I am completing a practice questions sheet for the topic "systems of linear of equations" and I've hit a roadblock on one of the questions.

1. Consider the system of equations $$\begin{aligned} x + 2y - z &= -3 \\\ \end{aligned}$$ $$\begin{aligned} 3x + 5y + kz &= -4 \\\ \end{aligned}$$ $$\begin{aligned} 9x + (k+13)y + 6z &= 9 \\\ \end{aligned}$$ a) Express these equations as an augmented matrix

which I think is (correct me if I'm wrong): $$ \left[\begin{array}{rrr|r} 1 & 2 & -1 & -3 \\ 3 & 5 & k & -4 \\ 9 & (k+13) & 6 & 9 \end{array}\right] $$

I am stuck on part (b) which is:

b) Show that this matrix can be row-reduced to

$$ \left[\begin{array}{rrr|r} 1 & 2 & -1 & -3 \\ 0 & 1 & -k-3 & -5 \\ 0 & 0 & k^2-2k & 5k+11 \end{array}\right] $$

  • $\begingroup$ What have you tried? $\endgroup$ – pitariver Jun 10 at 5:09
  • $\begingroup$ as suggested by siong below I performed R2−3R1 , R3−9R1, but the matrix I got had several differences to the one in the question. I'm not sure if I'm doing something wrong, missing a step, or if the answer in the question is incorrect and I just need to state that it is. $\endgroup$ – jakeymaths Jun 10 at 5:30


Perform $R_2-3R_1$, $R_3-9R_1$, $-R_2$, and you should be one step away from the solution.

  • $\begingroup$ so after performing the row operations, there were some parts that differ from the given matrix, such as -k+3 rather than -k-3. Would I, therefore, record the correct row echelon form and state that the given matrix is incorrect? $\endgroup$ – jakeymaths Jun 10 at 5:18
  • $\begingroup$ First, check if you have copied the question correctly. If it is correct, you might like to consider special values of $k$, for example let $k=0$. Check with a numerical solver if the matrices are row equivalent. $\endgroup$ – Siong Thye Goh Jun 10 at 5:32
  • $\begingroup$ sorry I just checked and I miss wrote the question. its my first time using mathjax $\endgroup$ – jakeymaths Jun 10 at 5:35
  • $\begingroup$ I have corrected the question, I apologize for that I should've noticed sooner $\endgroup$ – jakeymaths Jun 10 at 5:38
  • $\begingroup$ I redid it after fixing the errors however I still dont understand how you get the 0 in place of the (k+13) and where the (k^2-2k) and (5k+11) in the questions answer are coming from $\endgroup$ – jakeymaths Jun 10 at 5:53

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