| Line 60: | Line 60: | ||
<PRE> | <PRE> | ||
| − | [ x ] [ -2.5 ] | + | [ x ] [ -2.5 ] |
| − | [ y ]=[ 1.375 ] t | + | [ y ] = [ 1.375 ] t |
| − | [ z ] [ 1 ] | + | [ z ] [ 1 ] |
</PRE> | </PRE> | ||
| + | |||
| + | Thus, you can find the solution from the row echelon matrix, but I wouldn't say that you can read it off from there -- not without practice, at least. | ||
| + | |||
| + | |||
---- | ---- | ||
Revision as of 08:43, 5 September 2013
Homework 3 collaboration area
Question from James Down Under (Jayling):
For Page 329 Question 11. Am I meant to calculate all eigenvalues and eigenvectors or just calculate the eigenvector corresponding to the given eigenvalue of 3?
Answer from Steve Bell :
Yes, you are only supposed to find the eigenvector for lambda=3. (The idea here is to spare you from finding the roots of a rather nasty 3rd degree polynomial.)
Question from a student:
Let 3x+4y+2z = 0; 2x+5z= 0 be the system for which I have to find the basis.
When Row Reduced the above system gives [ 1 0 2.5 0 ; 0 1 -1.375 0].
Rank = no of non zero rows = 2 => Dim(rowspace) = 2 ; Nullity = # free variables = 1
Q1: Aren't [ 1 0 2.5] and [0 1 -1.375] called the basis of the system?
A1 from Steve Bell:
Those two vectors form a basis for the ROW SPACE.
The solution space is only 1 dimensional (since the number of free variables is only 1).
Q2: Why is that we get a basis by considering the free variable as some "parameter" and reducing further(and get 1 vector in this case). Isn't that the solution of the system?
A2 from Steve Bell :
If the system row reduces to
[ 1 0 2.5 0 ] [ 0 1 -1.375 0 ]
then z is the free variable. Let it be t. The top equation gives
x = -2.5 t
and the second equation gives
y = 1.375 t
and of course,
z = t.
So the general solution is
[ x ] [ -2.5 ] [ y ] = [ 1.375 ] t [ z ] [ 1 ]
Thus, you can find the solution from the row echelon matrix, but I wouldn't say that you can read it off from there -- not without practice, at least.
