([1x,1,0,b+c],[1,b,c+a],[1,c,a+b]|

If D=[[1,a,b],[1,b,c],[1,c,a]] , then [[a,b,c],[b,c,a],[1,1,1]] =

|{:(1,a,b+c),(1,b,c+a),(1,c,a+b):}|=0

Prove that |[1+a,1, 1], [1,1+b,1], [1, 1, 1+c]|=a b c(1+1/a+1/b+1/c)=a b c+b c+c a+a b

prove |[1, b c, a(b+c)],[ 1, c a, b(c+a)],[ 1, a b, c(a+b)]|=0

Show that |[1,b c, a(b+c)],[1,c a, b(c+a)],[1,a b ,c(a+b)]|=0 .

|{:(a-b,1,a),(b-c,1,b),(c-a,1,c):}|=|{:(a,1,b),(b,1,c),(c,1,a):}|

Using the property of determinants and without expanding, prove that: |[1,b c, a(b+c)],[1,c a, b(c+a)],[1,a b, c(a+b)]|=0

If A=[[1,0,1],[0,2,0],[1,-1,4]],A=B+C,B=B^(T) and C=-C^(T) , then C=

|{:(b+c,a,1),(c+a,b,1),(a+b,c,1):}|=0

If S=[(0,1,1),(1,0,1),(1,1,0)]a n dA=[(b+c,c+a,b-c),(c-b,c+b, a-b),(b-c,a-c,a+b)](a ,b ,c!=0),t h e nS A S^(-1) is