4 ms·
> Indeed one would not solve the scalar (n = 1) system 7x = 21 by computing x = 7^-1 × 21 but rather would carry out a division x = 21 / 7 I remember learning
by _gabe_ 4y ago
> Indeed one would not solve the scalar (n = 1) system 7x = 21 by computing x = 7^-1 × 21 but rather would carry out a division x = 21 / 7
I remember learning in algebra to solve this equation exactly the way he described and said we don't use. You multiply both sides by 1 / 7 which cancels the 7 on the left side, because 1 / 7 is the inverse of 7.
Now I just implicitly divide both sides by 7, but I'm still solving the equation by using the inverse of 7...
- eru 4y agoIn linear algebra finding a general inverse of a matrix is harder than solving a specific equation.
- flqn 4y agoI think the difference is that you did the inversion symbolically, not numerically. This is about numeric computation not applying algebraic manipulations.
- nh23423fefe 4y agook so what is sqrt(7)x = 21? well you just 1/sqrt(7) * sqrt(7)x = 1/sqrt(7) * 21 so x = sqrt(7)^2 * 3 / sqrt(7) = sqrt(7) * 3 but you didn't compute the inverse did you, and neither did i. you factored 21 and used the cancellation law (ax = ay => x = y)
- FabHK 4y agoBut that is sort of the point. In theory, it's the same. But that's not how you should actually compute it. The cheap way is to compute 21/7. The hard way is to compute 1/7 (one floating point operation), and then multiply it by 21 (another floating point operation). With matrices, the discrepancy in work and possibly precision between "solve Ax=b" and "compute x = A^-1 b" can be very large.