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I'm trying to discuss a system of equations that depends on a parameter $k$ using the method Gauss-Jordan but the more I calculate the more entangled it becomes. While the solution in the book seems so easy and simple.

\begin{equation} \begin{cases} kx-y-z+3t=0\\2x-kz+2t=k+4\\kx+y+t=-k\\x+y-z=2 \end{cases} \end{equation} My attempt is:

\begin{vmatrix} k & -1& -1& 3& 0 \\2 & 0& -k& 2& k+4 \\k & 1& 0& 1& -k \\ 1 & 1 & -1& 0& 2\end{vmatrix} $ \longrightarrow$ changing the order of first and fourth row \begin{vmatrix} 1 & 1& -1& 0& 2 \\2 & 0& -k& 2& k+4 \\k & 1& 0& 1& -k \\ k & -1 & -1& 3& 0\end{vmatrix}

$\xrightarrow{R_2 \to -2R_1+R_2,\hspace{2mm}R_3 \to -kR_1+R_3,\hspace{2mm}R_4 \to -kR_1+R_4}$

\begin{vmatrix} 1 & 1& -1& 0& 2 \\0 & -2& 2-k& 2& k \\0 & 1-k& k& 1& -3k \\ 0 & -1-k & -1+k& 3& -2k\end{vmatrix}

$\xrightarrow{R_2 \to R_2*(-1/2)}$

\begin{vmatrix} 1 & 1& -1& 0& 2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 1-k& k& 1& -3k \\ 0 & -1-k & -1+k& 3& -2k\end{vmatrix}

$\xrightarrow{R_1 \to -R_2+R_1,\hspace{2mm}R_3 \to -(1-k)R_2+R_3,\hspace{2mm}R_4 \to -(-1-k)R_2+R_4}$

\begin{vmatrix} 1 & 0& -k/2& 1& (k+4)/2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 0& (k^2-k+2)/2& -k& (-k^2-5k)/2 \\ 0 & 0 & (k^2+k-4)/2& 2-k& (-k^2-5k)/2\end{vmatrix}

$\xrightarrow{R_3 \to R_3*(2/(k^2-k+2)),\hspace{2mm}R_4 \to R_4*(2/(k^2+k-4))}$

\begin{vmatrix} 1 & 0& -k/2& 1& (k+4)/2 \\0 & 1& (k-2)/2& -1& -k/2 \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 1& 2(2-k)/(k^2+k-4)& (-k^2-5k)/(k^2+k-4)\end{vmatrix}

$\xrightarrow{R_1 \to R_3*(k/2)+R_1,\hspace{2mm}R_2 \to -(k-2/2)R_3,\hspace{2mm}R_4 \to -R_3+R_4}$

\begin{vmatrix} 1 & 0& 0& -k+2/k^2-k+2& -(k^2-k-1)/(k^2-k+2) \\0 & 1& 0& -k-2/k^2-k+2& (2k^2-6k)/(k^2-k+2) \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 0& 8(k-1)^2/(k^2-k+2)(k^2+k-4)& (2k^3+4k^2-30k)/(k^2-k+2)(k^2+k-4)\end{vmatrix}

$\xrightarrow{R_4 \to R_48(k-1)^2/(k^2-k+2)(k^2+k-4)}$

\begin{vmatrix} 1 & 0& 0& -k+2/k^2-k+2& -(k^2-k-1)/(k^2-k+2) \\0 & 1& 0& -k-2/k^2-k+2& (2k^2-6k)/(k^2-k+2) \\0 & 0& 1& -2k/(k^2-k+2)& (-k^2-5k)/(k^2-k+2) \\ 0 & 0 & 0& 1& 2k(k^2+2k-15)/8(k-1)^2 \end{vmatrix}

$\xrightarrow{R_1 \to R_4*(k-2/k^2-k+2)+R_1,\hspace{2mm}R_2 \to R_4*(k+2/k^2-k+2)+R_2,\hspace{2mm}R_3 \to R_4*(2k/k^2-k+2)+R_3}$

\begin{vmatrix} 1 & 0& 0& 0& (-6k^4+8*k^3-22k^2+36k+8)/8(k-1)^2*(k^2-k+2) \\0 & 1& 0& 0& (-14k^4+88*k^3-134k^2-12k)/8(k-1)^2*(k^2-k+2) \\0 & 0& 1& 0& (-4k^4-16*k^3+12k^2-40k)/8(k-1)^2*(k^2-k+2) \\ 0 & 0 & 0& 1& 2k(k^2+2k-15)/8(k-1)^2 \end{vmatrix}

