Consider the linear system

x+3y+kz=-5

kx+(3+3k)y+(6+k^2)z=-11

x+4y+k^2z=k-4

For which values of k does this system have one solution?

### 1 Answer | Add Yours

`A=[[1,3,k],[k,3+3k,6+k^2],[1,4,k^2]]`

det(A)=k^2-k-2

det(A)=(k-2)(k+1)

det(A)`!=0 ` if (k-2)(k+1)`!=0 `

`i.e x!=-1,2`

so rank of A =3

and rank of augument matrix =3 for all values of k.

if rank (A)=rank(A|b)=3 ,then system has unique solution .

system has unique solution for all values of k except -1,2 .

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