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Question
2.   For which values of k the following system of linear equations has
(a) no solutions,   (b) exactly one solution, and   (c) an infinite number of solutions

kx+y+z=1
x+ky+z=1
x+y+kz=1

I don't understand how to solve for k, when it's in each equation

Answer
Questioner:Vanessa
Country:Quebec, Canada
Category:Advanced Math
Private:No
Subject:Linear algebra
Question:

2.   For which values of k the following system of linear equations has
(a) no solutions,   (b) exactly one solution, and   (c) an infinite number of solutions

kx+y+z=1
x+ky+z=1
x+y+kz=1

I don't understand how to solve for k, when it's in each equation
..........................................
Suggestion: You have studied Cramer's rule?

D = the determinant:

| k   1   1   |
| 1   k   1   |
| 1   1   k   |

If this is nonzero, we have a unique solution.


= k^3 + 1 + 1 - k - k - k

= k^3 - 3k + 2

= (k - 1)(k^2 + k - 2)

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

So this is zero for  k = 1 and k = -2.

If k = 1, the equations are:

x+y+z=1
x+y+z=1
x+y+z=1

What do you think about these?



If k = -2, the equations are:

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

Try adding these up.  Do you like the result?  

What do you think about these?
------------------------------------------------
If you have NOT studied C's R, then take:

kx+y+z=1    A
x+ky+z=1    B
x+y+kz=1    C

1.1: Eliminate x from A,B:  Do A - kB, meaning: multiply B by -k and add that to A.  Call this D.

1.2: Eliminate x from A,C:  Do A - kC.  Call this E

Now you have two equations in y,z, called D,E.

2.1. Eliminate y from D,E.  (I leave that to you.)

2.2  You now have an equation in z, that looks like:

(k^3 - 3k + 2) z = (something with k's), so:

   (something with k's)
z = --------------------
    (k^3 - 3k + 2)

You can figure the rest.

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