This question got me thinking and confused on how to solve it. If $$\begin{align}k(x-a)+2x-z&=0\\k(y-a)+2y-z&=0 \\ k(z-a)-x-y+2z&=0\end{align}$$
Show that $$x = \frac{ak(k+3)}{k^2+4k+2}.$$
So far so good, I could only solve it simultaneously.
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Where's your work? If you could solve it simultaneously, that's a start. Let's see what you've done? – amWhy Oct 17 '18 at 20:52
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It's weird that that only $x,z$ are in the first equation and $y,z$ in the second. – Thomas Andrews Oct 17 '18 at 20:55
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This can be made a matrix equation: $$\begin{pmatrix}2+k&0&-1\0&2+k&-1 \-1&-1&2+k\end{pmatrix}\begin{pmatrix}x\y\z\end{pmatrix} = \begin{pmatrix}ka\ka\ka\end{pmatrix}$$ – Thomas Andrews Oct 17 '18 at 20:59
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@amwh, this question required proving using determinant method. Solving it using simultaneous method is not enough. So that's why your help will be appreciated – Michael Umande Oct 17 '18 at 21:03
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@thomas Andrew, which methods will be most suitable to use? Will determinant method do? – Michael Umande Oct 17 '18 at 21:06
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I don't know what you are calling the "determinant method," sorry. – Thomas Andrews Oct 17 '18 at 21:18
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I mean what method of solving system of equation like this should I employ? – Michael Umande Oct 17 '18 at 21:29
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@ThomasAndrews The “determinant method” might be Cramer’s Rule. – amd Oct 17 '18 at 23:01
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Note that the system remains the same if you interchange $x$ and $y$. Therefore $x = y$ for any solution that is unique. This means that there are only two equations: $$ k(x-a) + 2x - z = 0 \\ k(z-a) - 2x + 2z = 0 $$ Solve the first equation for $z$, substitute into the second equation, and solve for $x$.
Hans Engler
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Straightforward, using Cramer’s rule, see https://en.wikipedia.org/wiki/Cramer%27s_rule#Explicit_formulas_for_small_systems: $$\begin{align} {}&\det\begin{pmatrix} 2+k&0&-1\\ 0&2+k &-1\\ -1&-1&2+k \end{pmatrix}\\ &\qquad=(2+k)^2\det\begin{pmatrix} 1&0&-1/(2+k)\\ 0&1 &-1/(2+k)\\ -1&-1&2+k \end{pmatrix}\\ &\qquad=(2+k)^2\det\begin{pmatrix} 1&0&-1/(2-k)\\ 0&1 &-1/(2+k)\\ 0&-1&2+k-1/(2+k) \end{pmatrix}\\ &\qquad=(2+k)^2\det\begin{pmatrix} 1&0&-1/(2-k)\\ 0&1 &-1/(2+k)\\ 0&0&2+k-2/(2+k) \end{pmatrix}\\ &\qquad=(2+k)^2(2+k-2/(2+k))\\ &\qquad=(k+2)(k^2+4k +2)=:D. \end{align}$$ Now $$\begin{align} \det\begin{pmatrix} ka&0&-1\\ ka &2+k&-1\\ ka &-1&2+k \end{pmatrix} &=ka\det\begin{pmatrix} 1&0&-1\\ 1&2+k&-1\\ 1&-1&2+k \end{pmatrix}\\ &=ka\det\begin{pmatrix} 1&0&0\\ 1&2+k&0\\ 1&-1&3+k \end{pmatrix}\\ &=ka(2+k)(3+k)=:D_x. \end{align} $$ Hence $x=D_x/D$ in case $k\notin\{-2,-2\pm\sqrt2\}$.
Michael Hoppe
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