#$&*
course MTH-279
08/06 around 4:10 PM
Query 28 Differential Equations*********************************************
Question: Diagonalize the matrix [2, 3; 2, 3].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
just completed one problem here:
T^-1 AT = D
p(lambda) = det [ A-lambda I] = 0
---
det [2-lambda , 3; 2, 3 - lambda]
= (6- 2 lambda - 3lambda + lambda^2) - 6
= lambda^2 - 5lambda = 0
lambda(lambda - 5) = 0
lambda 1 = 0, lambda 2 = 5
---
use A - lambda I * x1vector = 0
---
we find x1 = -3 and x2 = 2, so x1vector = [-3;2]
---
repeat for lambda 2
--
we find x1 = 1 and x2 = 1, so x2vector = [1;1]
--
T = [-3, 1 ; 2, 1]
(-3 * 1) - (2 * 1) = -5
--
T^-1 = -1/5 [ 1, -1; -2, -3]
[-1/5, 1/5; 2/5, 3/5]
---
T^-1 AT = D
[-1/5, 1/5 ; 2/5, 3/5] * [2, 3 ; 2, 3] * [-3, 1; 2, 1] = D
---
simplify this into
[0, 0; 2, 3] * [-3, 1; 2, 1]
=[0, 0; 0, 5] = D
Confidence rating: 3
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): OK
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course MTH-279
08/06 around 4:10 PM
Query 28 Differential Equations*********************************************
Question: Diagonalize the matrix [2, 3; 2, 3].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
just completed one problem here:
T^-1 AT = D
p(lambda) = det [ A-lambda I] = 0
---
det [2-lambda , 3; 2, 3 - lambda]
= (6- 2 lambda - 3lambda + lambda^2) - 6
= lambda^2 - 5lambda = 0
lambda(lambda - 5) = 0
lambda 1 = 0, lambda 2 = 5
---
use A - lambda I * x1vector = 0
---
we find x1 = -3 and x2 = 2, so x1vector = [-3;2]
---
repeat for lambda 2
--
we find x1 = 1 and x2 = 1, so x2vector = [1;1]
--
T = [-3, 1 ; 2, 1]
(-3 * 1) - (2 * 1) = -5
--
T^-1 = -1/5 [ 1, -1; -2, -3]
[-1/5, 1/5; 2/5, 3/5]
---
T^-1 AT = D
[-1/5, 1/5 ; 2/5, 3/5] * [2, 3 ; 2, 3] * [-3, 1; 2, 1] = D
---
simplify this into
[0, 0; 2, 3] * [-3, 1; 2, 1]
=[0, 0; 0, 5] = D
Confidence rating: 3
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): OK
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
#$&*
course MTH-279
08/06 around 4:10 PM
Query 28 Differential Equations*********************************************
Question: Diagonalize the matrix [2, 3; 2, 3].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
just completed one problem here:
T^-1 AT = D
p(lambda) = det [ A-lambda I] = 0
---
det [2-lambda , 3; 2, 3 - lambda]
= (6- 2 lambda - 3lambda + lambda^2) - 6
= lambda^2 - 5lambda = 0
lambda(lambda - 5) = 0
lambda 1 = 0, lambda 2 = 5
---
use A - lambda I * x1vector = 0
---
we find x1 = -3 and x2 = 2, so x1vector = [-3;2]
---
repeat for lambda 2
--
we find x1 = 1 and x2 = 1, so x2vector = [1;1]
--
T = [-3, 1 ; 2, 1]
(-3 * 1) - (2 * 1) = -5
--
T^-1 = -1/5 [ 1, -1; -2, -3]
[-1/5, 1/5; 2/5, 3/5]
---
T^-1 AT = D
[-1/5, 1/5 ; 2/5, 3/5] * [2, 3 ; 2, 3] * [-3, 1; 2, 1] = D
---
simplify this into
[0, 0; 2, 3] * [-3, 1; 2, 1]
=[0, 0; 0, 5] = D
Confidence rating: 3
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary): OK
"