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If A^(-1)=[[1,-1, 2],[0, 3,1],[ 0 ,0,-1/...

If `A^(-1)=[[1,-1, 2],[0, 3,1],[ 0 ,0,-1/3]]` , then `|A|=-1` b. `adj A=[[-1, 1 ,-2],[ 0,-3,-1],[ 0, 0, 1/3]]` c. `A=[[1, 1/3, 7 ],[0, 1/3, 1],[0 ,0,-3]]` d. `A =[[1,-1/3,-7],[ 0,-3, 0],[ 0, 0, 1]]`

A

`|A|=-1`

B

adj `A=[(-1,1,-2),(0,-3,-1),(0,0,1//3)]`

C

`A=[(1,1//3,7),(0,1//3,1),(0,0,-3)]`

D

`A=[(1,-1//3,-7),(0,-3,0),(0,0,1)]`

Text Solution

Verified by Experts

`|A^(-1)|=-1 implies |A|=-1`
Now, use adj `A=|A|A^(-1)` and `A=(A^(-1))^(-1)`
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