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October 2

Announcements

  • Watch 2nd and 3rd 3blue1brown videos

  • Section 1.1, 1.2 due this Thursday

  • Worksheet 1 due Friday

  • Section 2.1, 2.2 due next Thursday

  • Office hours in Padelford C-8D

    • Wednesday 4:30 - 5:30

    • Thursday 12:00 - 1:00

Continue from last class

Justify the elementary row operations in class.

Gaussian elimination

Definition: The pivot positions are positions that contain a leading term. The pivot columns are columns that contain a pivot position. A pivot is the value of a pivot position.

Algorithm: Gaussian elimination is performed as follows:

  • find the pivot position in the first row

  • use elementary row operators to eliminate all value under the pivot position

  • continue

work out example in class

Reduced echelon form

Definition: A matrix is in reduced echelon form if

  • it is in echelon form

  • all pivot positions contain a 1

  • the only nonzero term in a pivot colum is in the pivot position

Algorithm: Gauss-Jordan elimination is performed as follows:

  • do Gaussian elimination

  • divide each row by the value of its pivot

  • eliminate all other values in pivot column.

work out example in class.

Homogenous linear systems

A linear system is homogenous if the numbers to the right of the equal sign are all zero. They always have the trivial solution

2.1 Vectors

A vector is a list of number with addition and scalar multiplication defined. Given vectors u=(u1,u2,,un)Rnu=(u_1, u_2, \ldots, u_n)\in \mathbb{R}^n, v=(v1,v2,,vn)Rnv=(v_1,v_2,\ldots,v_n)\in \mathbb{R}^n of equal dimension and a scalar cRc\in \mathbb{R}, we define

  • addition: u+v=(u1+v1,u2+v2,,un+vn)u+v=(u_1+v_1, u_2+v_2, \ldots, u_n+v_n),

  • scalar multiplication: cu=(cu1,cu2,,cun)cu=(cu_1,cu_2,\ldots,cu_n).

go over the geometry in class. tail to tip, parallelogram

Let a,ba,b be scalars and u,v,wRnu,v,w\in \mathbb{R}^n. Then

  • u+v=v+uu+v=v+u,

  • a(u+v)=au+ava(u+v)=au+av,

  • (a+b)u=au+bu(a+b)u=au+bu,

  • (u+v)+w=u+(v+w)(u+v)+w = u+(v+w),

  • a(bu)=(ab)ua(bu)=(ab)u,

  • u+(u)=0u+(-u)=0,

  • u+0=0+u=uu+0=0+u=u,

  • 1u=u1u=u.

Definition: The If u1,u2,,umu_1,u_2,\ldots,u_m are vectors and c1,c2,,cmc_1,c_2,\ldots,c_m are scalars, then c1u1+c2u2++cmum c_1u_1+c_2u_2+\ldots+c_mu_m is a linear combination of u1,,umu_1,\ldots,u_m. Note that the constants can be negative or zero.

give examples in class.