0
$\begingroup$

I've been told when we have a zero this means there is always many solutions. Is this correct?

Because in the following problem:

$$a= \left[ \begin{array}{c} 1\\ -2\\0 \end{array} \right] b= \left[ \begin{array}{c} 0\\1\\2 \end{array} \right] c= \left[ \begin{array}{c} 5\\-6\\8 \end{array} \right] $$

find whether $d$ is a linear combination of $a,b,c$ where $$d = \left[ \begin{array}{c} 2\\-1\\6 \end{array} \right] $$

The reduced row echelon form matrix I get is:

$$ \left[ \begin{array}{ccc|c} 1&0&5&2\\ 0&1&4&3\\0&0&0&0 \end{array} \right] $$

The idea I always got was a row with all zeros meant there was infinitely many solutions(even though in this case there are no values of $x_3$ that make the first and second equation hold in the corresponding linear system) and that in order for there to be no solutions we would have a row of zeros with a nonzero entry in the last column only.What have I missed?

I notice we do not have a pivot in every row, does that effect the solution set? Im sorry to be asking basic questions but our professor has assigned us confusing textbook.

$\endgroup$
3
  • 1
    $\begingroup$ Take $x_1=2,x_2=3$, and $x_3=0$. $\endgroup$ Oct 24 '15 at 23:58
  • $\begingroup$ @BrianM.Scott So Brian that implies that d is a linear combination right? $\endgroup$
    – Red
    Oct 25 '15 at 0:07
  • 1
    $\begingroup$ Yes: specifically, $d=2a+3b$. $\endgroup$ Oct 25 '15 at 0:09
1
$\begingroup$

There are infinitely many solutions. They are all of the form $(2-5t,3-4t, t)$.

** It is not possible for there to be a row of zeros and the system to be inconsistent because such a solution always has many solutions.

$\endgroup$
2
  • $\begingroup$ So that implies that d is a linear combination of the other vectors right? Because the textbook answers as no. $\endgroup$
    – Red
    Oct 25 '15 at 0:05
  • 1
    $\begingroup$ Yes, this would mean $d$ is a linear combination of $a,b$ and $c$. Either the book is mistaken, or you've made an error in the row reduction. $\endgroup$ Oct 25 '15 at 0:08
1
$\begingroup$

The pivots determine the rank of the matrix. The number of pivots determine the number of main unknowns; the other unknowns are considered parameters, upon which the main unknowns depend.

The general rule to have solutions is that the rank of the matrix is equal to the rank of the bordered matrix.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.