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My system of equations is: \begin{cases} x + 5y- 6z = 2 \\ kx + y - z = 3 \\ 5x - ky + 3z = 7 \end{cases}

So the augmented matrix is:

$$ \left[ \begin{array}{ccc|c} 1&5&-6&2\\ k&1&-1&3\\ 5&-k&3&7 \end{array} \right] $$

I reduced it to this (with the shown steps), but I'm not sure if i've done so correctly and i'm not sure if i need to do Gauss-Jordan elimination or just Gauss elimination:

\begin{align} r_2 -> r_2 - 2r_1 \end{align} \begin{align} r_3 -> r+3 - 5r_1 \end{align} \begin{align} r_2 -> r_2 * \frac{1}{-9} \end{align} \begin{align} r_1 -> r_1 - 5r_2 \end{align} \begin{align} r_3 -> r_3 +27r_2 \end{align}

$$ \left[ \begin{array}{ccc|c} 1-5(\frac{k-2}{-9})&0&\frac{1}{9}&\frac{13}{9}\\ \frac{k-2}{-9}&1&\frac{-11}{9}&\frac{1}{9}\\ 27(\frac{k-2}{-9})&-k+2&0&0 \end{array} \right] $$

From this reduced from that i got, i noticed that there is an infinite solution for k=2.

However, I believe that i need to get one of the values in the 4th column of the matrix in terms of k so that i can find all the solutions i need, but i am unsure on how to do this.

Could someone please help me and show full working, including the steps taken to reduce the matrix?

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  • $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for. $\endgroup$ – 5xum Oct 8 '18 at 13:42
  • $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. $\endgroup$ – 5xum Oct 8 '18 at 13:42
  • $\begingroup$ Hi, thank you for letting me know about that! Should I try to do more or is this okay? $\endgroup$ – Zorro Oct 8 '18 at 15:16
  • $\begingroup$ @user601850: Hints: The determinant is given by $$2 (k-2) (3 k-2)$$ I got the RREF as $$\begin{bmatrix} 1 & 0 & 0 & \dfrac{8}{3 k-2} \\ 0 & 1 & 0 & -\dfrac{24}{3 k-2} \\ 0 & 0 & 1 & \dfrac{-k-18}{3 k-2} \\ \end{bmatrix}$$ Can you use this to proceed? $\endgroup$ – Moo Oct 8 '18 at 15:29
  • $\begingroup$ Do you mind telling me what steps you used to get that RREF? $\endgroup$ – Zorro Oct 8 '18 at 15:50
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From my comments above, I always find the determinant first which gives us

$$\det \begin{bmatrix} 1&5&-6 \\ k&1&-1 \\ 5&-k&3 \end{bmatrix} = 2 (k-2) (3 k-2)$$

This tells us we may have to account for $$k = 2, k = \dfrac{2}{3}$$

The RREF is given by the steps

  • Swap $R_1$ and $R_2$

  • Set $R_2 \longleftarrow R_2 - \dfrac{1}{k} R_1$

  • Set $R_3 \longleftarrow R_3 - \dfrac{5}{k} R_1$

  • Swap $R_2$ and $R_3$

  • Set $R_3 \longleftarrow R_3 - \dfrac{5k-1}{-k^2-5} R_2$

  • Set $R_3 \longleftarrow - \dfrac{k^2+5}{2(3k^2-8k+4)} R_3$

  • Set $R_2 \longleftarrow R_2 - \dfrac{3k+5}{k} R_3$

  • Set $R_1 \longleftarrow R_1 + R_3$

  • Set $R_2 \longleftarrow \dfrac{k}{-k^2-5} R_2$

  • Set $R_1 \longleftarrow R_1 - R_2$

  • Set $R_1 \longleftarrow \dfrac{1}{k} R_1$

This results in the RREF of

$$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & \dfrac{8}{3 k-2} \\ 0 & 1 & 0 & -\dfrac{24}{3 k-2} \\ 0 & 0 & 1 & \dfrac{-k-18}{3 k-2} \\ \end{array} \right]$$

We can see that we have an issue for $k = \dfrac{2}{3}$.

Also, from the determinant, when $k = 2$, we have a RREF of

$$\left[ \begin{array}{ccc|c} 1 & 0 & \dfrac{1}{9} & \dfrac{13}{9} \\ 0 & 1 & -\dfrac{11}{9} & \dfrac{1}{9} \\ 0 & 0 & 0 & 0 \\ \end{array} \right]$$

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    $\begingroup$ Your first step is only valid when $k\ne0$. You also have to account for $k=0$ separately. $\endgroup$ – amd Oct 9 '18 at 0:42
  • $\begingroup$ @amd: That result is captured by the general case in the current answer with $k = 0$, as you get $$\begin{bmatrix} 1 & 0 & 0 & -4 \\ 0 & 1 & 0 & 12 \\ 0 & 0 & 1 & 9 \\ \end{bmatrix}$$ $\endgroup$ – Moo Oct 9 '18 at 1:52

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