# Stuck on a systems of linear equations question

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

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

• What have you tried? – pitariver Jun 10 at 5:09
• 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. – 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.
• 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. – Siong Thye Goh Jun 10 at 5:32