05aut-m1sols

05aut-m1sols - MATH 51 MIDTERM October 20, 2005 Name:...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 51 MIDTERM October 20, 2005 Name: Numeric Student ID: Instructors Name: I agree to abide by the terms of the honor code: Signature: Instructions: Print your name, student ID number and instructors name in the space provided. During the test you may not use notes, books or calculators. Read each question carefully and show all your work ; full credit cannot be obtained without sufficient justification for your answer unless explicitly stated otherwise. Underline your final answer to each question. There are ?? questions. You have 90 minutes to do all the problems. Question Score Maximum 1 10 2 15 3 10 4 10 5 10 6 10 7 10 8 10 Total 85 Question 1 of ?? , Page 2 of ?? Solutions 1. Solve the following system of equations in the variables x, y, z, w : x- y- z + w = 5 y- z + 2 w = 8 2 x- y- 3 z + 4 w = 18 If a solution exists, express your answer in parametric form. Solution: Writing this in matrix form, we obtain 1- 1- 1 1 1- 1 2 2- 1- 3 4 5 8 18 . We will solve this system of equations by performing Gaussian elimination. We proceed as follows. First, subtract twice the top row from the third row. This operation yields 1- 1- 1 1 1- 1 2 1- 1 2 5 8 8 . Now subtract the second row from the third row, which yields 1- 1- 1 1 1- 1 2 5 8 . At this point, we can read off the solution. First, the variables z and w are free. Next, y = z- 2 w and x = ( z- 2 w ) + z- w = 2 z- 3 w . Expressed parametrically, this is 2 z- 3 w z- 2 w z w z, w R or z 2 1 1 + w - 3- 2 1 z, w R . Question 2 of ?? , Page 3 of ?? Solutions 2. Let A be the matrix 1 1 1 4 1 2 1 1 6 1 1 1 3 2 2 1 7 . (a) Find a basis for the nullspace of A . Solution: For (a), we transform the matrix A to its RRE form: 1 1 1 4 1 2 1 1 6 1 1 1 3 2 2 1 7 7 1 1 1 4 1 1 2 1 1 1 3- 1- 1 7 1- 1 1 2 1 1 2 1 1- 1- 1 7 1- 1 1 1 1 2 1 1 . There are therefore two free variables x 3 and x 5 and three fixed variables x 1 , x 2 and x 4 . The homogeneous linear system associated to the RRE form is then: x 1- x 3 + x 5 = 0 x 2 + x 3 + 2 x 5 = 0 x 4 + x 5 = 0 ....
View Full Document

This note was uploaded on 01/12/2010 for the course MATH 51 at Stanford.

Page1 / 11

05aut-m1sols - MATH 51 MIDTERM October 20, 2005 Name:...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online