Consider the following system...Lin Alg x + y -z = 2 x + 2y + z = 3 x + y + (k 2

Question

Consider the following system...Lin Alg

x + y -z = 2

x + 2y + z = 3

x + y + (k2-5)z = k

For what values of k does the system have a unique solution. For what values of k does the system have infinitely many solutions? For what values of k does the system have no solutions?

I have tried making an augmented matrix with this problem and taking the rref of that matrix to obtain a unique solution. What happens however, is that I obtain values of x, y, and z that when plugged back into any three of the equations, will satisfy the equations but eliminate the variable k. For example, if I plug my values of x, y, and z into equation 1, I get 2 = 2.

My values:

x = 1 + (3/(k+2))

y = 1 - (2/(k+2))

z = 1 / (k+2)

Where am I going wrong?

Explanation / Answer

This is very simple. You can never solve them because you are dealing in 4 variable x,y,z and k.

So,this can only be done by simple procedure.

The system has no solutions when atleast 2 of these 3 equations have same coefficients of x,y and z but the constant term differs.

The system has infinite solutions when atleast 2 of these 3 equations have same coefficients of x,y,z and also the constant term.

For the given system,

x + y -z = 2 -----------------(1)

x + 2y + z = 3 -------------(2)

x + y + (k2-5)z = k ----------------(3)

let us equalize the co-efficients of equations (1) and (3)

which gives

k^2 - 5 = -1

k^2 = 4

k=2 or k=-2

Now, when k=2 the equations (1) and (3) are same. So there are 2 equations and 3 variables. So the system has infinite solutions

When k=-2 the equations (1) and (3) are

x + y -z = 2

x + y -z = -2

which makes no sense. So the system has no solutions.

For all other values of k (except 2 and -2), the equations will have unique solution.

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