ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Science
MCS 205 Basic Linear Algebra
1st Midterm
November 9, 2010
17:40-19:20
Surname
Name
ID #
Department
Section
Instructor
Signature
:
:
:
:
:
:
:
• The exam consists of 4 questions.
• Please read the questions carefully and write your answers under the corresponding questions. Be neat.
• Show all your work. Correct answers without sufficient explanation might not get full
credit.
• Calculators are not allowed.
GOOD LUCK!
Please do not write below this line.
Q1
Q2
Q3
Q4
TOTAL
25
25
25
25
100


1 2 3
Question 1. Let A =  2 5 3 .
1 0 8
(a) Find the determinant of A.
(b) Find A−1 by using elementary row operations on an identity matrix.
(c) Find A−1 by using cofactor expansion.
(d) Solve the system
x1 + 2x2 + 3x3 = 5
2x1 + 5x2 + 3x3 = 3
x1 + 8x3 = 17.
Answer 1.
(a)
1 2 3 2 3
|A| = 2 5 3 = 1 5 3
1 0 8 − 0 1 3
2 3
+ 8 1 2
2 5
= (6 − 15) + 8(5 − 4) = −9 + 8 = −1.
(b) Using the Gaussian elimination




1 2 3 1 0 0
1 0 0
1
2
3
1 +R2 →R2 
0
1 −3 −2 1 0  2R2 +R3 →R3
[A|I] =  2 5 3 0 1 0  −2R
−R1 +R3 →R3
1 0 8 0 0 1
0 −2
5 −1 0 1




1 2
3
1 0 0
1 2
3
1
0
0
3R3 +R2 →R2
 0 1 −3 −2 1 0  −R3 →R3  0 1 −3 −2
1
0  −3R
3 +R1 →R1
0 0 −1 −5 2 1
0 0
1
5 −2 −1




6
3
9
1 2 0 −14
1 0 0 −40 16
 0 1 0
13 −5 −3  −2R2 +R1 →R1  0 1 0
13 −5 −3  = [I|A−1 ]
0 0 1
5 −2 −1
0 0 1
5 −2 −1
we get

−40 16
9
=  13 −5 −3  .
5 −2 −1

A−1
(c) Note that


40 −13 −5
5
2 ,
cof(A) =  −16
−9
3
1
and
T 

40 −13 −5
40 −16 9
5
2  =  −13
5 3 
adj(A) =  −16
−9
3
1
−5
2 1


A−1

−40 16
9
adj(A) 
13 −5 −3  .
=
=
|A|
5 −2 −1
(d)

 

 

x1
−40 16
9
5
1
 x2  =  13 −5 −3   3  =  −1  ⇒ x1 = 1, x2 = −1, x3 = 2.
x3
5 −2 −1
17
2
Question 2. Consider the system of linear equations
x1 − 3x2 − 2αx3 = 0
−x1 + (α + 4)x2 + (2α + 1)x3 = −β
−x1 + (α + 4)x2 + (3α − 1)x3 = −β + 3
2x1 − 6x2 + (−6α + 4)x3 = −6.
(a) Find the values of α and β so that the system has
(i) no solution,
(ii) a unique solution,
(iii) infinitely many solutions.
(b) Find the solution for the special case α = 3 and β = 1.
Answer 2.
(a) Note that


1
−3
−2α
0
 −1 α + 4
−β 
2α + 1


 −1 α + 4
3α − 1 −β + 3 
2
−6 −6α + 4
−6

0
1
−3 −2α
 0 α+1
1 −β 
.

 0
0 α−2
3 
0
0
0
0

Clearly, if α 6= −1, α 6= 2 the system has a unique solution and if α = 2 the system has no
solution.
If α = −1, we continue the Gaussian elimination as follows




1 −3 2
1 −3
2
0
0
 0
 0
−β  .
0
1 −β 
0 1
0
0 −3
3
0
0 0 3 − 3β
Clearly, the system has no solution if β 6= 1 and infinitely many solutions if β = 1.
Therefore the system has
(i) no solution if α = 2; or α = −1, β 6= 1.
(ii) a unique solution if α 6= −1 and α 6= 2.
(iii) infinitely many solutions if α = −1, β = 1.
(b) If α = 3 and β = 1, we continue the Gaussian elimination as follows




1 −3 −6
0
1 0 0 15
 0
 0 1 0 −1  .
4
1 −1 
0
0
1
3
0 0 1
3
Therefore, the system has no solution x1 = 15, x2 = −1, and x3 = 3 if α = 3 and β = 1.
Question 3. Consider the linear system
4x1 − 3x2 − 11x3 + 4x4 − 6x5
−2x1 + 2x2 + 6x3 − 6x5
3x1 − 2x2 − 8x3 + 3x4 − 5x5
x1 − x2 − 3x3 + x4 − x5
−x1 + 2x2 + 4x3 + 3x4 − 17x5
= −9
= −6
= −7
= −2
= −19.
(a) Form the augmented matrix of the given system.
(b) Find the reduced row echelon form of the augmented matrix, which is found in part (a).
(c) Solve the system if it is consistent.
Answer 3.
(a) The augmented matrix of the

4
 −2

 3

 1
−1
(b) It is straightforward to show

4 −3 −11 4
 −2
2
6 0

 3 −2 −8 3

 1 −1 −3 1
−1
2
4 3
given system is
−3 −11 4 −6 −9
2
6 0 −6 −6
−2 −8 3 −5 −7
−1 −3 1 −1 −2
2
4 3 −17 −19



.


that
−6 −9
−6 −6
−5 −7
−1 −2
−17 −19




 rref 






1
0
0
0
0

0 −2 0
1
2
1
1 0 −2 −1 

0
0 1 −4 −5 
.
0
0 0
0
0 
0
0
0 0
0
(c) Using part (b) and putting x3 = s, x5 = t, we get x1 = 2 + 2s − t, x2 = −1 − s + 2t, and
x4 = −5 + 4t. Therefore the solution set is
x1 = 2 + 2s − t, x2 = −1 − s + 2t, x3 = s, x4 = −5 + 4t, x5 = t where s, t ∈ R.

1
 1
Question 4. Let A = 
 3
1
2
2
5
4
3
2
2
7

4
1 
. Find
1 
6
(a) |A|,
(b) |D| if A
2R1 +R2 →R2
7−→
B
3R3 →R3
7−→
C
R1 ↔R4
7−→
D,
(c) |AC −1 |,
(d) |A5 |,
(e) |(B T )−1 |,
(f) |2D|,
2 −1 (g) 3 A ,
(h) 45 C −2 .
Answer 4.
(a) By using the elementary row operations, it easy to see that
1
1 2 3 4 1
2
3
4
2
3
4
0
1 2 2 1 0
2
4
2 0 −1 −3 |A| = = − 0 −1 −7 −11 =
3 5 2 1 0 −1 −7 −11 0
1 4 7 6 0
0 −1 −3 2
4
2 1 2
1 2
3
4 3
4 0 2
0 2
4
2 4
2 =
−
= − 0 0 −5 −10 = −10.
0 0 −5 −10 0 0
0 0 −1 −3 0 −1 (b) |A| = |B| = −10, |C| = 3 |B| = −30 and |D| = −|C| = 30.
(c) |AC −1 | =
|A|
1
= .
|C|
3
(d) |A5 | = |A|5 = −105 = −100000.
(e) |(B T )−1 | =
1
1
1
=
=− .
T
|B |
|B|
10
(f) |2D| = 24 |D| = 480.
−1 2
1
1
81
(g) A
=−
.
= 2 = 2 4
3
| 3 A|
160
( 3 ) |A|
4
4
4
5 −2 5
5
5
1
625
−2
−1 2
(h) C =
|C | =
|C | =
=
.
2
4
4
4
4 |C|
3600
ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Science
MCS 205 Basic Linear Algebra
2nd Midterm
November 21, 2010
17:40-19:20
Surname
Name
ID #
Department
Section
Instructor
Signature
:
:
:
:
:
:
:
• The exam consists of 4 questions.
• Please read the questions carefully and write your answers under the corresponding questions. Be neat.
• Show all your work. Correct answers without sufficient explanation might not get full
credit.
• Calculators are not allowed.
GOOD LUCK!
Please do not write below this line.
Q1
Q2
Q3
Q4
TOTAL
25
30
34
21
110
Question 1. Consider the linear system:
x − 2y + z = 5
2x − 5y + 4z = −3
x − 4y + 6z = 10.
(a) Solve the system by using Cramer’s rule.


  
  
5
1
−2
1







−5 ,
4  if it is possible.
(b) Write −3 as a linear combination of 2 ,
10
1
−4
6
Answer 1.
(a) Let


1 −2 1
A =  2 −5 4  .
1 −4 6
Since
1 −2 1 |A| = 2 −5 4 = (−30 + 16) + 2(12 − 4) + (−8 + 5) = −14 + 16 − 3 = −1,
1 −4 6 5 −2 1 |A1 | = −3 −5 4 = 5(−30 + 16) + 2(−18 − 40) + (12 + 50) = −70 − 116 + 62 = −124,
10 −4 6 1
5
1
|A2 | = 2 −3 4 = (−18 − 40) − 5(12 − 4) + (20 + 3) = −58 − 40 + 23 = −75,
1 10 6 1 −2
5
|A3 | = 2 −5 −3 = (−50 − 12) + 2(20 + 3) + 5(−8 + 5) = −62 + 46 − 15 = −31,
1 −4 10 we have
x=
|A1 |
= 124,
|A|
y=
|A2 |
= 75,
|A|
and z =
|A3 |
= 31.
|A|
(b) By part (a)


 

 
5
1
−2
1
 −3  = 124  2  + 75  −5  + 31  4  .
10
1
−4
6

Question 2. Let

1
 2
A=
 3
−1
(a)
(b)
(c)
(d)
(e)
Find
Find
Find
Find
Find

−3 −1
2
2
−5
2
1
0 

−5
13 −6 −9 
−1 −15 10 13


2
 3 

and b = 
 0 .
4
a vector form of the general solution of the system Ax = b.
a basis for the row space of A.
a basis for the column space of A.
a basis for the null space of A.
the rank and the nullity of A.
Answer 2.
(a) Note that

1
 2

 3
−1


−3 −1
2
2 2

−5
2
1
0 3 
 rref 

−5
13 −6 −9 0 
−1 −15 10 13 4
1
0
0
0

0 11 −7 0 −21
1 4 −3 0 −9 
.
0 0
0 1 −2 
0 0
0 0
0
Putting x3 = s, x4 = t, we obtain x1 = −21 − 11s + 7t, x2 = −9 − 4s + 3t, and x5 = −2.
Therefore a vector form of the general solution of the system Ax = b is
 



 

7
−11
−21
x1
 3 
 −4 
 x2   −9 
 



 

 + t 0 .
 + s
=
0
x
1
x=
3
 



 

 1 

 x4  
0 
0 
0
0
−2
x5
(b) It is clear from part (a) that

 
 
2

1


 
 


 −3   −5  
 −1  ,  2  , 
 
 



 2   1  



0
2
3
−5
13
−6
−9
(c) It is clear from part (a) that

 
 
1
−3
2



  −5   0
2

 
 
 3  ,  −5  ,  −9


 −1
−1
13
(d) It is clear from part (a) that

 

−11
7



  3
−4


 

, 0
1

 



0   1



0
0
































is a basis for the row space of A.
is a basis for the column space of A.
is a basis for the null space of A.
(e) It is clear from part (a) that the rank of A is 3 and the nullity of A is 2.
Question 3. Let
p1 = 1 − 3x + x2 + 7x3 ,
p2 = −1 − 3x + 5x2 − x3 ,
p3 = 1 − 2x2 + 4x3 ,
p4 = 2 − 3x − x2 + 11x3 .
(a)
(b)
(c)
(d)
Find a basis S for the linear span of {p1 , p2 , p3 , p4 }.
Express each vector pi not in S as a linear combination of the vectors in S.
Find a basis T for P3 containing S.
Let p = −4 − 6x + 7x2 + 13x3 . Find [p]T .
Answer 3.
(a) Note that



1 0
1/2
3/2
1 −1
1
2


 −3 −3
0 −3 
 rref  0 1 −1/2 −1/2  .



 1
0 0
0
0 
5 −2 −1
0 0
0
0
7 −1
4 11
Therefore S = {p1 , p2 } form a basis for the linear span of {p1 , p2 , p3 , p4 }.
1
1
(b) It is clear from part (a) that p3 = p1 − p2 and p4 = 32 p1 − 12 p2 .
2
2
(c) Note that




1 −1
1
0
0
0
1 −1 1 0 0 0
 −3 −3 0 1 0 0  ref  0
1 −1/2 −1/6
0
0 
.



 0
 1
0
1
1/2 1/2
0 
5 0 0 1 0 
0
0
0
1 2/3 1/3
7 −1 0 0 0 1

Therefore T = {p1 , p2 , 1, x}
(d) Note that

1 −1
 −3 −3

 1
5
7 −1
Therefore
form a basis for P3 .
1
0
0
0

0 −4
1 −6 

0
7 
0 13

ref

1
 0

 0
0

2
 1 

[p]T = 
 −5  .
3
0
1
0
0
0
0
1
0

0
2
0
1 
.
0 −5 
1
3
Question 4.
(a) Is W =
a
a−b
a + 2b
b
where a, b ∈ R a subspace of M2×2 = R2×2 ? Explain.
(b) Is W = {(x, y, z) ∈ R3 where x + 2y + 3z = 1} a subspace of R3 ? Explain.

   

1
2
0 

(c) Is S =  −1  ,  1  ,  3  a basis of R3 ? Explain.

2
1
−3 
Answer 4.
(a) Let u =
a
a−b
a + 2b
b
, v=
c
c−d
c + 2d
d
∈ M2×2 and α ∈ R.
Since
c
c−d
u+v =
+
c + 2d
d
a+c
a−b+c−d
=
a + 2b + c + 2d
b+d
a+c
(a + c) − (b + d)
=
∈ M2×2 ,
(a + c) + 2(b + d)
b+d
a
a−b
a + 2b
b
and
αu = α
a
a−b
a + 2b
b
=
αa
α(a − b)
α(a + 2b)
αb
=
αa
αa − αb
αa + 2αb
αb
∈ M2×2 ,
W is a subspace of M2×2 .
(b) Note that (1, 0, 0) and (0, 21 , 0) belong to W but their sum (1, 12 , 0) do not belong to W .
Thus, W is not closed under addition and hence not a subspace of R3 .
(c) Since
1 2
0
−1 1
3
2 1 −3
S is not a basis of R3 .
= 0,
ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Science
MCS 205 - Basic Linear Algebra
FINAL EXAMINATION
17.01.2011
Question Grade Out of
STUDENT NUMBER:
1
25
2
15
3
20
DEPARTMENT:
4
20
INSTRUCTOR:
5
20
DURATION: 110 minutes
6
15
Total
115
NAME-SURNAME:
SIGNATURE:
SECTION:
IMPORTANT NOTES:
1) Please make sure that you have written your student number and name above.
2) Check that the exam paper contains 6 problems.
3) Show all your work. No points will be given to correct answers without reasonable
work.
Question 1. Let
q1 = 4 − 3x + x2
p1 = 5 + x − 3x2
2
q2 = 7 − 5x + 2x2
p2 = 7 − 2x + 4x
q 3 = 1 + x + x2
p 3 = 1 + x2
Given that S = {p1 , p2 , p3 } and T = {q1 , q2 , q3 } are both bases for P2 , perform the following
computations:


−5
(a) Find the polynomial p if [p]S =  3 .
1
(b) Find the transition matrix PT ←S from the S-basis to the T -basis.
Let q = 2 − 4x + 15x2 .
(c) Find [q]S .
(d) Find [q]T by using part (b) and (c).
(e) Find [q]T directly.
Answer 1.
(a) p = −5p1 + 3p2 + p3 = −5(5 + x − 3x2 ) + 3(7 − 2x + 4x2 ) + (1 + x2 ) = −3 − 11x + 28x2 .
(b) Note that




1 0 0
4
7 1
5
7 1
76 −9 −5
rref
 0 1 0 −44
 −3 −5 1
1 −2 0 
6
3 .
1
2 1 −3
4 1
0 0 1
9
1
0


76 −9 −5
6
3 .
Therefore PT ←S =  −44
9
1
0
(c) Note that




1 0 0 −2
5
7 1
2
 1 −2 0 −4  rref  0 1 0
1 .
−3
4 1 15
0 0 1
5


−2

1 .
Therefore [q]S =
5
(d)


 

76 −9 −5
−2
−186
6
3   1  =  109  .
[q]T = PT ←S [q]S =  −44
9
1
0
5
−17
(e) Note that




4
7 1
2
1 0 0 −186
 −3 −5 1 −4  rref  0 1 0
109  .
1
2 1 15
0 0 1 −17


−2

1 .
Therefore [q]S =
5
Question 2. Describe the linear transformation L : R3 → R2 which is represented by the matrix
−9 1 4
A=
3 1 7
4
−3
3
with respect to the standard basis for R and the basis
,
for R2 .
−1
1
Answer 2. Since


|
|
|
A =  [L(e1 )]T [L(e2 )]T [L(e3 )]T 
|
|
|
we conclude that
[L(e1 )]T =
−9
3
,
1
1
−3
1
[L(e2 )]T =
,
[L(e3 )]T =
4
7
and hence
L(e1 ) = −9
L(e2 ) =
4
−1
4
−1
4
−1
+3
+
−3
1
−3
1
=
=
1
0
−45
12
,
,
−5
3
L(e3 ) = 4
+7
=
.




x1
x1
Now, let  x2  ∈ R3 . Since  x2  = x1 e1 + x2 e2 + x3 e3 , we have
x3
x3


x1
L  x2  = x1 L(e1 ) + x2 L(e2 ) + x2 L(e3 )
x3
−45
1
−5
= x1
+ x2
+ x3
12
0
3
−45x1 + x2 − 5x3
=
.
12x1 + 3x3
,
Question 3. Let L : R2 → P2 be the linear transformation such that
−1
−2
2
L
= t − 3, and L
= t + 1.
4
9
a
(a) Find L
.
b
7
(b) Find L
.
−2
a1
(c) Find u =
∈ R2 such that L(u) = 3t2 − 2t − 11.
a2
Answer 3.
−1
4
(a) Firstly, find c1 and c2 such that c1
+ c2
−1 −2 a rref 1 0 −9a − 2b
⇒
4
9 b
0 1 4a + b
−2
9
=
a
b
:
c1 = −9a − 2b, c2 = 4a + b.
So,
L
a
b
−1
−2
= L c1
+ c2
4
9
−2
−1
+ c2 L
= c1 L
9
4
= (−9a − 2b)(t2 − 3) + (4a + b)(t + 1)
= (−9a − 2b)t2 + (4a + b)t + 31a + 7b.
(b) Putting a = 7, b = −2 and using part (a), we get
7
L
= −59t2 + 26t + 203.
−2
(c) Using (a), it is seen clearly that we should find a1 and a2 such that
−9a1 − 2b1
4a1 + b1
31a1 + 7b1




