{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math_294Prelim2solnew.08

# Math_294Prelim2solnew.08 - Math 294 Prelim 2 solution...

This preview shows pages 1–2. Sign up to view the full content.

Math 294 Prelim 2 solution Spring 2008 1. The matrix A of a linear transformation T from 2 3 R R is 1 1 1 0 0 1 A = 1a. Find a basis for Im A , what is the dimension of Im A T ? 1b. Find a basis for Ker A T 1c Is the equation Ax b = G G consistent where (3,3,3) T b = G , if not, find the least square solution of Ax b = G G . Solution : (This problem is a modification of a HW in 5.4, pb20) 1a: the two columns of A are linear independent vectors, so Im A is the span of these two column vectors. Since the rankA T = rankA, dim Im A T =dim Im A = 2. 1b: The Ker A T is the solution of the equation 1 1 0 1 0 1 0 x = G G It is also the vector subspace of 3 R which is orthogonal to Im A . By inspection, the vector ( 1,1,1) T is orthogonal to the columns of A which spans the ImA. Since the KerA T = ( ) Im A and the dim(Im A) = 2, the dim( KerA T ) = 1, so ( ) ( 1,1,1) T T KerA span = . 1c. A simple calculation using RREF shows that the system Ax b = G G is inconsistent. The solution of 1c is a HW problem, pb. 20, section 5.4. 2. Solution : 2a: [ ] 1 1 2 3 1 1 2 1 1 1 2 1 0 2 2 M f f f M     = = + + =       B where 1 2 3 1 0 1 1 0 0 , , 0 0 0 0 0 1 f f f = = =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}