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    $$\begin{bmatrix} 1 & 0 & 0 & 0 & \dfrac{-3 k-8}{2 (k-2)} \\ 0 & 1 & 0 & 0 & \dfrac{7 k}{2 (k-2)} \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & \dfrac{k^2+5 k}{2 (k-2)} \\ \end{bmatrix}$$ – 2017-02-06
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    What specific problem are you having? are you having trouble understanding the question or Gauss-Jordan method or something else? – 2017-02-06
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    it seems all ok with Gauss-Jordan method because I can solve all the other systems but this one is terrible!!! – 2017-02-06
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    @Sum-Meister I can 't manage to semplify properly – 2017-02-06
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    give me a few minutes to work an example out to clarify. – 2017-02-06
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    @Sum-Meister Look at my answer Where am I making mistakes? – 2017-02-08
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    You're missing some fraction simplifications. – 2017-02-08
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    @egreg Can you show me where please??? – 2017-02-08
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    @Anne I can confirm Moo's result; your is much different and I guess it's due to computation errors. I've checked it with Pari-GP. – 2017-02-08
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    @egreg I agree with my book, Moo, you and Sum-Meister . But can you explain where I made mistakes? how do Moo and you get the final answer? – 2017-02-08
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    @Moo Please can you explain your solution? – 2017-02-21
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    Does the book solution look like mine? – 2017-02-21
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    @Moo: Everybody can do this. My computer algebra system (MAPLE), of course, gives the same outcome as yours. But the OP's question is: how to do it, with the Gauss-Jordan method, by hand, without getting lost in cumbersome algebra. IMO that part has not been addressed yet in the current answers. – 2017-02-22
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    @HandeBruijn: Okay, well that was nasty, but I added the steps to a solution. I did not want to write out all those intermediate matrices! :-) – 2017-02-22
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    @Moo: Alright .. No bad (+1) presentation! – 2017-02-23

2 Answers 2

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Here is one solution (did it two other ways also - one using Cramer's Rule and one using a different set of steps for Gaussian Elimination than is shown below). $R_x$ means Row - $x$ and these are the row steps in order.

  • $R_3 = R_3 - R_1$
  • $R_1 = \dfrac{R_1}{k}$
  • $R_2 = R_2 - 2 R_1$
  • $R_4 = R_4 - R_1$
  • $R_2 = \dfrac{k}{2} R_2$, the matrix after this step is

$$\begin{bmatrix} 1 & -\dfrac{1}{k} & -\dfrac{1}{k} & \dfrac{3}{k} & 0 \\ 0 & 1 & -\dfrac{k^2}{2} + 1 & k - 3 & \dfrac{k}{2} (k + 4) \\ 0 & 2 & 1 & -2 & -k \\ 0 & \dfrac{1}{k}(k + 1) & \dfrac{1}{k} (-k + 1) & -\dfrac{3}{k} & 2 \end{bmatrix}$$

  • $R_1 = R_1 - \left(-\dfrac{1}{k} R_2\right)$
  • $R_3 = R_3 - 2 R_2$
  • $R_4 = R_4 - \left(\dfrac{1}{k}(k+1)\right)R_2$
  • $R_3 = \left(\dfrac{1}{k^2-1}\right)R_3$
  • $R_1 = R_1 - \left(-\dfrac{k}{2}\right)R_3$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & \dfrac{2k-1}{k^2-1} &-\dfrac{k^2+k+4}{2k^2 -2} \\ 0 & 1 & -\dfrac{k^2}{2} + 1 & k - 3 & \dfrac{k}{2} (k + 4) \\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & \dfrac{k^2}{2} + \dfrac{k}{2}-2 & -k+2 & -\dfrac{k}{2}(k+5) \end{bmatrix}$$

  • $R_2 = R_2 -\left(-\dfrac{k^2}{2} + 1 \right)R_3$
  • $R_4 = R_4 -\left(\dfrac{k^2}{2} + \dfrac{k}{2} - 2 \right)R_3$
  • $R_4 = \left(\dfrac{k^2-1}{k^2-5k+6}\right)R_4$
  • $R_1 = R_1 -\left(\dfrac{2k-1}{k^2-1}\right)R_4$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & 0 &-\dfrac{3k+8}{2(k-4)} \\ 0 & 1 & 0 & -\dfrac{k^2+k-1}{k^2-1} & \dfrac{k(-k^2+k+6)}{2k^2-2}\\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2(k-2)} \end{bmatrix}$$

