Find values of k in this linear system

[Robb]

Homework Statement

Find the values of k so that each of the following systems in unknowns x, y, and z has (i) a unique solution, (ii) no solution, (iii) an infinite number of solutions.

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

Homework Equations

The Attempt at a Solution

I really don't know where to begin with this. Is there a systematic way to solve? It seems empirical to me but that doesn't make sense either.

 

[BvU]

This isn't calculus. Did you learn about matrices ? Inverting them and determinants and such ?

(Please check the double = sign in the first eqn)

 

[Robb]

Yes, I know about matrices and determinants.
z +y + kz = 1
x + ky +z = 1
kx + y + z = 1

 

[fresh_42]

Yes, I know about matrices and determinants.
z +y + kz = 1
x + ky +z = 1
kx + y + z = 1

So what's the matrix of the system and what the determinant?

 

[scottdave]

So what's the matrix of the system and what the determinant?

You may find this useful http://www.ulaff.net/LAFF-N2LW.html
I took the edX course as a Linear Algebra refresher, and it was worth the effort.

For simple 2 or 3 variable systems, Cramer's rule should work for you.

 

[Robb]

A = 1 1 1 1
1 1 1 1
1 1 1 1

detA = 0

 

[scottdave]

A = 1 1 1 1
1 1 1 1
1 1 1 1

detA = 0

With Det(A) = 0, there are two possibilities: there is No Solution, or Infinite Solutions. Which one applies here? When k does not equal 1, what can you say about the system?

 

[Robb]

When k does not equal 1 the determinant is nonzero (implying a unique solution) and when k equals 1 the equations are equal so there are infinitely many solutions.

 

[scottdave]

Actually, A is a square matrix (3x3) rather than the (3x4) you showed.

 

[Robb]

Actually, A is a square matrix (3x3) rather than the (3x4) you showed.

Right, I used the augmented matrix...sorry.

 

[BvU]

Hi Robb,

You have me wondering about your matrix: you should realize $k$ is an unknown coefficient and your matrix in fact looks like $$\begin{pmatrix}
1 & 1 & k \\
1 & k & 1 \\
k & 1 & 1
\end{pmatrix}$$and the equation like
$$\begin{pmatrix}
1 & 1 & k \\
1 & k & 1 \\
k & 1 & 1
\end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} =\begin{pmatrix} 1\\1\\1 \end{pmatrix} $$

 

[Ray Vickson]

Right, I used the augmented matrix...sorry.

IMHO using matrices on this problem is a great waste of time; better to just do what people did a hundred years before matrices were invented. In other words, just apply Gaussian elimination: use the first equation to solve for x in terms of y and z, then substitute that expression for x into the second and third equations. That will give you two equations in y and z, and if you actually write them out in detail the solution will become---if not blindingly obvious---at least straightforward.