If`|(a,1,1),(1,b,1),(1,1,c)|=0` then show that `1/(1-a)+1/(1-b)+1/(1-c)=1`

The inverse of the matrix [(1,1,1),(1,0,2),(3,1,1)] is a) (1)/(4)[(-2,0,2),(5,-1,2),(1,-1,-2)] b) (1)/(4)[(-2,0,2),(5,-2,-1),(1,2,-1)] c) (1)/(4)[(-2,0,2),(2,5,-1),(-2,-1,1)] d) (1)/(4)[(-2,0,2),(5,-1,1),(1,-2,-1)]

If A=[(a,0),(0,(1)/(b))] , then A^(-1) =

If A=[(1,-1,1),(0,2,-3),(2,1,0)],B=(adjA) and C=5A, then (|adjB|)/(|C|)=

If the product of the matrix B=[(2,6,4),(1,0,1),(-1,1,-1)] with a matrix A has inverse C=[(-1,0,1),(1,1,3),(2,0,2)] then A^-1=

If A[(1,2,2),(-2,-1,-1),(1,-4,-4)] , then the sum of the elements of A^(-1) is...a)0 b)1 c)-1 d) Cannot be found

The inverse of the matrix A[(1,1,1),(6,7,8),(6,7,-8)] using adjoint method is ...a) -(1)/(16)[(-112,1,15),(96,-14,-2),(0,-1,1)] b) -(1)/(16)[(-112,15,1),(96,-2,-14),(0,-1,1)] c) -(1)/(16)[(-112,15,1),(96,-14,-2),(0,-1,1)] d) -(1)/(16)[(112,15,1),(96,-14,-2),(-1,0,1)]

If A=[(1,-1,0),(1,0,0),(0,0,-1)] , then A^(-1) is...a) A^T b) A^2 c) A d) I

A=[(2,-2),(-2,2)], B=[(1,1),(1,1)] then a) A^(-1)=B b) B^(-1) does not exist c) A^(-1) does not exist d)Both (b) and (c)

If {:A=[(a,0,0),(0,b,0),(0,0,c)]:}," then "A^(-1) , is..... a) [(a,0,0),(0,(1)/(b),0),(0,0,c)] b) [((1)/(0),0,0),(0,(1)/(b),0),(0,0,c)] c) [(a,0,0),(0,(1)/(b),0),(0,0,(1)/(c))] d) [((1)/(a),0,0),(0,(1)/(b),0),(0,0,(1)/(c))]