In linear algebra we have been talking a lot about the three elementary row operations. What I don't understand is why we can't multiply by any column by a constant? Since a matrix is really just a grouping of column vectors, shouldn't we be able to multiply a whole column by a constant but maintain the same set of solutions for the original and resulting matrices?
Answer
Let's say we have some relation, like
$\left( \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z\end{array}\right) = \left( \begin{array}{ccc} 10 \\ 11 \\ 12\end{array} \right)$
This has a lot of information here. For instance, it says that $x + 2y + 3z = 10$. In fact, one can uniquely solve for $x, y, z$ from this relationship. But suppose we multiply the first column by 3:
$\left( \begin{array}{ccc} 3 & 2 & 3 \\ 12 & 5 & 6 \\ 21 & 8 & 9 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z\end{array}\right) = \left( \begin{array}{ccc} 10 \\ 11 \\ 12\end{array} \right)$
Both of these cannot be correct (you can fully solve it out if you want, in fact, I encourage it). For instance, knowing that $x + 2y + 3z = 10$ and that $3x + 2y + 3z = 10$ tells us that $x = 0$. That's not so good - clearly, it is not always the case (and it's not here, either). Further, this would result in 3 non-dependent equations in 2 variables - no solution exists.
The main idea is that matrices hide linear equations, and the way in which they hide them is dependent on the laws of matrix multiplication. It's linear with respect to row operations, because that's the same as redundant operations. But it's not the same with respect to column operations. If we wanted to change rows, then we could use matrices... if we multiply on the right instead of the left. Again, it's an artifact of the way in which matrices multiply.
No comments:
Post a Comment