Math 18D
March 1, 2017
Midterm Exam 2 ver. A Name:
(Total Points: 25 )
PID:
Instructions
1. No calculators, tablets, phones, or other electronic devices are allowed during this exam.
2. You may use one handwritten page of notes, but no books or other assistance during this exam.
3. Read each question carefully and answer each question completely.
4. Show all of your work. No credit will be given for unsupported answers, even if correct.
5. Write your Name at the top of each page.
(1 point) 0. Carefully read and complete the instructions at the top of this exam sheet and any additional
instructions written on the chalkboard during the exam.


1 −2 −2
0
 4 −8 −9
0 
.
(5 points) 1. Compute the determinant of A = 
 −4 7
6
−2 
8
3 −10 0
1 −2 −2 0
1 −2 −2 4 −8 −9 0
3+4
det A = =
(−1)
·(−2)·
4 −8 −9 −4 7
6 2
8 3 −10
8 3 −10 0
−8 −9 4 −9 1+1
1+2
= 2 ·[(−1) ·1· +(−1) ·(−2)· +
3 −10
8 −10
4 −8
1+3
(−1) · (−2) · ]
8 3 = 2 ·[(80 + 27) + 2 · (−40 + 72) − 2 · (12 + 64)]
= 2 ·(107 + 64 − 152)
= 38
ver. A (page 2 of 4 )
    

−1
2 
 1
2. Let B = −4 ,  3  ,  −7  .


−5
7
−13
(3 points)
(3 points)
Name:
(a) Explain why B is a basis of R3 .


0
(b) Let p(t) =  4 . Compute its coordinate relative to the basis B.
−11
Sketched Solution.
dim R3 = 3.


r2 → r2 + 4 r1 
1 −1
2
0
1 −1 2
0
r3 → r3 + 5 r1
 −4 3
 0 −1 1
−7
4  ∼pivot at (1,1)
4 
−5 7 −13 −11
0 2 −3 −11


1 −1 2
0
r3 → r3 + 2 r2
 0 −1 1
4 
∼pivot at (2,2)
0 0 −1 −3


1 −1 2 0
 0 −1 1 4 
0 0 1 3

∼r3 →−r3
r1 → r1 − 2 r3 
1 −1
r2 → r2 − r3

0 −1
∼pivot at (3,3)
0 0


1 −1 0 −6
∼r2 →−r2  0 1 0 −1 
0 0 1 3

1 0 0
r1 → r1 + r2
 0 1 0
∼pivot at (2,2)
0 0 1
So [b1 b2 b3 ] ∼ I3 , and B is hence a basis.
 
−7

The desired coordinate vector is −1
3

0 −6
0 1 
1 3

−7
−1 
3
ver. A (page 3 of 4 )

Name:

−5
11 
 is row equivalent to B =
−9 
8
1
0
2
−2 −1
 −1 −1 −2 −1
2
3. Given that A = [a1 a2 · · · a6 ] = 
 5 −2 9 −16 −2
−2 −1 −1
1
1


1 0 0 −2 0 −1
 0 1 0 3 0 −2 
4

[b1 b2 · · · b6 ] = 
 0 0 1 0 0 0 . Let H be the subspace of R spanned by {a1 , a2 , a3 }
0 0 0 0 1 4
and K be the subspace of R4 spanned by {a4 , a5 , a6 }. Find a basis for each of the following
subspaces.
(1 point)
(a) Col(A), the column space of A.
(3 points)
(b) Nul(A), the null space of A.
(1 point)
(c) H + K, the sum of the two subspaces H and K.
T
(d) H K, the intersection of the two subspaces H and K.
(2 points)
• Version A.
3(a). From B, we know that a1 , a2 , a3 , a5 are pivot columns of A. So the
basis of column space of A is
       
1
0
2
−1 


       

−1 −1 −2  2 

{a1 , a2 , a3 , a5 } =   ,   ,   ,  
(1)
5
−2
9
−2 





−2
1
−1
1
3(b). Nul(A)={x|Ax = 0} = {x|Bx = 0}. By solving Bx = 0, we get
x4 , x6 are free variables and
 
 
 
1
x1
2
2
x2 
−3
 
 
 
 
x3 
 
  = x4  0  + x6  0 
(2)
0
x4 
1
 
 
 
−4
x5 
0
1
0
x6
So basis of Nul(A) is
   
1 
2






−3  2 








   
0
0
 , 






 1   0 



  −4



 0


1
0
3(c). H + K=col(A). So its basis is same with col(A).
       
