InterviewSolution
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InterviewSolution
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सारणिकों के गुणधर्मों का प्रयोग करके निम्नलिखित को सिद्ध कीजिएः (i) `|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|` `=(a+b+c)^(3)` (ii) `|(x+y+2z,x,y),(z,y+z+2x,y),(z,x,z+x+2y)|` `=2(x+y+z)^(3)` |
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Answer» (i) L.H.S `=|(a-b-c,2a,2a),(2b,b-c-a,2b),(2c,2c,c-a-b)|` `=|(-(a+b+c),0,2a),(a+b+c,-(a+b+c),2b),(0,a+b+c,c-a-b)|` `(C_(1)toC_(1)-C_(2),C_(2)toC_(2)-C_(3))` `=(a+b+c)^(2)|(-1,0,2a),(1,-1,2b),(0,1,c-a-b)|` `=(a+b+c)^(2)|(-1,0,2a),(0,-1,2b+2a),(0,1,c-a-b)|` `(R_(2)toR_(2)+R_(1))` `=(a+b+c)^(2).(-1)|(-1,2b+2a),(1,c-a-b)|` (`C_(1)`से विस्तार करने पर) `=(a+b+c)^(2)(-a)(-c+a+b-2a-2b)` `=(a+b+c)^(2)(-1)(-a-b-c)` `=(a+b+c)^(2)(a+b+c)` `=(a+b+c)^(3)=R.H.S` (ii) `L.H.S=|(x+y+2z,x,y,),(z,y=z+2x,y),(z,x,z+x+2y)|` `=|(2x+2y+2z,x,y),(2x+2y+23z,y+z+2x,y),(2x+2y+2z,x,z+x+2y)|` `(C_(1)toC_(1)+C_(2)+C_(3))` `=(2x+2y+2z)|(1,x,y),(1,y+z+2x,y),(1,x,z+x+2y)|` `=2(x+y+z)|(1,x,y),(0,x+y+z,0),(0,0,x+y+z)|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `=2(x+y+z).1|(x+h+z,0),(0,x+y+z)|` (`C_(1)`से विस्तार करने पर) `=2(x+y+z)^(3)=R.H.S` |
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