−9 −2
3
1 0
1
rref
 4
 0 1 −6  ⇒
1 −2 
31
7 −11
0 0
0
=3
= −2
= −11.
a1 = 1, a2 = −6,
and u =
1
−6
.
Question 4. Let L be the linear operator on R3 defined by

 

x1
x1 − x2 + x3
L  x2  =  x1 + 2x2 − x3  .
x3
−x1 + 4x2 − 3x3


a
(a) What must be the value of a if  −2a  ∈ Ker(L)?
1


a
(b) What must be the value of a if  −2a  ∈ Range(L)?
1
(c) Find a basis for Ker(L).
(d) Find nullity(L) and rank(L).
(e) Is L invertible? Justify your answer.
Answer 4.
(a)


 
  

a
3a + 1
0
a
 −2a  ∈ Ker(L) ⇔ L  −2a  =  −3a − 1  =  0  ⇔ a = − 1 .
3
1
−9a − 3
0
1
(b)



1 −1
1
a
a
 −2a  ∈ Range(L) ⇔  1
2 −1 −2a  is consistent.
1
−1
4 −3
1
Note that




1 0
1/3
1 −1
1
a
0
rref
 0 1 −2/3
 1
2 −1 −2a 
−a  .
−1
4 −3
1
0 0
0 4a + 1
Therefore


a
 −2a  ∈ Range(L) ⇔ 4a + 1 = 0 ⇔ a = − 1 .
4
1

(c) Note that




1 −1
1 0
1 0
1/3 0
rref
 1
 0 1 −2/3 0  .
2 −1 0 
−1
4 −3 0
0 0
0 0


 −1/3 
Therefore  2/3  is basis for Ker(L).


1
(d) It follows from (c) that nullity(L) = 1 and rank(L) = 2.
 
 0 
(e) No, because Ker(L) 6=  0  .


0
Question 5. Let

 


1
3
1
 1 
 1 
 1 

 


v1 = 
 1  , v2 =  −1  , v3 =  3  .
3
1
1
4
(a) Find an orthogonal basis S for the subspace W of R spanned by {v1 , v2 , v3 }.
(b) Find an orthonormal basis T for the subspace W of R4 spanned by {v1 , v2 , v3 }.


0
 −1 

(c) Find [v]S if v = 
 4 .
5

Answer 5.
(a) Applying the Gram-Schmidt orthogonalization process, we get


 

  
2
1
3
1


 1 

  
 , w2 = v2 − <v2 ,w12> w1 =  1  − 4  1  =  0  ,
w 1 = v1 = 
kw1 k
 1 
 −1  4  1   −2 
0
1
1
1

 
 
 

1
0
1
2
 1  8  1  −4  0   −1 
< v3 , w1 >
< v3 , w2 >
.
=
 −  −

w 3 = v3 −
w
−
w
=
1
2
 3  4 1 
kw1 k2
kw2 k2
8  −2   0 
1
3
1
0
Therefore
  
 
2
0
1


  
  −1
0
1
 
 
S= 
 1  ,  −2  ,  0


 1
0
1
is an orthogonal basis for W .
√
√
(b) Note that kw1 k = 2, kw2 k = 2 2, and kw3 k = 2.