−1 
2
0
1



       
−2
−1
−1
, , , 2 
{a1 , a2 , a3 , a5 } = 
 5  −2  9  −2





1
−1
1
−2
(3)
(4)
3(d). If y ∈ H ∩ K, then there are c1 , c2 , c3 , d1 , d2 , d3 such that
−y
y
= c1 a1 + c2 a2 + c3 a3
= d1 a4 + d2 a5 + d3 a6
Since y − y = 0, we know [c1 , c2 , c3 , d1 , d2 , d3 ]T ∈ Nul(A) implying that
   
 
d1
0 
 1
d2  ∈ span 0 , −4
(5)


d3
0
1
Then we consider y1 = 1a4 + 0a5 + 0a6 ,y2 = 0a4 − 4a5 + 1a6 . y1 , y2 are
linearly independent and thus basis of H ∩ K is {y1 , y2 }.


 
−2
−1
 −1 
3




y1 = 
,y =
(6)
−16 2 −1
1
4
ver. A (page 4 of 4 )
Name:
4. We known that A is a 6 by 8 matrix. We also know that Nul(A) can be written as span{b1 , b2 , b3 , b4 },
but we do not know if b1 , . . . , b4 are linearly independent or not.
(3 points)
(a) Find all possible value(s) of dim Nul(A), the dimension of the null space of A. Be sure to
state the formula you used.
(3 points)
(b) Find all possible value(s) of rank(A), the rank of A. Be sure to state the formula you used.
4(a) By the rank-nullity theorem we have that rank(A) + dim(null(A)) = 8. Since rank(A) = dim(row(A)) ≤ 6,
we know that dim(null(A)) ≥ 8 - 6 = 2. Because null(A) is the span of 4 vectors, dim(null(A)) ≤ 4. Therefore 2 ≤
dim(null(A)) ≤ 4.
4(b) This in turn implies (via the rank-nullity theorem) that 4 ≤ rank(A) ≤ 6.
Math 18D
March 1, 2017
Midterm Exam 2 ver. B Name:
(Total Points: 25 )
PID:
Instructions
1. No calculators, tablets, phones, or other electronic devices are allowed during this exam.
2. You may use one handwritten page of notes, but no books or other assistance during this exam.
3. Read each question carefully and answer each question completely.
4. Show all of your work. No credit will be given for unsupported answers, even if correct.
5. Write your Name at the top of each page.
(1 point) 0. Carefully read and complete the instructions at the top of this exam sheet and any additional
instructions written on the chalkboard during the exam.


4
2
2
3
 −10 −2 −6 −9 

(5 points) 1. Compute the determinant of A = 
 −4 −2 −3 −1 .
0
2
0
0
4
2
2
3
4
2 3
−10 −2 −6 −9
4+2
det A = = (−1) ·2·−10 −6 −9
−4 −2 −3 −1
−4 −3 −1
0
2 0 0
−6 −9
−10 −9
1+1
1+2
= 2 ·[(−1) · 4 · + (−1) · 2 · −3 −1
−4 −1
−10 −6
1+3
(−1) · 3 · ]
−4 −3
= 2 ·[4 · (6 − 27) − 2 · (10 − 36) + 3 · (30 − 24)]
= 2 ·(−84 + 52 + 18)
= -28
ver. B (page 2 of 4 )
     
2
0 
 1





5 , 11 , 2 .
2. Let B =


−4
−4
7
(3 points)
(3 points)
Name:
(a) Explain why B is a basis of R3 .


2
(b) Let p(t) =  8 . Compute its coordinate relative to the basis B.
−14
Sketched Solution. dim R3 = 3.


r2 → r2 − 5 r1 
1
2 0
2
1 2 0 2
r3 → r3 + 4 r1
 5 11 2
 0 1 2 −2 
8  ∼pivot at (1,1)
−4 −4 7 −14
0 4 7 −6


1 2 0
2
r3 → r3 − 4 r2
 0 1 2 −2 
∼pivot at (2,2)
0 0 −1 2


1 2 0 2
∼r3 →−r3  0 1 2 −2 
0 0 1 −2

1 2
r2 → r2 − 2 r3

0 1
∼pivot at (3,3)
0 0

1 0
r1 → r1 − 2 r2

0 1
∼pivot at (2,2)
0 0

So [b1 b2 b3 ] ∼ I3 , and B is hence a basis.
 
