1111_08_t3sol

1111_08_t3sol - MATH1111/24Oct08/Test3 1 Algebra - Test 3...

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

View Full Document Right Arrow Icon
MATH1111/24Oct08/Test3 1 MATH1111 Linear Algebra - Test 3 Solution Outline 1. (10 marks) Let M = 1 - 1 0 - 2 3 1 1 0 1 1 2 3 . (a) Find the nullspace of M . (b) Find a basis for the column space of M . Ans . (a) (5 marks) Apply the elementary row operations: 2 R 1 + R 2 , - R 1 + R 3 , - R 1 + R 4 , M is reduced to 1 - 1 0 0 1 1 0 1 1 0 3 3 . Then it is further reduced to 1 - 1 0 0 1 1 0 0 0 0 0 0 . Solutions of Mx = 0 are α ( - 1 - 1 1) T where α is any real number. Therefore, N ( M ) = span(( - 1 - 1 1) T ). [So easy. Don’t blame on anybody if you lose these marks.] (b) (5 marks) From the row echelon form in part (a), the first two columns (1 - 2 1 1) T and ( - 1 3 0 1) T form a basis for the column space of M . [Easy!? See the textbook or lecture 19 slides.] 2. (15 marks) Let P 2 be the vector space of all linear polynomials (i.e. of degree 1). Consider the two ordered bases
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/06/2010 for the course MATH MATH101 taught by Professor Chan during the Spring '09 term at HKUST.

Page1 / 4

1111_08_t3sol - MATH1111/24Oct08/Test3 1 Algebra - Test 3...

This preview shows document pages 1 - 2. Sign up to view the full document.

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