  • $R_2 = R_2 -\left(\dfrac{-k^2+k-1}{k^2-1}\right)R_4$, the matrix after this step is

$$\begin{bmatrix} 1 & 0 & 0 & 0 &-\dfrac{3k+8}{2(k-4)} \\ 0 & 1 & 0 & 0 & \dfrac{7k}{2k^2-2}\\ 0 & 0 & 1 & \dfrac{2(-k+2)}{k^2-1)} & -\dfrac{k(k+5)}{k^2-1} \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2(k-2)} \end{bmatrix}$$

  • $R_3 = R_3 - \left(\dfrac{2(-k+2)}{k^2-1}\right)R_4$, , the matrix after this step is the final result and is

$$\begin{bmatrix} 1 & 0 & 0 & 0 & -\dfrac{3 k+8}{2 (k-2)} \\ 0 & 1 & 0 & 0 & \dfrac{7 k}{2 (k-2)} \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & \dfrac{k(k+5)}{2 (k-2)} \\ \end{bmatrix}$$

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    please put some matrices !!! From the tenth passage i loose you!!! – 2017-02-23
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    ok but I don't understand all the last passages for the final risult – 2017-02-23
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    Yes I don't know how to calculate from the passage before the third zero in the first row to the last. I understand why you multiply e sum for that numbers but in the calculation I forget something perhaps – 2017-02-23
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    the results are $x=-(3k+8)/(2k-4)$ $y=7k/(4k-4)$ $z=0$ $t=k(k+5)/(4k-4)$ – 2017-02-23
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    ok and but the second not – 2017-02-23
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    thanks for the extra matrices but I don't see how I could obtain your final results – 2017-02-23
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    Please compare my results to [*Wolfram Alpha*](http://www.wolframalpha.com/input/?i=rref+%7B%7Bk,-1,-1,3,0%7D,%7B2,0,-k,2,k%2B4%7D,%7Bk,1,0,1,-k%7D,%7B1,1,-1,0,2%7D%7D). Looks to me like your book has a couple of typos as I doubt the others would match. They lost a factor of $2$ in two of the results. You can actually verify my result works by subbing in $x, y, z, t$ and making sure you get the RHS in the original system. – 2017-02-23
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    I'm afraid today i have no more time to dedicate to this problem. In any case i give your 100 point and thank you. – 2017-02-24
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$ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} $ $\xrightarrow{R_3 \to aR_3-gR_1,\hspace{2mm}R_2 \to aR_2-dR_1}\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & h' & i' \end{bmatrix} $

$ \begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & h' & i' \end{bmatrix} $ $\xrightarrow{R_3 \to e'R_3-h'R_1}\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & 0 & i'' \end{bmatrix}$

$\begin{bmatrix} a & b & c \\ 0 & e' & f' \\ 0 & 0 & i'' \end{bmatrix}$ $\xrightarrow{R_2 \to i''R_2-f'R_3,\hspace{2mm}R_1 \to i''R_1-cR_3}$ $\begin{bmatrix} a' & b' & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix}$

$ \begin{bmatrix} a' & b' & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix} $ $\xrightarrow{R_1 \to e''R_1-b'R_2}\begin{bmatrix} a'' & 0 & 0 \\ 0 & e'' & 0 \\ 0 & 0 & i'' \end{bmatrix}$

Now, using these Row operations, you can get a 3*3 matrix into the required from. Try forming the operations for 4*4 and see if it helps.

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    that's all clear – 2017-02-07
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    could you attach a scheme of how you solved the exercise? – 2017-02-07
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    Sure. I'll explain for 3*3, and 4*4 is just similar. Start by focusing on the 2nd and 3rd element of first column. we have to do some row operations to get zeros in that position. So we use some multiple (think of which multiple will suit here) of 1st row, and subtract it from 3rd and 2nd row. So now, we have zeros in the $a_{2,1}$ and $a_{3,1}$ position. Next, we move on to element $a_{3,2}$ (Use some multiple of 2nd row here.) Now, go on to $a_{2,3}$ and $a_{1,3}$. And finally $a_{1,2}$. I can explain even further if you still do not see it clearly. – 2017-02-07
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    my solution for t is $2*k(k^2+2k-15)/8*(k-1)^2$ and I've tried for hours...please can you post your row operations ? – 2017-02-07
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    You want me to show you row operations for 4*4 matrix given in the question? – 2017-02-09
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    yes if you want – 2017-02-10