I've decided to try to teach myself linear algebra, and I'm into the second chapter of the textbook I've pirated, and I'm stumped on this problem concerning 3x3 systems of linear equations.

I've learned a system of equations is singular, having no solution, when the third row of the matrix the system forms is a linear combination of the first two rows.

So I'm doing a problem from the textbook that gives me two equations and has asked me to form a third equation that would make the three a system with no solution.

The first two equations are x+y+z=0 and x-2y-z=1. In order to create a third equation that'd lead to no solution, I created a linear combination of the first two, specifically 2(eq1)-(eq2)

This yielded the third equation x+4y+z=-1. The only problem is I checked and it gave a solution for the system (1/3,-1/3,0) which I thought wasn't supposed to happen with linear combinations.

To make matters worse, I then formed another linear combination through (eq1)-4(eq2) which yielded -3x+9y+5z=-4 and this one actually doesn't have a solution.

So what gives? I figure I'm misunderstanding something important here.

  • cosecantphi [he/him]
    ·
    3 years ago

    Thanks for the correction! I wrote no solutions, but you're right I should have wrote I'm looking for a third equation that would simply not give a single solution. Specifically the question in the book asks for a third equation that would not yield a single solution, not specifically that there should be no solutions.

    And definitely the matrix is what is singular here, not the equations. I realized while correcting my answer that may be the case. The answer guide briefly mentioned that the right side of the equation does not need to follow the relations you used to create the linear combination in order to just make a singular matrix from the first two equations. I was still a bit confused until you reiterated that, so thanks very much!