Elementary linear algebra?

2015-01-14 3:07 pm
Consider the matrices
[2 1 2] [x1]
A = [2 2 -2] and x= [x2]
[3 1 1] [x3]

(a)Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use the result to solve Ax = x for x.
(b)Solve for Ax= 4x.

I have no idea how to figure this question out.
please help.

回答 (2)

2015-01-14 4:24 pm
(a) AX=X=>AX=IX=>(A-I)X=0=>
. || 2 1+2 ||...|| 1 0 0 ||....|| x1 ||
( || 2 2 -2 || - || 0 1 0 || )*|| x2 ||=0
. || 3 1+1 ||...|| 0 0 1 ||....|| x3 ||
=>
|| 1 1 2 ||..|| x1||
|| 2 1-2 ||*|| x2 ||=0
|| 3 1 0 ||..|| x3 ||
=>
x1+x2+2x3=0
2x1+x2-2x3=0
3x1+x2+0x3=0
=>x1=0, x2=0, x3=0
=>X=0 [since del (A-I) =/=0]

(b) AX=4X=>(A-4I)X=0=>
del(A-4I)=
| -2+1+2 |=| -2+1+2 |=(-1)| 0+0 |
|+2 -2 -2 |..|+0 -1+0 |........|+3-3 |
|+3+1 -3 |..|+3+1 -3 |
=>del(A-4I)=0
-2x1+x2+2x3=0
2x1-2x2-2x3=0
3x1+x2-3x3=0
=>
2x1-2x3=0
3x1-3x3=0
=>
x1=x3; x2=0
=>
X=|| x1 || (infinitely many solutions)
.....||+0 ||
.....|| x3 ||
2015-01-14 3:23 pm
a) Ax = x
Ax - x = 0
(A-I)*x = 0

A-I =
[2 1 2]-[1 0 0]
[2 2 -2]-[0 1 0]
[3 1 1] -[0 0 1]

A-I =
[1 1 2]
[2 1 -2]
[3 1 0]

System of equations looks like
x + y + 2z = 0
2x + y -2z = 0
3x + y = 0

http://www.solvemymath.com/online_math_calculator/algebra_combinatorics/system_of_equations/index.php
x = 0
y = 0
z = 0

b) Ax = 4x
Ax - 4x = 0
(A-4I)*x = 0

A-4I =
[2 1 2]-[4 0 0]
[2 2 -2]-[0 4 0]
[3 1 1]-[0 0 4]

A-4I =
[-2 1 2]
[2 -2 -2]
[3 1 -3]

Our system of equations loks like this:
(A-4I)*x = 0
-2x+y+2z = 0
2x-2y-2z = 0
3x+y-3z = 0

The determinant of (A-4I) is 0, so there is no solution.


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