सिद्ध करे कि ` |{:(3a,b-a,c-a),(a-b,3b,c-b),(a-c,b-c,3c):}|= 3 (a+b+c) (ab+bc+ca)`

|{:(" "3a," "-a+b,-a+c),(a-b," "3b," "c-b),(a-c," "b-c," "3c):}|=3(a+b+c)(ab+bc+ca)

Prove that |(3a, -a+b, -a+c),(a-b, 3b, c-b),(a-c, b-c, 3c)| = 3(a+b+c)(ab+bc+ca) .

{:|( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) |:} = 3( a+b+c) ( ab+bc+ca)

Using properties of determinants, prove the following |(3a,-a+b,-a+c),(a-b,3b,c-a),(a-c,b-c,3c)|=3(a+b+c)(ab+bc+ca)

3a,-a+b,-a+ca-b,3b,c-ba-b,3b,c-ba-c,b-c,3c]|=

Show that |(3a,-a+b,-a+c),(-b+a,3b,-b+c),(-c+a,-c+b,3c)|=3(a+b+c)(ab+bc+ca)

Using properties of determinants in Exercise 11 to 15 prove that |{:(3a,-a+b,-a+c),(-b+a,3b,-b+c),(-c+a,-c+b,3c):}|=3(a+b+c)(ab+bc+ca)

solve {:[( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) ]:} = 3( a+b+c) ( ab+bc+ca)

Prove {:[( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) ]:} = 3( a+b+c) ( ab+bc+ca)

Using the property of determinants prove that {:|( 3a,-a+b,-a+c),( -b+a, 3b,-b+c) ,( -c+a,-c+b,3c) |:} = 3( a+b+c) ( ab+bc+ca)