Solutions 2 - Math 601 Solutions to Homework 2 1 Consider...

Info iconThis preview shows pages 1–3. 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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 601 Solutions to Homework 2 1. Consider the following matrix: A = 1 2 1 2 5 4 1- 1- 4 (a) Compute A- 1 . (b) Use A- 1 to solve A x = 2- 1 2 . Answer: (a) First we compute A- 1 : we create the augmented matrix ( A | I ) and row reduce. 1 2 1 1 0 0 2 5 4 0 1 0 1- 1- 4 0 0 1 -→ 1 0 0- 16 7 3 0 1 0 12- 5- 2 0 0 1- 7 3 1 Thus, A- 1 = - 16 7 3 12- 5- 2- 7 3 1 (b) We can solve for x by multiplying both sides of the matrix equation by A- 1 . We get x = A- 1 2- 1 2 = - 16 7 3 12- 5- 2- 7 3 1 2- 1 2 = - 33 25 15 1 2. Consider the following vectors: u = 3 1 2 v = 1 1 2- 3 w = 2- 2 5 x = 6 2 4 y = 1 4 (a) Find a basis for the subspace S = Span { u , v , w , x , y } . (b) Is the vector 1 2- 1 in S ? Answer: (a) Here are two methods to find a basis for S . Method 1: We create a matrix with the vectors as columns, and then row reduce: 3 1 2 6 1 1 1 2 2- 2 2- 3 5 4 4 -→ 1 1 2 1- 1 1 In the row reduced matrix, we see that the columns 1, 2, and 5 are linearly independent (these are the columns containing piv- ots), and that adding any other column to this list would create...
View Full Document

{[ snackBarMessage ]}

Page1 / 6

Solutions 2 - Math 601 Solutions to Homework 2 1 Consider...

This preview shows document pages 1 - 3. Sign up to view the full document.

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