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Text Solution

2`1/2`3`1/3`

Answer :

DSolution :

Let ` B = [[alpha_(1),beta_(1) ,gamma_(1)],[alpha_(2),beta_(2) ,gamma_(2)],[alpha_(3),beta_(3) ,gamma_(3)]]` <br> `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)]]` <br> `rArr AC_(1) = [[alpha_(1)],[alpha_(1)+alpha_(2)],[3alpha_(1)+alpha_(2)+alpha_(3)]] = [[1],[0],[0]]` <br> `rArr alpha _(1) = 1 , alpha _(2) = -2, alpha _(3) = 1` <br> `rArr AC_(2) = [[beta_(1)],[2beta_(1)+beta_(2)],[3beta_(1)+2beta(2)+beta_(3)]] = [[2],[3],[0]]` <br> `beta _(1) = 2, beta_(2) = -1, beta_(3) = -4` <br> and `AC_(3) = [[gamma_(1)],[2gamma_(1)+gamma_(2)],[3gamma_(1)+2gamma_(2)+gamma_(3)]] = [[2],[3],[1]]` <br> `rArr gamma_(1)= 2, gamma_(2) = -1, gamma_(3) = -3` <br> `therefore B = [[1,2,2],[-2,-1,-1],[1,-4,-3]]` <br> `rArr det B = [[1,2,2],[-2,-1,-1],[1,-4,-3]]` <br> ` = 1 (3-4) - 2 (6+1) + 2(8+1) = 3` <br> and `C=1/3 [[1,0,0],[2,1,0],[3,2,1]] [[1, 2, 2 ],[-2,-1, -1],[1, -4, -3]]` <br> `= 1/ 3 [[1,2,2],[0,3,3],[0,0,1]]` <br> `therefore det C= abs[[1/3,2/3,2/3],[0,1,1],[0,0,1/3]] = 1/9` <br> `det (B^(-1)) = 1/(det B) = 1/3`