1. No category

MA1101R
Linear Algebra I
AY 2009/2010 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Chang Hai Bin
MA1101R Linear Algebra I
AY 2009/2010 Sem 1
Question 1
(a)
(i) Assume
the number of orange, grapefruit, and mango bought be x, y, z respectively.
x
+y
+z
= 100
0.5x +1.1y +1.5z = 200
(ii) Note that the most expensive item is mango, and even if Dr. Ng spend all of his $ 200 on
mango, he still could not spend all of his money. So, it is not possible for Dr. Ng to make his
purchases.
In terms of mathematics, if x, y, z ∈ N, x + y + z = 100, then 0.5x + 1.1y + 1.5z ≤ 1.5x +
1.5y + 1.5z = 1.5(x + y + z) = 1.5 × 100 = 150 < 200 which is a contradiction.




k 1 1 0
1 2k 1 1
row−swap
 0 1 1 1 
(b)  1 2k 1 1 
→
0 1 1 1
k 1 1 0


1
1 2k
1

 0 1
1
1
2
2
0 0 2k − k 2k − k − 1

1
1
2k
1
 0
1
1
1 
2
0 1 − 2k 1 − k −k

R3 −kR1
→
R3 −(1−2k2 )R2
→
(i) The system is inconsistent when 2k 2 − k = 0, because from the third row of the REF, we will
get the equation 0x1 + 0x2 + 0x3 = −1.
So, the system has no solution when k = 0 or k = 21
(ii) For k 6= 0, 21 , there will be 3 pivot-columns, and hence there is a unique solution.
(iii) There is no k ∈ R such that the linear system has infinitely many solutions, since it is either
inconsistent (k = 0, 21 ) or has a unique solution (k 6= 0, 12 )


k 1 1
Alternatively, det  1 2k 1  = 2k 2 − k.
0 1 1
1
Case 1: k 6= 0 or 2 . Then the inverse of the matrix exists and is unique, and the system has a

 
−1  
x1
k 1 1
0
unique solution of  x2  =  1 2k 1  ·  1 
x3
0 1 1
1

x2 + x3 = 0

x1 +
+ x3 = 1 which is
Case 2: k = 0. Then the original linear system becomes

x2 + x3 = 1
inconsistent.
 1
 2 x1 + x2 + x3 = 0
x1
+ x2 + x3 = 1 Substituting Row 3 into Row 1 and 2, we
Case 3: k = 21 . Then

x2 + x3 = 1
1
2 x1 +1 = 0 which is inconsistent.
get:
x1
+1 = 1
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
(c)
1
(i)  −1
2
So, the
AY 2009/2010 Sem 1



1 1 1
1 1
GaussianElimination 
1 2 
0 2 3 
→
1 3
0 0 52
linear system is inconsistent.
(ii) AT Av = AT b


 
1
1 1 1 −1 2 
x
1 −1 2  
−1 1 
2
⇒
=
1 1 1
y
1 1 1
2 1
3
6 2
x
5
⇒
=
.
2 3
y
6
x
3/14
⇒
=
.
y
13/7
(iii) if y is a least square solution to Ax = b
⇒ AT Ay = AT b
⇒ rAT Ay = rAT b for non-zero r.
⇒ AT A(ry ) = AT (rb)
⇒ ry is a least square solution to Ax =rb
Question 2

(a)







1 2 0
1 0 0
0 0 1
1 0 0
(i) E 1 =  0 1 0 , E 2 =  0 1 0 , E 3 =  0 1 0 , E 4 =  0 1 0 
0 0 1
0 0 2
1 0 0
0 −1 1
(ii) So, A−1 = E 4 E 3 E 2 E 1
To solve Ax =b
x = A−1 b = E 4 E 3 E 2 E 1 b




 
1 0 0
0 0 1
1 0 0
1 2 0
1









0 1 0
0 1 0
0 1 0
0 1 0
2 
=
0 −1 1
1 0 0
0 0 2
0 0 1
3


6

2 
= ... =
−4
(b)
(i)
2
1
2
1
T
=T 2
−3
=2T
−3T
3
2
3
2



  
3
2
0
= 2 5  − 3 3  =  1 
−1
−1
1
0
1
2
1
2
T
=T 2
−
=2T
− T
1
2
3
2
3

 
 

