# If every row of a 2x3 matrix is a pivot position, how can it span R3?

In my text there is a T/F statement: If every row of an $m \times n$ matrix A contains a pivot position, then the matrix equation $Ax=b$ is consistent for every b in $R^n$

This is listed as true.

I thought about a $2 \times 3$ matrix...

Doesn't this require that since $b$ will be a $2\times1$ matrix, $Ax=b$ would be consistent for every b in $R^m$ ($R^2$ in my example)?

For it to span $R^n$ wouldn't it be required that the columns of A span $R^3$ in my example (impossible)?

Your number of $b$ entries is always the same as the number of $rows$ you have. So when you consider the $2 \times 3$ matrix, we will have, actually an infinite number of solutions since there will be a column without a pivot. As you said, in order to span $R^2$, we need $2$, linearly indepenedent vectors. In the $2 \times 3$ matrix, we have a third "additional" column. So, it's not needed, and so it will make your system have infinite solutions. (If you don't know infinite solutions), wait until next class or so, I'm sure you'll learn it soon.
• Yes this makes sense. Do you think this is an error in the book? How would a 2x3 matrix span $R^3$? Feb 24, 2017 at 3:26
• It is impossible for a $2 \times 3$ matrix to span $R^3$ and here's why. Suppose you have this matrix: $\begin{bmatrix} 1 & 2 & 5\\ 3& 4 & 6 \end{bmatrix}$. When you row reduce it, the $max$ amount of pivots you will get is $2$ (since there are only two rows! To span $R^3$, you need three linearly independent vectors, which we cannot get since we are in $R^2$ right now. Feb 24, 2017 at 3:49
• I believe your book meant to say $R^2$, you should ask for confirmation to your teacher. Feb 24, 2017 at 3:49