In Exercises 1017, evaluate the determinant of the given matrix by reducing the matrix to row echelon form.

MATH 10B Lecture #10 Notes Ivan Lopez October 25 2016 End of Section 6.1: Line Integrals 3 De▯nition: A simple curve in the image (range) C of a map c : [a;b] ! R (or R ), which is 1-1 (one-to-one, which is to say that C(t ) =1 6 C(t 2 if t16= t2and does not have any self-intersections except possibly at endpoints). In addition, the curve is a simple closed curve if c(a) = c(b). Remark: A curve is called simple if it is both x-simply and y-simple. Let us now focus on how line integrals are independent of reparametrization. From the this image, we can make the following statements: We have a curve C, where we have c : [a;b], we also have h : [a ;b ] 1hi1h can be parametrized to follow c. We can simplify this with P, which can be written as P = c ▯ h : [a1;b1] ! R (or R ) 2 Theorem: A vector line integral is independent of the choice of the parametriza- tion of C compatible with the orientation of C (i.e., dh > 0 8t 2 [a ;b ]) dt 1 1 The same is true for a scalar line integral but we simply need dh 6= 0 8t (We do dt not need compatibility with the orientation of C) Proof. Say we are in 2D, to make notation simpler. Assume F is C ~ 1 R Rb R b F ▯ ds = F(x(t);y(t))s (t)dt = F(c(t)) ▯ c (t)dt, F(x(t);y(t)) = F(c(