hey how are you getting rank(A|B)=rank(A)??

6 votes

The equations.

- $x_{1}+2x_{2}+3x_{3}=1$
- $x_{1}+4x_{2}+9x_{3}=1$
- $x_{1}+8x_{2}+27x_{3}=1$

have

- Only one solution
- Two solutions
- Infinitely many solutions
- No solutions

7 votes

Best answer

This is non homogeneous equation ...for such type of questions we have to check following 3 cases :

1)if rank(A)<rank(A|B) then AX=B has no solution

2)if rank(A|B)=rank(A)=no of unknowns then AX=B has unique non-zero solution

3)rank(A|B)=rank(A)<no of unknowns then infinite no of solution

here I'm getting rank(A|B)=rank(A)=no of unknowns =3 so only one solution..

1)if rank(A)<rank(A|B) then AX=B has no solution

2)if rank(A|B)=rank(A)=no of unknowns then AX=B has unique non-zero solution

3)rank(A|B)=rank(A)<no of unknowns then infinite no of solution

here I'm getting rank(A|B)=rank(A)=no of unknowns =3 so only one solution..