Q: The system of equations:




b) Find the vector $v$ such that there are an infinite number of solutions to the system of equations of the form $x^{*} + kv$


I've attached a photo of the question with my attempts to the post.

I've reduced the matrix to echelon form to find solutions for $x_{1}, x_{2}, x_{3}$ and also using the assumption that $b_{1}= 2b_{3}+b_{2}$, arrived to the confirmation that all 3 equations are independent, and the existance of more free variables than non zero rows leads to the conclusion of infinite solutions. However, I ended up going full circle achieving the equations that were given, and still unsure how to put in the required vector notation.

Revised solution for part A, and full question

Solution for part 2


Since you were able to reduce the matrix to the identity matrix, the solution you found was unique. Note that this reduction can be done regardless of the values of $b_1,b_2,b_3$ (as you did it), so you can always find a unique solution. Since the solution is always unique, the question asking to describe infinite solutions doesn't make sense.

Perhaps you should double-check your row reduction. You should get a solution with a free parameter (perhaps, say, x_3), and then you can rewrite the solution as $w+x_3v$ for some known vectors $w,v$, which answers the question.

Also, I think part (b) is following up on part (a), i.e., you are assuming $b_1=2b_3+b_2$. Otherwise there may not be a solution.

  • $\begingroup$ Hi there, thank you so much for your response! You were right, I did make a mistake in my reduction, but using the assumption you mentioned, I managed to get a sensible result at the end. However, I ended up going full circle and produced the equations they originally gave us! In the description is my updated attempt. Although I understand the mechanics of row reduction, I don't really understand the translation to vector form $\endgroup$ – Reece Aug 14 at 18:26
  • $\begingroup$ @ReeceBuckle When there is dependency in your equations, you need to assign one (or more) variable to be a free parameter. It's easiest to use the columns without pivots as the free variables, so in this case $x_3$. Then you want to solve for the other variables in terms of $x_3$, and finally write the whole solution in vector form. $\endgroup$ – BallBoy Aug 14 at 18:36
  • $\begingroup$ Thank you :) I have arrived at a solution I am satisfied with now. $\endgroup$ – Reece Aug 14 at 19:19
  • $\begingroup$ @Reece See this for a detailed description of how to read a basis for the null space of a matrix directly from its RREF. $\endgroup$ – amd Aug 14 at 20:27
  • $\begingroup$ @amd thank you that's very insightful! $\endgroup$ – Reece Aug 15 at 10:48

Hint: Multiplying the third equation by $(-2)$ and adding to the first we get $$x_2+x_3=b_1-2b_3$$ and the second equation is given by $$x_2+x_3=b_2$$ Can you proceed?

  • $\begingroup$ Thank you for your help! I updated my solution to part A as this is a much more solid solution $\endgroup$ – Reece Aug 14 at 18:23

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