1
$\begingroup$

I am having difficulty with true/false statements and their justifications regarding systems of linear equations.

(a) A linear system of three equations in five unknowns is always consistent (i.e. it has at least one solution)

(b) A linear system of five equations in three unknowns cannot be consistent

(c) If a linear system in echelon form is triangular then the system has the unique solution

(d) If a linear system of n equations in n unknowns has two equations that are multiples of one another, then the system is inconsistent.

So far, for (a) I have said False, as it will always be consistent if it is homogeneous, but not if it is non-homogeneous.

For (b) I have said false, but am having difficulty justifying this assertion

(c) I know to be true.

(d) I believe may be false as having equations that are multiples could result in free variables and hence infinite solutions?

I am rather unsure on what I have done so far.

Any assistance is greatly appreciated.

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

for part $b$ , you may consider below counter example :

$x_1+x_2+x_3=1$

$x_1+x_2+x_3=1$

$x_1+x_2+x_3=1$

$x_1+x_2+x_3=1$

$x_1+x_2+x_3=1$

for part $c$ , look @ Is it possible for a triangular matrix in echelon form to not have a unique solution and how?

for part $d$ :

you're correct about the answer but your reasoning is only partially correct - the system can also be inconsistent :

$x_1+x_2+x_3 = 1$

$2x_1+2x_2+2x_3=2$

$x_1+x_2+x_3 = 2$

$\endgroup$
5
  • $\begingroup$ Thanks very much for that, it's greatly appreciated. How is my Part a response? $\endgroup$
    – user2768428
    Aug 11, 2014 at 13:39
  • $\begingroup$ np :) your justification for part a looks perfect to me ! $\endgroup$
    – AgentS
    Aug 11, 2014 at 13:42
  • $\begingroup$ Could I possibly ask you to explain part C? I wasn't able to understand fully what was discussed in the link you provided $\endgroup$
    – user2768428
    Aug 11, 2014 at 14:25
  • $\begingroup$ here is a triangular matrix in echelon form $\left[\begin{array}{ccc} 1&0&0 \\ 0&1&0 \\ 0&0&0\\ \end{array}\right]$ Clearly for this system, the solutions cannot be unique as the determinant is 0. depending on the right side, it will have infinitely many or 0 solutions... $\endgroup$
    – AgentS
    Aug 11, 2014 at 14:40
  • $\begingroup$ basically a triangular matrix need not have non zero elements in the diagonal. definition @ mathworld.wolfram.com/UpperTriangularMatrix.html $\endgroup$
    – AgentS
    Aug 11, 2014 at 14:48
0
$\begingroup$

for part(1), I can give one example a+b+c+d+e=1,2a+b+c+d+e=2,3a+b+c+d+e=5, this system of three equations in 5 variables has no solution or it is inconsistent.Hence the given statement is false. for part(2), we can have one example a+b+c=3, 2a+2b+c=5, 2a+b+2c=5, a+2b+2c=5, a+b+2c=4. This is a system of five equations in 3 variables with solution a=b=c=1. Hence the given statement is false.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .