Download Linear Algebra Problem Set 1 Solutions

Linear Algebra Problem Set 2 Solutions
38 14 
 25 / 54  7 / 27
1
1. PQ  
,
(
PQ
)


 32 / 27 19 / 27 
64 25


4
/
27
1
/
27
1
/
2

3
/
2




 25 / 54  7 / 27
1
1 1
,
,
P 1  
Q

Q
P


 1
 32 / 27 19 / 27 
4 
 7 / 27 5 / 27



2. Assume that there is an inverse and then try and find it. If there is an inverse then:
 0 B   X Y   I 0
C D W Z   0 I 


 

If we can find X, Y, W and Z then we have found the inverse. Do the
multiplication on the left and set it equal to the right.
 0 X  BW 0Y  BZ   I 0
CX  DW CY  DZ   0 I 

 

0 X  BW  I  BW  I  W  B 1
0Y  BZ  0  BZ  0  Z  0
CX  DW  0  CX  DB 1  0  CX   DB 1  X  C 1 DB 1
CY  DZ  I  CY  D0  I  CY  I  Y  C 1
So the inverse is:
1
B
 C 1 DB 1 C 1 



D 
B 1
0 

C and B must be invertible.
0
C

  3 4  x   2
 5 / 31 4 / 31
3. 
   , A 1  



,
 4 5   y  1 
 4 / 31 3 / 31
 6 / 31
A 1b  x  
  x  6 / 11, y  11 / 31
 11 / 31 
4.
Check by showing AA1  I . Details are below.
5.
a) Let A be a square matrix with an ith row of zeros. Let B  A1 . Then AB  I . The
ith row of AB  ith row of A * B = [0 0 0 ….. 0]. Since AB  I then ( AB) ii  1 ,
but ( AB) ii  0 . Thus a matrix with a row of all zeros is not invertible.
b) A similar argument is used to show if a matrix has a column of zeros it is not
invertible. Use the fact that BA  I and reach a contradiction.
6.
a) False: The zero matrix is a counterexample. Any non-invertible matrix will not
be able to be written as the product elementary matrices.
3 0
1 0
b) False: Let E1  
and E 2  

 are both elementary matrices, but
0 1 
3 1
3 0
E1 E 2  
 is not an elementary matrix.
3 1
c) True: Multiplying a row of matrix A by a constant and adding it to another row is
an elementary row operation and can be written as E1 A . If A is invertible then this
product is invertible because an elementary matrix is invertible and the product of
invertible matrices is invertible.
d) True: AB  0  A 1 AB  A 1 0  IB  0  B  0
7.
 1  2  1 b1 
 4 5
2 b2 

 4 7
4 b3 
After Gauss Elimination the matrix becomes:
1 0 0  3b1  1 / 2b2  3 / 2b3 
0 1 0

 4b1  b3


0 0 1 4b1  1 / 2b2  3 / 2b3 
The system is consistent for all b’s so no restrictions.
x1  3b1  1 / 2b2  3 / 2b3
x 2  4b1  b3
x3  4b1  1 / 2b2  3 / 2b3