If a, b, c≠0 and a + b + c = 0, then the matrix 1+1a1111+1b1111+1c is

1+1/a1111+1/b1111+1/c=1abca+1bcab+1cabc+1Using C1→C1+C2+C3, we get|A|=1abc(a+b+c+1)1bc1b+1c1bc+1Using R2→R2-R1, R3→R3-R1, we get|A|=1abc(a+b+c+1)1bc010001      =1abc(a+b+c+1)≠0