0
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

Guide:

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.