If `A=[(3,1),(-1,2)]` then `A^2=` (A) `[(8,-5),(-5,3)]` (B) `[(8,-5),(5,3)]` (C) `[(8,-5),(-5,-3)]` (D) `[(8,5),(-5,3)]`

The adjoint of the matrix [(2,5),(3,1)] is (A) [(2,5),(3,1)] (B) [(1,-5),(-3,2)] (C) [(1,-3),(-5,2)] (D) [(-1,3),(5,-2)]

Simplify : (i) (5/8)^(-7)×\ (8/5)^(-5) (ii) ((-2)/3)^(-2)×\ (4/5)^(-3)

If 2(2n+5)=3\ (3n-10), then n= (a)5 (b) 3 (c)7 (d) 8

If A=[{:(,2,-1),(,3,4):}]=B=[{:(,5,2),(,7,4):}],C=[{:(,2,5),(,3,8):}] and AB-CD=0 find D.

Solve for X : [(1,2,-3),(2,1,3)]-X=[(5,1,8),(-6,0,5)] .

If C= [{:(,1,4,6),(,7,2,5),(,9,8,3):}] [{:(,0,2,3),(,-2,0,4),(,-3,-4,0):}] [{:(,1,7,9),(,4,2,8),(,6,5,3):}] Then trace of C+C^(3)+C^(5)+……+C^(99) is

(2/3)^(-5) is equal to (a) ((-2)/3)^5 (b) (3/2)^5 (c) (2^-5)/3 (d) 2/(3^5)

((-1)/5)^3-:((-1)/5)^8 is equal to (a) (-1/5)^5 (b) (-1/5)^(11) (c) (-5)^5 (d) (1/5)^5

Simplify and write the answer in the exponential form. (i) (2^5 -: 2^8)^5 xx 2^(-5) (ii) (-4)^(-3) xx (5)^(-3) xx (-5)^(-3) (iii) 1/8 xx (3)^(-3) (iv) (-3)64 xx (5/3)^4

Simplify (i) {(1/3)^(-2) - (1/2)^(-3)} -: (1/4)^(-2). (ii) (5/8)^(-7) xx (8/5)^(-5).