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.
 Spring '13
 EricCytrynbaum
 Differential Equations, Geometry, Equations

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