+A^(-1) .

Statement 1: If the matrices,A,B,(A+B) are non-singular,then [A(A+B)^(-1)B]^(-1)=B^(-1)+A^(-1). Statement 2:[A(A+B)^(-1)B]^(-1)=[A(A^(-1)+B^(-1))B]^(-1)=[(I+AB^(-1))B]^(-1)=[(B+AB^(-1))B]^(-1)=[(B+AI)]^(-1)=[(B+A)]^(-1)=B^(-1)+A^(-1)

For non-singular square matrix A ,\ B\ a n d\ C of the same order (A B^(-1)C)^(-1)= A^(-1)B C^(-1) (b) C^(-1)B^(-1)A^(-1) (c) C B A^(-1) (d) C^(-1)\ B\ A^(-1)

If A=[1-12-1],B=[a1b-1] and (A+B)^(2)=A^(2)+B^(2), find a and b

A=[[1,-12,-1]],B=[[a,1b,-1]] and (A+B)^(2)=A^(2)+B^(2), find a and b

(1) ((1-cos B)(1+cos B))/((1-sin B)(1+sin B))=(1)/(3)

i) If tan (A+B) = n tan (A-B), prove that: (n+1)/(n-1) = (sin2A)/(sin2B)

In the arithmetic mean of a and b is (a^(n)+b^(n))/(a^(n-1)+b^(n-1)) then n= ?

If tan(A+B)=ntan(A_B) then prove that (n+1)/(n-1)=(sin2A)/(sin2B)

If the total number of relations form set A to set B is 1024, and n(A)gt1, n(B)gt1, then (A) n(A)=2, n(B)=5 (B) n(A)=5, n(B)=2 (C) n(A)=10, n(B)=1 (D) n(A)=1, n(B)=10