Prove that a system of equations has infinitely many solutions

Question

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?

Answer 1

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.

Answer 2

You performed the second row operation wrong. If you carry out the row operation $R_3=R_1\cdot 2 + R_3$ on the matrix

$$\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\0&2&1&|&0\end{vmatrix} $$

then you should end up with the matrix

$$\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\\color{red}{-2}&0&-1&|&0\end{vmatrix}$$

Not the matrix

$$\begin{vmatrix}-1&-1&-1&|&0\\0&2&1&|&0\\\color{red}{0}&0&-1&|&0\end{vmatrix}$$

This is just a simple arithmetic error. If you go back and do your calculations again, you will find a different echelon form and RREF, showing you that the rank is not $3$.

Answer 3

It's sometimes worth a few seconds looking for simple relationships between rows in such problems. Observe with your original matrix $2r_1+r_3=r_2$ so by inspection the rows are linearly dependent and the matrix cannot be rank 3.