Lecture 6 Notes

And each equation describes a plane in this case only

This preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 0 1 00 and −1/3 5/3 0 5 x2 + x3 = 0 3 (because it doesn’t have a leading one). and Each equation describes a plane. In this case, only two of them really matter. x3 can be whatever Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and Ax = 0 . 5 x2 + x3 = 0 3 • Thus, the solution can be written as and x3 can be whatever. . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = x3 3 • Thus, the solution can be written as . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = x3 3 5 x2 = − x3 3 • Thus, the solution can be written as . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = x3 3 5 x2 = − x3 3 x3 = C • Thus, the solution can be written as . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 x3 = C • Thus, the solution can be written as . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as . Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 x3 = C Ax = 0 . 5 x2 + x3 = 0 3 and 1 x1 = C 3 5 x2 = − C 3 x3 can be whatever. Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 C −5 . x= 3 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1 − x3 = 0 3 and 1 x1 = x3 3 5 x2 = − x3 3 Ax = 0 . 5 x2 + x3 = 0 3 and x3 can be whatever. 1 x1 = C 3 5 x2 = − C 3 x3 = C • Thus, the solution can be written as 1 ￿ . x = C −5 3 Solutions to homogeneous matrix equations • Example 1. Solve the equation • so 1 x1...
View Full Document

This note was uploaded on 02/12/2014 for the course MATH 256 taught by Professor Ericcytrynbaum during the Spring '13 term at UBC.

Ask a homework question - tutors are online