ST512
1.
Quiz1
SSII-2010
(40 pts.) Below are two matrices A and B. Find, if possible, each requested item. If not
possible, briefly explain why:
 4 1
A   2 1
 1 1
 3 2 1
B

 1 0 1 
a) (8 pts)
A + B  NP, A is 3 × 2 and B is is 2 × 3
b) (8 pts)
 4 1
 3 2 1 
17 0 
BA  
2 1  



 1 0 1  
 5 0


1
1


c) (8 pts)
 4 1  3 1  7 0 
 3 2 1 
 
 

A  B  
  2 1   2 0    4 1 

1
0
1


 1 1  1 1   2 2
d) (8 pts)
B1  NP, since B is not square.
e) Rank of the matrix B is 2 C1= (-1)*(C2+C3))
2. (24 pts.) Here are two new matrices A and B.
1
1
A
1

1
0 0 3
1 0 0 
1 1 0

1 1 1
 0.2 1.2 0.6 0.6 
 0.2 0.2 0.6 0.6 

B= 
 0
1
1
0 


 0.4 0.4 0.2

a) (8 pts.) Fill in the missing element in B to make B the inverse of A. (Note, AA1  I )
July 8, 2010
Page 1
ST512
Quiz1
SSII-2010
Answer 1. Product of first row and last column should be 0, thus missing value is
0  1   0.6    0    0.6    0    0    3   x 
given by
 0.6  3x  0
x
0.6
 0.2
2
Or
1  1   0.6   1   0.6   1   0   1   x 
 0.6  0.6  x  1
x  0.2
If you use the other alternatives,
second_row of A times fourth_column of B equal 0,
1   0.6  1   0.6   0   0   0   0  0
And similarly for third_row of A times fourth_column of B equal 0,
in any case, you are using the properties of the product of a matrix and its inverse.
b) (8 pts.) Compute, if possible, the rank of matrix B. rank  B   NP, since missing
value.
c) (8 pts.) Compute, if possible, from the given information, B1  A or NP based on
answer to 1. Since we do not have information about the missing value.
d) Using row and matrices operations, B turns to be nonsingular when missing
value is -12/10, since x=1 and x4=3 , C1+3C4=0. And, when missing value is -0.2
B has an inverse.
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ST512
Quiz1
SSII-2010
3. (36 pts.) Let Gi denote the weight gain for the firstborn male piglet in litter i over a given
period of time and let Pi denote the number of other piglets (siblings) in the same litter. I
observed n litters of pigs and fit a least squares line relating first born male weight gain G
to number of siblings P so my model was Gi  0  1Pi  ei . The estimated model came
out to Gˆ  80  0.5P . My n 2 X matrix had a column of 1s and a column of Pi values as
i
i
usual. I obtained
 n
 1
i 1
XX   n

  Pi
 i 1
 50 110 
XX  

110 260

n
P 
 n

 Sum _ Pi
Pi 2  

i 1

i 1
i
n
Sum _ Pi 
Sum _ Pi 2 
Answer these questions below, if possible from this information (if not possible, put NP)
(a) (7 pts.) How many observations (litters) did I have in my experiment? n = _50___
(b) (7 pts.) Compute, if possible, the sum of the n weight gains in this experiment
n
G
i 1
i
 3945
Use fact: intercept = G_mean – slope*P_mean
b0  G  b1 P
110
50
110
G  80  0.5 
50
90
110 

Gi  50   80  0.5 

  3945
50 

i 1
80  G   0.5  
or
80 = G_mean – (-0.5)*(110/50) ,
thus, G_mean = 80 - 0.5*(110/50)
G_SUM = G_Mean*50= 80*50 - 0.5*110=4000-55=3945
Alternative solution
July 8, 2010
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ST512
Quiz1
SSII-2010
b   XX  XY
1
 XX  b   XX  XX 
1
XY  XY
Note that in our case P takes the role of X, explanatory variable and G takes the role of Y,
dependent variable.
 50 
 Yi
 50 110   80  3945  i 1 
XY  


   50 
110 260   0.5 8670  Y 2 
 i 
 i 1 
n
Thus,
G
i 1
i
 3945
(c) (9 pts.) Compute 90  XX  
1
1
 50 110 
 260 110 
1
1  260 110   260 / 900 110 / 900 



 XX   



 50  260  110 110   110 50   900   110 50   110 / 900 50 / 900 
110 260 
 0.288889 0.122222 


 0.122222 0.055556 
 260 / 900 110 / 900 
1
90  XX   90 

 110 / 900 50 / 900 
1
 26 11


 11 5 
(d) (13 pts.) We have seen that the estimated intercept b 0 and slope b1 together form a random
vector that varies from sample to sample with variance-covariance matrix  XX   2 . Assuming
1
 2 is known to be 90, compute the covariance between b 0 and b1 , as well as the variance of W ,
where W  2b0  3b1  18
July 8, 2010
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ST512
Quiz1
SSII-2010
cov(b)   XX   2
1
 2  90
 26 11
cov(b)  

 11 5 
var  b0   26
var  b0   5
co var  b0 , bq   11
W  2b0  3b1  18
b 
W   2 3  0   18
 b1 
2
var(W )   2 3 cov(b)  
3
 26 11  2 
2
  2 3 
 19 7   



 11 5   3 
3
 17
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