Given the system $$ \left[ \begin{array}{ccc|c} x_1&x_2&x_3&k\\ x_1&x_2&kx_3&1\\ x_1&kx_2&x_3&1\\ kx_1&x_2&x_3&1\\ \end{array} \right] $$ I tried to solve this...It looks simple but I found a problem at the end... $$ \left[ \begin{array}{ccc|c} 1&1&1&k\\ 1&1&k&1\\ 1&k&1&1\\ k&1&1&1\\ \end{array} \right] $$ $$ \left[ \begin{array}{ccc|c} 1&1&1&k\\ 0&0&k-1&1-k\\ 0&k-1&0&1-k\\ k-1&0&0&1-k\\ \end{array} \right] $$ $$ \left[ \begin{array}{ccc|c} 1&1&1&k\\ 0&0&1& \frac{1-k}{k-1}\\ 0&1&0& \frac{1-k}{k-1}\\ 1&0&0& \frac{1-k}{k-1}\\ \end{array} \right] $$ Finally, $$ \left[ \begin{array}{ccc|c} 1&0&0& -1\\ 0&1&0& -1\\ 0&0&1& -1\\ 0&0&0&k+3\\ \end{array} \right] $$ There is no way to get infinitely many solutions. $$$$ However, I tried to put the system into other online calculator... Click me$$$$ There is infinitely many solutions when k=4 or other numbers. What's wrong of my work...
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In step 1, looks like you put $x_1=x_2=x_3=1.$ But they might not even be all equal, certainly not all $1.$ – coffeemath Dec 11 '17 at 14:42
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1You have divided by $k-1$, hence if it is $0$, you have a problem. For $k=1$, you have obviously infinitely many solutions. – Kelenner Dec 11 '17 at 14:47
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Hummm...It only means the coefficient of x_1, x_2 and x_3 are 1 – Dennis Dec 11 '17 at 14:48
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I was so careless...@Kelenner Thanks for your notice!:) – Dennis Dec 11 '17 at 14:51
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You are welcome ! – Kelenner Dec 11 '17 at 14:52
2 Answers
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Let us add the last three equations to gather to get: $$(k+2)(x_1+x_2+x_3) =3$$ Then we use the first one to get, $$(k+2)(k)=3 \implies k=1, -3$$
So, ?
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The system has solutions only if
$\det(A|B)=0$, otherwise $\text{rank }(A|B)=4>\text{rank }(A)$
that is $k^4-6 k^2+8 k-3=0\to (k-1)^3 (k+3)=0$
for $k=-3$ and $k=1$
for $k=-3$ we get the system
$\begin {cases} x+y+z=-3\\ x+y-3 z=1\\ x-3 y+z=1\\ -3x+y+z=1\\ \end{cases} $
which has solution $(-1,-1,-1)$
for $k=1$ we get
$\begin {cases} x+y+z=1\\ x+y+ z=1\\ x+y+z=1\\ x+y+z=1\\ \end{cases} $
that is
$x+y+z=1$ which gives infinite solutions
$(t,u,1-t-u)$
Hope this helps
Raffaele
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