if `A=[{:(0,1,-1),(2,1,3),(3,2,1):}]`then (A(adjA)`A^-1`)A is equal to

Options

A. `2{:[(3,0,0),(0,3,0),(0,0,3)]:}`
B. `{:[(-6,0,0),(0,-6,0),(0,0,-6)]:}`
C. `{:[(0,1//6,-1//6),(2//6,1//6,3//6),(3//6,2//6,1//6)]:}`
D. none of these

Answer

A

Explanation

We have,
`{:absA=[(0,1,-1),(2,1,3),(3,2,1)]:}=0+7-1=6`
`:. (A(adj A)A^(-1))A=(A(adjA))(A^(-1)A)`
`rArr(A(adj A)A^(-1))A=(absAI)I`
`rArr(A(adj A)A^(-1))A=absAI={:[(6,0,0),(0,6,0),(0,0,6)]=2[(3,0,0),(0,3,0),(0,0,3)]:}`
Hence, option (a) is correct.