### 2 Answers | Add Yours

The augmented matrix has no trivial solution. ie.

1 c-1 1

D= 0 c 4

1 c c+1

By elementry row operation

1 c-1 1

D = 0 c 4

0 1 1

If we expand this then det (d)=C^2-4

for non trivial solution we have Det(D)=0

c=2,-2

If the augmented matrix has no trivial solution then we cannot find the value of c in the real field. As it is coming c^2+4=0 i.e. c=+-(2i).

If we prepare the augmented matrix and reduce by elementary operations we get 4-c^2=0 or c=+2 and c=-2.

For above values of c, the system has no trivial solutions. As we are getting a 2by2 matrix for three variables.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes