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LetA= [[1,0,0],[2,1,0],[3,2,1]] be a squ...

Let`A= [[1,0,0],[2,1,0],[3,2,1]]` be a square matrix and ` C_(1), C_(2), C_(3)` be three
column matrices satisfying `AC_(1) = [[1],[0],[0]], AC_(2) = [[2],[3],[0]] and AC_(3)= [[2],[3],[1]]` of matrix B. If the matrix `C= 1/3 (AcdotB).`
The value of `det(B^(-1))`, is

A

2

B

`1/2`

C

3

D

`1/3`

Text Solution

Verified by Experts

The correct Answer is:
D
Let ` B = [[alpha_(1),beta_(1) ,gamma_(1)],[alpha_(2),beta_(2) ,gamma_(2)],[alpha_(3),beta_(3) ,gamma_(3)]]`
`therefore C_(1)=[[alpha_(1)],[alpha_(2)],[alpha_(3)]], C_(2) = [[beta_(1)],[beta_(2)],[beta_(3)]]and C_(3) = [[gamma_(1)],[gamma_(2)],[gamma_(3)]]`
`rArr AC_(1) = [[alpha_(1)],[alpha_(1)+alpha_(2)],[3alpha_(1)+alpha_(2)+alpha_(3)]] = [[1],[0],[0]]`
`rArr alpha _(1) = 1 , alpha _(2) = -2, alpha _(3) = 1`
`rArr AC_(2) = [[beta_(1)],[2beta_(1)+beta_(2)],[3beta_(1)+2beta(2)+beta_(3)]] = [[2],[3],[0]]`
`beta _(1) = 2, beta_(2) = -1, beta_(3) = -4`
and `AC_(3) = [[gamma_(1)],[2gamma_(1)+gamma_(2)],[3gamma_(1)+2gamma_(2)+gamma_(3)]] = [[2],[3],[1]]`
`rArr gamma_(1)= 2, gamma_(2) = -1, gamma_(3) = -3`
`therefore B = [[1,2,2],[-2,-1,-1],[1,-4,-3]]`
`rArr det B = [[1,2,2],[-2,-1,-1],[1,-4,-3]]`
` = 1 (3-4) - 2 (6+1) + 2(8+1) = 3`
and `C=1/3 [[1,0,0],[2,1,0],[3,2,1]] [[1, 2, 2 ],[-2,-1, -1],[1, -4, -3]]`
`= 1/ 3 [[1,2,2],[0,3,3],[0,0,1]]`
`therefore det C= abs[[1/3,2/3,2/3],[0,1,1],[0,0,1/3]] = 1/9`
`det (B^(-1)) = 1/(det B) = 1/3`
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