Solutions 2

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

This preview shows pages 1–3. Sign up to view the full content.

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

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

View Full Document
2. Consider the following vectors: u = 3 1 0 2 v = 1 1 2 - 3 w = 2 0 - 2 5 x = 6 2 0 4 y = 1 0 0 4 (a) Find a basis for the subspace S = Span { u , v , w , x , y } . (b) Is the vector 1 0 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 0 2 0 0 2 - 2 0 0 2 - 3 5 4 4 -→ 1 0 1 2 0 0 1 - 1 0 0 0 0 0 0 1 0 0 0 0 0 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 a dependency. Since row operations preserve the dependencies among the columns, we know that columns 1, 2, and 5 from the
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern