Using the property of determinants and without expanding in questions 1 to 7 prove that ,
`|{:(x,a,x+a),(y,b,y+b),(z,c,z+c):}|=0`

4 views

1 Answers

`L.H.S.=|{:(x,a,x+a),(y,b,y+b),(z,c,z+c):}|=|{:(x,a,0),(y,a,0),(z,c,0):}|`
`(C_(3)toC_(3)-C_(1)-C_(2))`
`=" "(therefore" all elements of "C_(3)" are zero")`
=R.H.S.

4 views

Related Questions