If one of the eigenvalues of a square matrix a order `3xx3` is zero, then prove that det `A=0`.

Explanation

Let `A=[(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))]`
`implies A-lambdaI=[(a_(1)-lambda,b_(1),c_(1)),(a_(2),b_(2)-lambda,c_(2)),(a_(3),b_(3),c_(3)-lambda)]`
`implies` det. `(A-lambdaI)=(a_(1)-lambda)[(b_(2)-lambda)(c_(3)-lambda)-b_(3)c_(2)]`
`-b_(1) [a_(2) (c_(3)-lambda)-c_(2)a_(3)]+c_(1)[a_(2)b_(3)-a_(3)(b_(2)-lambda)]`
Now if one of the eigenvalues is zero, one root of `lambda` should be zero. Therefore, constant term in the above polynominal is zero.
`implies a_(1) b_(2)c_(3)-a_(1)b_(3)c_(2)-b_(1)a_(2)c_(3)+b_(1)c_(2)a_(3)+a_(1)a_(2)a_(3)-c_(1)a_(3)b_(2)=0`
But in above equation L.H.S. is the value of determinant of A. Hence,
det. `A=0`