# Row of zeros and inconsistency possible?

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.

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

There are infinitely many solutions. They are all of the form $(2-5t,3-4t, t)$.
• 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. Oct 25 '15 at 0:08