Constrained positive deÖnite matrixes
But we must calculate 3 1 =2 determinants: Deleting the last row and
the last columndet0
@
0 1 1
1 2 0
1 0 2
1
A= 4.
As both determinants are negative,p= 1 ,so the sign for the minimum is
( 1 )p= 1 this is a local minimum.