13.quad [3a,a+b,a+1],[-b+a,3b,b+c],[-c+a,-c+b,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. Prove that |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]|=3(a+b+c)(a b+b c+c a)

Show that: |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]|=3(a+b+c)(a b+b c+c a)dot

Show that: |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]|=3(a+b+c)(a b+b c+c a) .

Show that: |[3a,-a+b,-a+c],[-b+a,3b,-b+c],[-c+a,-c+b,3c]|=3(a+b+c)(a b+b c+c a)dot

Using properties of determinants. Prove that |[3a,-a+b,-a+c], [-b+a,3b,-b+c], [-c+a,-c+b,3c]|=3(a+b+c)(a b+b c+c a)

Show 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)

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