Usingproperties of determinants, prove the following:`|3"a"-"a"+"b"-"a"+"c""a"-"b"3"b""c"-"b""a"-"c""b"-"c"3"c"|=3("a"+"b"+"c")("a b"+"b c"+"c a")`

Usingproperties of determinants, prove the following:`|3"a"-"a"+"b"-"a

1.	Usingproperties of determinants, prove the following:`\|3"a"-"a"+"b"-"a"+"c""a"-"b"3"b""c"-"b""a"-"c""b"-"c"3"c"\|=3("a"+"b"+"c")("a b"+"b c"+"c a")`
Answer» Let the value of the given determinant be `Delta`. Then, `Delta =\|{:(" "3a,-a+b, -a+c),(a-b, " "3b, c-b),(a-c, b-c, 3c):}\|` `=\|{:(a+b+c,-a+b, -a+c),(a+b+c, " "3b, c-b),(a+b+c, b-c, " "3c):}\| [C_(1) to (C_(1) + C_(2) + C_(3))]` `=(a+b+c)\|{:(1,-a+b, -a+c),(1, " "3b, c-b),(1, b-c, 3c):}\| ["taking (a+b+c) common from"R_(1)]` `=(a+b+c)\|{:(1,-a+b, -a+c),(1, 2b+a, a-b),(0, a-c, 2c+a):}\| [R_(2) to (R_(2) -R_(1)) "and"(R_(3) -R_(1))]` `=(a+b+c)1\|{:(2b+a, a-b),(a-c, 2c+a):}\| ["expanded by"C_(1)]` ` = (a+b+c)[(2b+a)(2c+a)-(a-c)(a-b)]` ` =(a+b+c)[(4bc+2ab+2ac+a^(2))-(a^(2)-ab-ac +bc)]` `=3(a+b+c)(ab+bc+ca).` Hence, `Delta = 3(a+b+c) (ab+bc+ca)`