−2

The desired coordinate is 2 
−2

0 2
0 2 
1 −2

0 −2
0 2 
1 −2
ver. B (page 3 of 4 )

Name:

−1 1
−2
−3 1
−8 
 is row equivalent to B =
1 −3
0 
−6 5 −15
1 0 −1
 3 1 −5
3. Given that A = [a1 a2 · · · a6 ] = 
 −2 1 −1
2 1 −8


1 0 0 0 0 0
 0 1 0 2 0 −2 
4

[b1 b2 · · · b6 ] = 
 0 0 1 1 0 1 . Let H be the subspace of R spanned by {a1 , a2 , a3 }
0 0 0 0 1 −1
and K be the subspace of R4 spanned by {a4 , a5 , a6 }. Find a basis for each of the following
subspaces.
(1 point)
(a) Col(A), the column space of A.
(3 points)
(b) Nul(A), the null space of A.
(1 point)
(c) H + K, the sum of the two subspaces H and K.
T
(d) H K, the intersection of the two subspaces H and K.
(2 points)
• Version B.
3(a). From B, we know that a1 , a2 , a3 , a5 are pivot columns of A. So the
basis of column space of A is
       
1
0
−1
1 


       

3  1 −5  1 

{a1 , a2 , a3 , a5 } =   ,   ,   ,  
(7)
−2
1
−1
−3 





2
1
−8
5
3(b). Nul(A)={x|Ax = 0} = {x|Bx = 0}. By solving Bx = 0, we get
x4 , x6 are free variables and
 
 
 
0
x1
0
2
x2 
−2
 
 
 
 
x3 
 
  = x4 −1 + x6 −1
(8)
0
x4 
1
 
 
 
1
x5 
0
1
0
x6
So basis of Nul(A) is
   
0 
0






−2  2 








   
−1
−1
 , 






 1   0 



   1 



 0


1
0
3(c). H + K=col(A). So its basis is same with col(A).
       
1 
−1
0
1



       
−5
1
3
, , , 1 
{a1 , a2 , a3 , a5 } = 
−2 1 −1 −3





5
−8
1
2
(9)
(10)
3(d). If y ∈ H ∩ K, then there are c1 , c2 , c3 , d1 , d2 , d3 such that
−y
y
= c1 a1 + c2 a2 + c3 a3
= d1 a4 + d2 a5 + d3 a6
Since y − y = 0, we know [c1 , c2 , c3 , d1 , d2 , d3 ]T ∈ Nul(A) implying that
   
 
d1
0 
 1
d2  ∈ span 0 , 1
(11)


d3
0
1
Then we consider y1 = a4 +0a5 +0a6 ,y2 = 0a4 +a5 +a6 .y1 , y2 are linearly
independent and thus basis of H ∩ K is {y1 , y2 }.
 


−1
−1
−3
 −7 



y1 = 
(12)
 1  , y2 =  −3 
−6
−10
ver. B (page 4 of 4 )
Name:
4. We known that A is a 8 by 10 matrix. We also know that Nul(A) can be written as span{b1 , b2 , b3 , b4 },
but we do not know if b1 , . . . , b4 are linearly independent or not.
(3 points)
(a) Find all possible value(s) of dim Nul(A), the dimension of the null space of A. Be sure to
state the formula you used.
(3 points)
(b) Find all possible value(s) of rank(A), the rank of A. Be sure to state the formula you used.
(a) By the rank-nullity theorem we have that rank(A) + dim(null(A)) =
10. Since rank(A) = dim(row(A)) ≤ 8, we know that dim(null(A)) ≥ 10
- 8 = 2. Because null(A) is the span of 4 vectors, dim(null(A)) ≤ 4. So 2 ≤ dim(null(A)) ≤ 4.
(b) This in turn implies (via the rank-nullity theorem) that 6 ≤ rank(A) ≤ 8.
Math 18D
March 1, 2017
Midterm Exam 2 ver. C Name:
(Total Points: 25 )
PID:
Instructions
1. No calculators, tablets, phones, or other electronic devices are allowed during this exam.
2. You may use one handwritten page of notes, but no books or other assistance during this exam.
3. Read each question carefully and answer each question completely.
4. Show all of your work. No credit will be given for unsupported answers, even if correct.
5. Write your Name at the top of each page.
(1 point) 0. Carefully read and complete the instructions at the top of this exam sheet and any additional
instructions written on the chalkboard during the exam.


−2 −1 −1 0
 4
3
3 2 

(5 points) 1. Compute the determinant of A = 
 −4 −6 −7 0 .
−1 −5 −9 0
−2 −1 −1 0 −2 −1 −1
4 3 3 2
det A = = (−1)2+4 · 2 · −4 −6 −7
−4 −6 −7 −0
−1 −5 −9
−1 −5 −9 0 −6 −7
1+1
= 2 ·[(−1)
· ((−2) · + (−1)1+2 · (−1) ·
−5 −9
−4 −6
−4 −7
1+3
+ (−1) · (−1) · ]
−1 −5
−1 −9
= 2 ·[−2 · (54 − 35) + (36 − 7) − (20 − 6)]
= 2 ·(−38 + 29 − 14)
= -46
ver. C (page 2 of 4 )
     
−2
2 
 1





2 , −5 , 5  .
2. Let B =


1
2
−1
(3 points)
(3 points)
Name:
(a) Explain why B is a basis of R3 .
 
−3

(b) Let p(t) = −8. Compute its coordinate relative to the basis B.
−2
Sketched Solution. dim R3 = 3.


r2 → r2 − 2 r1 
1 −2 2 −3
1 −2 2 −3
r → r3 − r1
 2 −5 5 −8  ∼ 3
 0 −1 1 −2 
pivot at (1,1)
1 2 −1 −2
0 4 −3 1


1 −2 2 −3
r3 → r3 + 4 r2
 0 −1 1 −2 
∼pivot at (2,2)
0 0 1 −7

r1 → r1 − 2 r3 
1
r2 → r2 − r3
 0
∼pivot at (3,3)
0

1 −2 0
r2 →−r2 
0 1 0
∼
0 0 1

1
r1 → r1 + 2 r2

0
∼pivot at (2,2)
0
So [b1 b2 b3 ] ∼ I3 , and B is hence a basis.
 
1

The desired coordinate is −5
−7

−2 0 11
−1 0 5 
0 1 −7

11
−5 
−7

0 0 1
1 0 −5 
0 1 −7
ver. C (page 3 of 4 )

Name:

1 −2 4 −2 −4
0
1
1
1
1 
 is row equivalent to B =
1
0
5
0 −1 
−2 3 −6 2
6
0
 1
3. Given that A = [a1 a2 · · · a6 ] = 
 1
2


1 0 0 1 0 −1
 0 1 0 4 0 0 
4

[b1 b2 · · · b6 ] = 
 0 0 1 0 0 4 . Let H be the subspace of R spanned by {a1 , a2 , a3 }
0 0 0 0 1 −2
and K be the subspace of R4 spanned by {a4 , a5 , a6 }. Find a basis for each of the following
subspaces.
(1 point)
(a) Col(A), the column space of A.
(3 points)
(b) Nul(A), the null space of A.
(1 point)
(c) H + K, the sum of the two subspaces H and K.
T
(d) H K, the intersection of the two subspaces H and K.
(2 points)
• Version C.
3(a). From B, we know that a1 , a2 , a3 , a5 are pivot columns of A. So the
basis of column space of A is
       
0
1
−2
−2 


       

1  0   1   1 

{a1 , a2 , a3 , a5 } =   ,   ,   ,  
(13)
1
1
0
0 





2
−2
3
2
3(b). Nul(A)={x|Ax = 0} = {x|Bx = 0}. By solving Bx = 0, we get
x4 , x6 are free variables and
 
 
 
1
x1
−1
0
x2 
−4
 
 
 
 
x3 
 
  = x4  0  + x6 −4
(14)
0
x4 
1
 
 
 
2
x5 
0
1
0
x6
So basis of Nul(A) is
   
1 
−1



   


−4


  0 


 0  −4
 , 
   


 1   0 



   2 



 0


1
0
3(c). H + K=col(A). So its basis is same with col(A).
       
