SOLUTION: Please help me solve this problem. Determine all the value of k such that the following system of linear equations has (i) Unique solution (ii) No solution and (iii) More than o

Calculate the determinant of the matrix A = .

If you are unfamiliar with it or want to refresh your knowledge, see the lesson

    - Determinant of a 3x3 matrix 

in this site.


The determinant is equal to  = .

The general theory says:


     If the determinant is not zero, then the system has a unique solution.

     The degenerate cases are those when the determinant is zero:

     d(k) =  = 0.


One root of this equation is k = 1.

Hence, the polynomial is multiple of (k-1), and the ratio  is .

Then the two other roots of the determinant are the zeroes of this quadratic polynomial . 

You can easily find these zeroes using the quadratic formula:

 =   and   = .


Next, if k = 1, your original system consists of three identical equations.
Hence, the system has infinitely many solutions at k = 1.

If k =  or k = , then the system has infinitely many solutions, too.

   Indeed, if you add all three equations at these values of k, you will get the equation 

   (k+2)x + (k+2)y + (k+2)z = 3,  or

   x + y + z = .   (*)

   Now distract      the equation (1) from (*), and you will find the solution for x.
   Distract then     the equation (2) from (*), and you will find the solution for y.
   Finally, distract the equation (3) from (*), and you will find the solution for z.

   Thus the system has at least one solution for k =  and/or k = , having the determinant equal to zero at these values of k.
    It is enough for the system to have infinitely many solutions then.