Then l0 l1 l2 0, otherwise 0 will be the solution of Ax 0, what does the contradiction mean

1) Here 0 is a special solution of the non-homogeneous linear equation ax=b, and b is not 0 (otherwise, it is a homogeneous linear equation).

0 is not a solution for ax=0.

2)(l0+l1+l2) 0+l1 1+l2 2=0, a[(l0+l1+l2) 0+l1 1+l2 2] = a0=0 .

And 1, 2 is a basic solution system of the derived group ax=0: a[l1 1+l2 2]=al1 1+al2 2=l1a 1+l2a 2=0

Substituting (*) formula:

a(l0+l1+l2) 0=0=(l0+l1+l2)a 00 is not a solution of ax=0, only l0+l1+l2=0 thank you].

If l0+l1+l2 is not 0, then 0 can be a linear combination of 1 and 2 ( 0=-l1 (l0+l1+l2) 1-l2 (l0+l1+l2) 2), then 0 is the solution of ax=0, but 0 is the solution of ax=b, which is contradictory (b is not 0, because ax=b is a system of nonhomogeneous linear equations).

then l0+l1+l2=0, otherwise 0 will be the solution of ax=0, contradictory. So l1 1+l2 2=00 is not a solution of ax=0 2)(l0+l1+l2) 0+l1 1+l2 2=0, a

At. Standard normal distribution.

, the graph is about x

Axial symmetry. When the comma is taken to zero, the probability of a good branch is 1 2, which is half the area of the whole graph.

Trouble, thanks!

1) Here 0 is a special solution of the non-homogeneous linear equation ax=b, and b is not 0 (otherwise, it is a homogeneous linear equation).

0 is not a solution for ax=0.

2)(l0+l1+l2) 0+l1 1+l2 2=0, a[(l0+l1+l2) 0+l1 1+l2 2] = a0=0 .

And 1, 2 is a basic solution system of the derived group ax=0: a[l1 1+l2 2]=al1 1+al2 2=l1a 1+l2a 2=0

Substituting (*) formula:

a(l0+l1+l2) 0=0=(l0+l1+l2)a 0 0 is not a solution of ax=0, only l0+l1+l2=0

l2x-1l=-a

According to the title. a<0

a>0 is what you want.

If l0+l1+l2 is not 0, then 0 can be a linear combination of 1 and 2 ( 0=-l1 (l0+l1+l2) 1-l2 (l0+l1+l2) 2), then 0 is the letting solution of ax=0, but 0 is the solution of ax=b, which is contradictory (b is not a segment of sliding 0, because ax=b is a non-homogeneous linear square grip tancheng group).