Therefore

 

√  
0
1/
2
1/2




√

 1/2  
  −1/ 2 
0






√
,
T = 
,
1/2   −1/ 2  
0√ 



 1/2

1/ 2
0
is an orthonormal basis for W .
(c) Since S is orthogonal, we have
< v, w2 >
< v, w3 >
< v, w1 >
v=
w1 +
w2 +
w3
2
2
kw1 k
kw2 k
kw3 k2
and hence
8
−8
6
= w1 +
w2 + w3 = 2w1 − w2 + 3w3 ,
4
8
2


2
[v]S =  −1  .
3
Question 6. In each part, determine whether the given transformation is linear or not.
(a) L : R3 → R2 defined by


a1
a
−
a
1
2
L  a2  =
.
a1 a3
a3
(b) L : P3 → P3 defined by
L (p(x)) = 2p(x) + (1 − x)p0 (x).
Answer 6.
(a) The transformation is not linear because
  
 
 
1
2
1
0
0
0
L 2  1  = L  2  =
6=
=2
= 2 L  1  .
4
2
1
1
2
1
(b) Note that for any p(x), q(x) ∈ P3
L(p(x) + q(x)) = 2(p(x) + q(x)) + (1 − x)(p(x) + q(x))0
= 2p(x) + 2q(x) + (1 − x)(p0 (x) + q 0 (x))
= 2p(x) + (1 − x)p0 (x) + 2q(x) + (1 − x)q 0 (x)
= L(p(x)) + L(q(x)),
and for any p(x) ∈ P3 and α ∈ R
L(αp(x)) = 2(αp(x)) + (1 − x)(αp(x))0
= 2αp(x) + (1 − x)(αp0 (x))
= α(2p(x) + (1 − x)p0 (x))
= αL(p(x)).
Therefore, the transformation is linear.
ÇANKAYA UNIVERSITY
Department of Mathematics and Computer Science
MCS 205 Basic Linear Algebra
Make-up
January 24, 2011, 10:00-11:50
Questions
(1) Let
p1 = 5 + x − 3x2
q1 = 4 − 3x + x2
2
p2 = 7 − 2x + 4x
q2 = 7 − 5x + 2x2
2
p3 = 1 + x
q3 = 1 + x + x2
Given that S = {p1 , p2 , p3 } and T = 
{q1 , q2 ,q3 } are both bases for P2 , perform the following computations:
−5
(a) Find the polynomial p if [p]S =  3 .
1
(b) Find the transition matrix PT ←S from the S-basis to the T -basis.
Let q = 2 − 4x + 15x2 .
(c) Find [q]S .
(d) Find [q]T by using part (b) and (c).
(e) Find [q]T directly.
(2) Describe the linear transformation L : R3 → R2 which is represented by the matrix
−2
3 1
A=
5 −6 0
8
3
with respect to the standard basis for R3 and the basis
,
for R2 .
3
1
(3) Let L : R3 → R2 be the linear transformation such that




2
1
−1
1
,
, and L  9  =
L  2  =
0
1
0
1


a
(a) Find L  b .
c


7
(b) Find L  13 .
7


a1
3
(c) Find u =  a2  ∈ R3 such that L(u) =
.
0
a3

3
0
.
and L  3  =
1
4

(4) Let L : R4 → R3 be a linear operator on defined by




x1
2x1 − 3x2 + 4x3
 x2 

  x1 + 2x3 − x4  .
L
 x3  =
−3x2 + 2x4
x4
(a) Find a basis β for Ker(L).
(b) Extend β to basis for R4 .
(c) Find a basis for Range(L).


15
(d) Is v =  −2  ∈ Range(L)?
19
(e) Find nullity(L) and rank(L).
(5) Let






5
0
3
v1 =  2  , v2 =  1  , v3 =  −7  .
−1
−1
1
(a) Find an orthogonal basis S for the subspace W of R4 spanned by {v1 , v2 , v3 }.
(b) Find an orthonormal basis T for the subspace W of R4 spanned by {v1 , v2 , v3 }.


7
(c) Find [v]S if v =  14 .
−6
(6) In each part, determine whether L is a linear transformation or not.
(a) L : R3 → R2 defined by


a1
2a1 − a2 + a3
L  a2  =
.
a2 − 4a3
a3
(b) L : R2×2 → R defined by
a b
L
= a 2 + b2 .
c d