−2 
−2
1
0



       
1
0
1
, , , 1 
{a1 , a2 , a3 , a5 } = 
1  1   0   0 





2
3
−2
2
(15)
(16)
3(d). If y ∈ H ∩ K, then there are c1 , c2 , c3 , d1 , d2 , d3 such that
−y
y
= c1 a1 + c2 a2 + c3 a3
= d1 a4 + d2 a5 + d3 a6
Since y − y = 0, we know [c1 , c2 , c3 , d1 , d2 , d3 ]T ∈ Nul(A) implying that
   
 
d1
0 
 1
d2  ∈ span 0 , 2
(17)


d3
0
1
Then we consider y1 = a4 + 0a5 + 0a6 ,y2 = 0a4 + 2a5 + a6 . y1 , y2 are
linearly independent and thus basis of H ∩ K is {y1 , y2 }.
 
 
4
−8
1
3



y1 =   , y2 =  
(18)
5
−1
−6
10
ver. C (page 4 of 4 )
Name:
4. We known that A is a 7 by 10 matrix. We also know that Nul(A) can be written as span{b1 , b2 , b3 , b4 },
but we do not know if b1 , . . . , b4 are linearly independent or not.
(3 points)
(a) Find all possible value(s) of dim Nul(A), the dimension of the null space of A. Be sure to
state the formula you used.
(3 points)
(b) Find all possible value(s) of rank(A), the rank of A. Be sure to state the formula you used.
(a) By the rank-nullity theorem we have that rank(A) + dim(null(A)) =
10. Since rank(A) = dim(row(A)) ≤ 7, we know that dim(null(A)) ≥ 10
- 7 = 3. Because null(A) is the span of 4 vectors, dim(null(A)) ≤ 4. So 3 ≤ dim(null(A)) ≤ 4.
(b) This in turn implies (via the rank-nullity theorem) that 6 ≤ rank(A) ≤ 7.
Math 18D
March 1, 2017
Midterm Exam 2 ver. D Name:
(Total Points: 25 )
PID:
Instructions
1. No calculators, tablets, phones, or other electronic devices are allowed during this exam.
2. You may use one handwritten page of notes, but no books or other assistance during this exam.
3. Read each question carefully and answer each question completely.
4. Show all of your work. No credit will be given for unsupported answers, even if correct.
5. Write your Name at the top of each page.
(1 point) 0. Carefully read and complete the instructions at the top of this exam sheet and any additional
instructions written on the chalkboard during the exam.


2 −1 −1
2
 3 −4 −3
4 

(5 points) 1. Compute the determinant of A = 
 −3 −1 4 −14 .
0
2
0
0
2 −1 −1 2 2 −1 2 3 −4 −3 4 4+2
det A = =
(−1)
·2·
3 −3 4 −3 −1 4 −14
−3 4 −14
0 2 0
0 4 −3 4 3
1+1
1+2
= 2 ·[(−1) ·2·
+(−1) ·(−1)·
+
4 −14
−3 −14
3 −3
(−1)1+3 · 2 · ]
−3 4 = 2 ·[2 · (42 − 16) + (−42 + 12) + 2 · (12 − 9)]
= 2 ·(52 − 30 + 6)
= 56
ver. D (page 2 of 4 )
     
−2
2 
 1





−1 , 1 , −4 .
2. Let B =


4
−7
11
(3 points)
(3 points)
Name:
(a) Explain why B is a basis of R3 .
 
4

(b) Let p(t) = 3. Compute its coordinate relative to the basis B.
6
Sketched Solution. dim R3 = 3.


r2 → r2 + r1 
1 −2 2 4
1 −2 2
4
r → r3 − 4 r1
 −1 1 −4 3  ∼ 3
 0 −1 −2
7 
pivot at (1,1)
4 −7 11 6
0 1
3 −10


1 −2 2
4
r3 → r3 + r2
 0 −1 −2 7 
∼pivot at (2,2)
0 0
1 −3

r1 → r1 − 2 r3 
1
r2 → r2 + 2 r3
 0
∼pivot at (3,3)
0

1 −2 0
r2 →−r2 
0 1 0
∼
0 0 1

1
r1 → r1 + 2 r2

0
∼pivot at (2,2)
0
So [b1 b2 b3 ] ∼ I3 , and B is hence a basis.
 
1

The desired coordinate is −5
−7

−2 0 10
−1 0 1 
0 1 −3

10
−1 
−3

0 0 8
1 0 −1 
0 1 −3
ver. D (page 3 of 4 )

Name:

−2
6 
 is row equivalent to B =
−3 
−6
1 −1 1
1 −1
 1
0
2
1
1
3. Given that A = [a1 a2 · · · a6 ] = 
 −5 4 −7 −6 1
0 −4 1
5
3


1 0 0 −1 0 −1
 0 1 0 −1 0 3 
4

[b1 b2 · · · b6 ] = 
 0 0 1 1 0 3 . Let H be the subspace of R spanned by {a1 , a2 , a3 }
0 0 0 0 1 1
and K be the subspace of R4 spanned by {a4 , a5 , a6 }. Find a basis for each of the following
subspaces.
(1 point)
(a) Col(A), the column space of A.
(3 points)
(b) Nul(A), the null space of A.
(1 point)
(c) H + K, the sum of the two subspaces H and K.
T
(d) H K, the intersection of the two subspaces H and K.
(2 points)
• Version D.
3(a). From B, we know that a1 , a2 , a3 , a5 are pivot columns of A. So the
basis of column space of A is
       
1
−1
1
−1 


       

1 0 2 1

{a1 , a2 , a3 , a5 } =   ,   ,   ,  
(19)
−5
4
−7
1 





0
−4
1
3
3(b). Nul(A)={x|Ax = 0} = {x|Bx = 0}. By solving Bx = 0, we get
x4 , x6 are free variables and
 
 
 
1
x1
1
−3
x2 
1
 
 
 
 
x3 
 
  = x4 −1 + x6 −3
(20)
0
x4 
1
 
 
 
−1
x5 
0
1
0
x6
So basis of Nul(A) is
   
1 
1






 1  −3








   
−3
−1
 , 






 1   0 



  −1



 0


1
0
3(c). H + K=col(A). So its basis is same with col(A).
       
−1 
1
−1
1



       
2
0
1
, , , 1 
{a1 , a2 , a3 , a5 } = 
−5  4  −7  1 





3
1
−4
0
(21)
(22)
3(d). If y ∈ H ∩ K, then there are c1 , c2 , c3 , d1 , d2 , d3 such that
−y
y
= c1 a1 + c2 a2 + c3 a3
= d1 a4 + d2 a5 + d3 a6
Since y − y = 0, we know [c1 , c2 , c3 , d1 , d2 , d3 ]T ∈ Nul(A) implying that
   
 
d1
0 
 1
d2  ∈ span 0 , −1
(23)


d3
0
1
Then we consider y1 = a4 + 0a5 + 0a6 ,y2 = 0a4 − a5 + a6 . y1 , y2 are
linearly independent and thus basis of H ∩ K is {y1 , y2 }.
 
 
1
−1
1
5



y1 =   , y2 =  
(24)
−6
−4
5
−9
ver. D (page 4 of 4 )
Name:
4. We known that A is a 6 by 9 matrix. We also know that Nul(A) can be written as span{b1 , b2 , b3 , b4 },
but we do not know if b1 , . . . , b4 are linearly independent or not.
(3 points)
(a) Find all possible value(s) of dim Nul(A), the dimension of the null space of A. Be sure to
state the formula you used.
(3 points)
(b) Find all possible value(s) of rank(A), the rank of A. Be sure to state the formula you used.
(a) By the rank-nullity theorem we have that rank(A) + dim(null(A)) =
9. Since rank(A) = dim(row(A)) ≤ 6, we know that dim(null(A)) ≥ 9 - 6
= 3. Because null(A) is the span of 4 vectors, dim(null(A)) ≤ 4. So 3 ≤ dim(null(A)) ≤ 4.
(b) This in turn implies (via the rank-nullity theorem) that 5 ≤ rank(A) ≤ 6.