Find k so that there is only one solution for set of equations, and k also satisfy $x1+x3=k+5$ The set of equations: $$x1+x2+kx3=k$$ $$kx2=k$$ $$kx1+x2+x3=1$$
I remember we solved similar exercises, but I kind of confused how to procced...
My way of solution, by writing in matrix form:
$\begin{pmatrix} 1 &1 &k &k \\ 0& k &0 &k \\ k& 1 &1 &1 \end{pmatrix}r3-kr1\rightarrow r3$
$\begin{pmatrix} 1 &1 &k &k \\ 0& k &0 &k \\ 0& 1-k &1 -k^2 &1-k^2 \end{pmatrix}kr3-r2(1-k)\rightarrow r3$
$\begin{pmatrix} 1 &1 &k &k \\ 0& k &0 &k \\ 0& 0 &k(1-k^2) &k(1-k^2)-k(1-k)=-k^3+k^2 \end{pmatrix}$
I've written the 4th column as the solution because I don't know how to add the dividing line.
after reaching the reduced form, I have no clue how the condition: $x1+x3=k+5$ helps for solution.
how do I continue?
answer: $k\neq 0,1$
If $k=0$, the second of your equations (in your final form) is just $0=0$. This is a trivial equation, and therefore you can forget it. What's left is a system of two linear equations in three unknows. Such a system either has no solutions or it has infinitely many.
Suppose now that $k=1$. Then what I wrote above about the second equation holds if “second” gets replaced by “third”.
In the remaining cases, there is one and only one solution. You get from your system that$$x_3=\frac{-k^3+k^2}{k(1-k^2)},$$and that $x_2=1$. Finally, from this information and from the fact that $x_1+x_2+kx_3=k$, you get the value of $x_1$.