If the linear system $$ \begin{align*} ax+by&=k\\ cx+dy&=l\\ ex+fy&=m \end{align*} $$ is consistent, why at least one equation can be discarded from the system without altering the solution set? (Anton Elementary Linear Algebra 9th edition, Exercise Set 1.1, problem 13)

My Attempt

If the linear system is consistent, then it has at least one solution (exactly one solution or infinitely many solutions). If each of the linear equations in the system correspond to lines $l_1$, $l_2$, and $l_3$, then:

  • If we have infinitely many solutions, then the three lines are coincident. Two of the lines are coincident so the third line can be discarded without altering the solution set.
  • If we have exactly one solution, then the three lines intersect in exactly one point. Two of the lines intersect in one point so the third line can be discarded without altering the solution set.
  • $\begingroup$ Yes, your attempt seems correct. $\endgroup$ – Jimmy R. Mar 29 '16 at 8:52

Your attempt is correct. (Just be careful about the geometric interpretation of equations as lines; as long as you have clearly defined how you are converting an algebra problem into a geometry problem and back, nobody can question you.)

Another [algebraic] approach is to show that two of the equations are sufficient to solve for $x$ and $y$ in terms of the coefficients (you can use matrices, the standard substitution method, or whatever else to do this). As you have assumed nothing about the actual coefficients, you can choose any two of the given equations and you will always be able to find the solution. This method requires some special handling of the infinite solutions or no solutions cases, but does not require geometry.


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