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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.
