Lecture 6 Notes

And each equation describes a plane in this case only

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

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