I am given a system of equations (in which all equations are equal to zero) and I have to prove it has an infinite number of solutions. The augmented matrix looks like this:
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\2&4&3&|&0\end{vmatrix}
I was taught that I need to find the rank of the matrix, and in order to do that I reduce it to row-echelong form. So I perform row operations as follows:
r3= r1(2)+r3 and I'm left with
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\0&2&1&|&0\end{vmatrix}
Then I do r3= r1(2)+r3 again and I'm left with:
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\0&0&-1&|&0\end{vmatrix}
At this point I should look at what the number of independent rows is. And I see that there isn't any row of zeros-only, so the rank must be three. Yet when I use a calculator, I get that the rank is two. If the rank was indeed two, the system would have infinite solutions. What am I doing wrong?
Best Answer
Your mistake:
after
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\0&2&1&|&0\end{vmatrix}
you should do $r3=r3-r2$ to get
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\0&0&0&|&0\end{vmatrix}
If you did $r3=r1(2)+r3$, you would get
\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\ \color{red}{-2}&0&-1&|&0\end{vmatrix}, which doesn't help.