2
3
1
= 2 3  −  5  =  1 
−1
−1
−1


0 1
So the standard matrix for T is  1 1 
1 −1
1
0
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  
 0
(ii) A basis for the range of T is  1  , 

1

 

2
3






3
5
Note: We can use
,


−1
−1
AY 2009/2010 Sem 1

1

1 

−1
as well.
(iii) rank(T) = 2, 
0 1
Since  1 1  has 2 columns,
1 −1
null(T ) = no. of columns - rank(T ) = 0
(c) Note that (1, 1, 1, 0) · (0, x, −x, 0) = 0 = (0, −1, −1, −1) · (0, x, −x, 0)
So,ifwe define

  
 Ssuch that:
a
0 0 0 0
a
0
 b   1 0 0 0   b   a 

  
  
S 
 c  =  −1 0 0 0   c =  −a 
0
d
0 0 0 0
d
Then the range, which is span {(0, 1, −1, 0)} , is orthogonal to span {(1, 1, 1, 0), (0, −1, −1, −1)}
Note: There are many other ways to make your S such that it satisfies the condition in the question.
Question 3
λ − 1 −2
0 2 = . . . = (λ − 1)(λ − 3)(λ + 3)
(a) det(λI − A) = −2 λ + 1
0
2
λ−1 So the eigenvalues are1, 3, −3.



λ − 1 −2
0
0 −2 0
2  v =  −2 2 2  v = 0
For the eigenvalue 1,  −2 λ + 1
0
2
λ−1
0
2 0
1
1
√ , 0, √
So E 1 = span{(1, 0, 1)} = span
2
2



λ − 1 −2
0
2 −2 0
2  v =  −2 4 2  v = 0
For the eigenvalue 3,  −2 λ + 1
0
2
λ−1
0
2 2
1 1 −1
√ ,√ ,√
So E 3 = span{(1, 1, −1)} = span
3 3 3




λ − 1 −2
0
−4 −2 0
2  v =  −2 −2 2  v = 0
For the eigenvalue −3,  −2 λ + 1
0
2
λ−1
0
2 −4
−1 2 1
√ ,√ ,√
So E −3 = span{(−1, 2, 1)} = span
6 6 6


1
1
√
√
0
 2
2 
 1
1
−1 
 √
√
√ 
So let P = 

 3
3
3 
 −1
2
1 
√
√
√
6
6
6


1 0 0
AP=P  0 3 0 
0 0 −3
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AY 2009/2010 Sem 1


1 0 0
So, for this matrix P, P T AP =  0 3 0 
0 0 −3
(b)
(i) B 2 =


k 1 0
k 1
, B3 =  0 k 1 
0 k
0 0 k
(ii) There is only one eigenvalue: k. This is because that B n is upper triangular, and hence the
eigenvalues are the diagonal entries.
(iii) If B n has an eigenvector, its corresponding eigenvalue
eigenvalue k is span{(1,0,0,...,0)}, since:

 
k 0 0 ··· 0
 0 k 0 · · · 0  

 

 
..



.
0 
(kI − B n )v =  0 0 k
−
 .. .. .. . .
 
 . . .
. 0  
0 0 0 ··· k


0 1 0 ··· 0
 0 0 1 ··· 0 




.
.
. 0 
=
v = 0
 0 0 0

 .. .. .. . .
 . . .
. 1 
0 0 0 ··· 0
must be k. The eigenspace of the
k 1 0 ···
0 k 1 ···
.
0 0 k ..
.. .. .. . .
.
. . .
0 0 0 ···

