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Last years exam Q3

Posted by sherylm 
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Last years exam Q3
October 17, 2006 02:16PM
Does anyone have a solution for Question3?
Re: Last years exam Q3
October 18, 2006 10:08AM
(right-click/view source for formatted text)

b = [1 1 0 1]T; f = [0 2 0 1]T
(a)
|1/2 0 0 0||1| |1/2|
| 0 1 0 0||1|=| 1 |
| 0 0 1 0||0| | 0 |
| 0 0 0 1||1| | 1 | and

|1/2 0 0 0||0| |0|
| 0 1 0 0||2|=|2|
| 0 0 1 0||0| |0|
| 0 0 0 1||1| |1|

(b)
sin(45) = 1/sqrt(2) but positive rotation is anti-clockwise and we're asked for
clockwise rotation: sin(360-45) = sin(360)cos(45)-cos(360)sin(45) = -sin(45)
= -1/sqrt(2)
cos(360-45) = sin(360)sin(45)+cos(360)cos(45) = +1/sqrt(2)
(so the hint is actually a trick - if you'd just remembered the signs wrong you
would have the calculated the answer correctly).

|1/sqrt(2) 0 -1/sqrt(2) 0||1/2| |1/(2sqrt(2))|
| 0 1 0 0|| 1 |=| 1 |
|1/sqrt(2) 0 1/sqrt(2) 0|| 0 | |1/(2sqrt(2))|
| 0 0 0 1|| 1 | | 1 | and

|1/sqrt(2) 0 -1/sqrt(2) 0||0| |0|
| 0 1 0 0||2|=|2|
|1/sqrt(2) 0 1/sqrt(2) 0||0| |0|
| 0 0 0 1||1| |1|

(c)
|1 0 0 0||1/(2sqrt(2))| | 1/(2sqrt(2)) |
|0 1 0 0|| 1 |=| 1 |
|0 0 1 2||1/(2sqrt(2))| |1/(2sqrt(2)) +2|
|0 0 0 1|| 1 | | 1 | and

|1 0 0 0||0| |0|
|0 1 0 0||2|=|2|
|0 0 1 2||0| |2|
|0 0 0 1||1| |1|
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