Let for `A=[(1,0,0),(2,1,0),(3,2,1)]`, there be three row matrices `R_(1), R_(2)` and `R_(3)`, satifying the relations, `R_(1)A=[(1,0,0)], R_(2)A=[(2,3,0)]` and `R_(3)A=[(2,3,1)]`. If B is square matrix of order 3 with rows `R_(1), R_(2)` and `R_(3)` in order, then
The value of det. `(2A^(100) B^(3)-A^(99) B^(4))` is

Options

A. `-2`
B. `-1`
C. 2
D. -27

Answer

D

Explanation

`overset(B)([(-,R_(1),-),(-,R_(2),-),(-,R_(3),-)])overset(A)([(1,0,0),(2,1,0),(3,2,1)])=overset(C)([(1,0,0),(2,3,0),(2,3,1)])` (1)
`:.` (det. B) (det. A)=3
`:.` (det. B)=3 [as det. A=1]
det. `(2A^(100)B^(3)-A^(99)B^(4))`
= det. `(A^(99) (2A-B)B^(3))`
`=("det. A")^(99)xxdet. (2A-B)xx("det B")^(3)`
Now from (1), we get
`B=A^(-1) C=[(1,0,0),(-2,1,0),(1,-2,1)][(1,0,0),(2,3,0),(2,3,1)]`
`=[(1,0,0),(0,3,0),(-1,-3,1)]`
`:. 2A-B=[(1,0,0),(4,-1,0),(7,7,1)]`
`:.` det. `(2A-B)=-1`
`:.` det. `(2A^(100) B^(3)-A^(99) B^(4))=(1)^(99) (-1) (3)^(3)=-27`