0

0 



0 
 v

1 
k
So v ∈ span{(x1 , x2 , . . . , xn )|x2 = x3 = . . . = xn = 0} = span{(1, 0, . . . , 0)}
Hence, B n has at most one linearly independent eigenvectors, but not n linearly independent
eigenvectors.
B n is not diagonalizable for n ≥ 2
(c) No. We need to show that whenever C is symmetric, there exists an x ∈ R, such that det(C )−xI =
0, and hence the rows of C −xI is not a basis for Rn
By theorem 6.3.4, in the real number field R, a square matrix is orthogonally diagonalizable if and
only if it is symmetric.
So, if C is symmetric, then it is (orthogonally) diagonalizable.
In particular, there exists λ ∈ R which is an eigenvalue for C , with a corresponding eigenvector v
such that C v = λv = λI v
And so, (C − λI )x = 0 has a non-trivial solution v.
Hence, the row-space of (C − λI ) does not form a basis for Rn .
Question 4
(i) u 1 · u 2 = (1, 2, −1, 0) · (−1, 1, 1, 3) = −1 + 2 + (−1) + 0 = 0
u 1 · u 3 = (1, 2, −1, 0) · (2, −1, 0, 1) = 2 + (−2) + 0 + 0 = 0
u 2 · u 3 = (−1, 1, 1, 3) · (2, −1, 0, 1) = −2 + (−1) + 0 + 3 = 0
(ii) Using the hints given, u 4 = (−2)u 1 + (1)u 2 + (−1)u 3
u 5 = (1)u 1 + (−1)u 2 + (1)u 3
u 6 = (−2)u 1 + (−2)u 2 + (5)u 3
(iii) Since u 6 is a linear combination of u 1 , . . . , u 5 (or to be specific, a linear combination of
u 1 , u 2 , u 3 ),
span{u 1 , . . . , u 6 } = span{u 1 , . . . , u 5 }
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AY 2009/2010 Sem 1
Similarly, span{u 1 , . . . , u 5 } = span{u 1 , . . . , u 4 } = span{u 1 , . . . , u 3 }
Since S ={u 1 , u 2 , u 3 } is orthogonal, and hence linearly independent,
So S is a basis for span{u 1 , . . . , u 6 }
(iv) 
If we express {u 1
,.
. . , u 6 }as row

 vectors,
u4
−2 1 −1
u1
 1 −1 1   u 2  =  u 5 
u6
−2 −2 5
u3



−1
−2 1 −1
−2 1 −1
Since det  1 −1 1  = 3 6= 0,  1 −1 1  exists, and
−2 −2 5
−2 −2 5

−1 
 

−2 1 −1
u4
u1
 1 −1 1   u 5  =  u 2 
−2 −2 5
u6
u3
So u 1 , u 2 , u 3 can be expressed in linear combination of u 4 , u 5 , u 6
By using similar argument in part (iii), we get span{u 1 , . . . , u 6 } = span{u 4 , u 5 , u 6 }
Since span{u 1 , . . . , u 6 } has dimension 3 (from part (iii)), and span{u 1 , . . . , u 6 } = span{u 4 , u 5 , u 6 },
So T is also a basis for span{u 1 , . . . , u 6 }
(v) The transition matrix from T to
 S is:

−2 1 −2
P= [u 4 ]S [u 5 ]S [u 6 ]S =  1 −1 −2 
−1 1
5
(vi) If we are given the assumption that (1, 0, 0, 1) and (0, 1, 0, 1) is in span{u 1 , . . . , u 6 },
then we can choose (a, b, c, d) = (−1, 1, 1, 3) = u 2
If we observe the third entry, we know that (−1, 1, 1, 3) is not a linear combination of (1, 0, 0, 1)
and (0, 1, 0, 1),
Hence, {(−1, 1, 1, 3), (1, 0, 0, 1) and (0, 1, 0, 1)} is linearly independent, and has 3 elements.
By part (iii), span{u 1 , . . . , u 6 } has dimension 3,
So, {(−1, 1, 1, 3), (1, 0, 0, 1) and (0, 1, 0, 1)} is a basis for span{u 1 , . . . , u 6 }.
In particular, {(−1, 1, 1, 3), (1, 0, 0, 1) and (0, 1, 0, 1)} spans span{u 1 , . . . , u 6 }
If we are not given the assumption that (1, 0, 0, 1) and (0, 1, 0, 1) is in span{u 1 , . . . , u 6 }, then
we need to prove that (1, 0, 0, 1) and (0, 1, 0, 1) is a linear combination of {u 1 , . . . , u 6 }.
We can either solve for it through tedious computations, or make some clever observation/guesses.
Hint: u 1 + u 2 = (1, 2, −1, 0) + (−1, 1, 1, 3) = (0, 3, 0, 3),
11 · u 3 − u 6 = (22, −11, 0, 11) − (10, −11, 0, −1) = (12, 0, 0, 12).
After that, we can follow the arguments above.
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