Dot Product, Cross Product, Determinants
... A2 = (a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2 . Hence A = k~xk with the vector ~x defined by ~x := (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ). Note that x j = ak b` − a` bk where 1. j, k, ` are different 2. k, ` are in “positive order” if we arrange 1,2,3 on a circle. This vec ...
... A2 = (a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2 . Hence A = k~xk with the vector ~x defined by ~x := (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ). Note that x j = ak b` − a` bk where 1. j, k, ` are different 2. k, ` are in “positive order” if we arrange 1,2,3 on a circle. This vec ...
ACP Algebra II Summer Review Packet
... 8. The formulas to find the area of an equilateral triangle are ...
... 8. The formulas to find the area of an equilateral triangle are ...
Universal Identities I
... matrices come from the 4 matrix entries on both sides of the identity ...
... matrices come from the 4 matrix entries on both sides of the identity ...
Math 416 Midterm 1. Solutions. Question 1, Version 1. Let us define
... But when we write it this way, it is clear that we can: adding the second row to the first puts this in RREF, and we will have three pivots. (We could work harder for the actual solution, but that is not needed here.) So now we need to find a subset that is a basis. Since we know that any basis of P ...
... But when we write it this way, it is clear that we can: adding the second row to the first puts this in RREF, and we will have three pivots. (We could work harder for the actual solution, but that is not needed here.) So now we need to find a subset that is a basis. Since we know that any basis of P ...
