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Using the property of determinants and without expanding in questions 1 to 7 prove that , `|{:(a-b,b-c,c-a),(b-c,c-a,a-b),(c-a,a-b,b-c):}|=0` |
Answer» `L.H.S.=|{:(a-b,b-c,c-a),(b-c,c-a,a-b),(c-a,a-b,b-c):}|=|{:(0,b-c,c-a),(0,c-a,a-b),(0,a-b,b-c):}|` `((C_(1)toC_(1)+C_(2)+C_(3))` `=0 " "(because" all elements of "C_(1)" are zero)` =R.H.S. |
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