# Can this matrix be put into reduced row echelon form?

I am given this matrix and the directions are "For the matrix A​ below, find a nonzero vector in Nul A and a nonzero vector in Col A."

$$\begin{bmatrix}2&-4\\-1&2\\-3&6\\-4&8\end{bmatrix}$$

My first step was to try and put this into rref using my TI-84, and no matter what I was given the "Error" output. I tried to augment the matrix (below) and I was still given an "Error" message.

$$\begin{bmatrix}2&-4&0\\-1&2&0\\-3&6&0\\-4&8&0\end{bmatrix}$$

When I tried to put it onto an online program it gave me the answer $$\begin{bmatrix}1&-2&0\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

And the online homework assignment gave me $$\begin{bmatrix}1&2&0\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$

Which one of these is correct? Can this matrix be put into rref? I thought maybe that my calculator won't do it because it has more rows than columns, but I don't remember ever reading about that being a condition, and I couldn't find anything online.

• Observe that the rows of this matrix are multiples of each other, so you know that the rref is going to have only one non-zero row—the first. The upper-left element is positive, so this row in the rref will have the same pattern of signs as in the original matrix. – amd Nov 20 '16 at 1:06

The middle one seems to be correct: $$\left[ \begin{array}{rr} 2 & -4 \\ -1 & 2 \\ -3 & 6 \\ -4 & 8 \end{array} \right] \to \left[ \begin{array}{rr} 1 & -2 \\ 2 & -4 \\ -3 & 6 \\ -4 & 8 \end{array} \right] \to \left[ \begin{array}{rr} 1 & -2 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{array} \right]$$

• Do you know why my calculator won't solve it? It tells me invalid dimensions. I can't find any information online about this. – matryoshka Nov 20 '16 at 0:55
• Sorry, I have no manual for that device. – mvw Nov 20 '16 at 0:57