ma 351: introduction to linear algebra and its fall 2021

12
MA 351: Introduction to Linear Algebra and Its Applications Fall 2021, Test Two Instructor: Yip This test booklet has FIVE QUESTIONS, totaling 100 points for the whole test. You have 75 minutes to do this test. Plan your time well. Read the questions carefully. This test is closed book, closed note, with no electronic device. In order to get full credits, you need to give correct and simplified answers and explain in a comprehensible way how you arrive at them. Name: (Major: ) Question Score 1.(20 pts) 2.(20 pts) 3.(20 pts) 4.(20 pts) 5.(20 pts) Total (100 pts) 1 Answer key

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MA 351: Introduction to Linear Algebra and Its

Applications

Fall 2021, Test Two

Instructor: Yip

• This test booklet has FIVE QUESTIONS, totaling 100 points for the whole test. You

have 75 minutes to do this test. Plan your time well. Read the questions

carefully.

• This test is closed book, closed note, with no electronic device.

• In order to get full credits, you need to give correct and simplified answers and

explain in a comprehensible way how you arrive at them.

Name: (Major: )

Question Score

1.(20 pts)

2.(20 pts)

3.(20 pts)

4.(20 pts)

5.(20 pts)

Total (100 pts)

1

Answer key

1. Fill in blanks: (Show your computations.)

0

B@2

�1

1

1

CA = ?

0

B@1

1

0

1

CA+ ?

0

B@1

0

1

1

CA+ ?

0

B@0

1

1

1

CA

2 + x+ 3x2= ? (1 + x+ x2

) + ? (2� x+ x2) + ? (x� x2

)

3

�1

0

�2

!= ?

2

�1

�1

0

!+ ?

2

�1

0

�1

!+ ?

1

�1

�1

1

!

2

a

I sit lo

1 1 itl

1 of e f eat

You can use this blank page.

3

b 21 329 I txt x 9 2 xxx E x x

2 C 2Ca t OGI C Ca t G3 C t Cz Cz

1 111 1O O 4 4

t.it ii l.iIi it

2 4 3 3 211 txt x2 to E xxx X X

You can use this blank page.

4

e

It l p

0 1 I 2

d Y 89 o 90 1 0 0 2 4 30 1 1 2 0 l l 2

1 1I lit

F sp af fi'D

2. Can you find a linear transformation T from R2to R2

that maps the triangle ABC to

triangle eA eB eC, where

A =

�1

�4

!, B =

1

3

!, C =

�2

4

!,

and

eA =

4

4

!, eB =

�3

�1

!, eC =

�4

�28

!?

(You can assume that T maps A to eA, B to eB, and C to eC.) If so, find the explicit

formula for this linear transformation.

5

1 4 4a 43 4e 4d 4

s 3at be 3Ct 3d 1

2 ga 43 42C t Hd 28

1 E C se1 2 2 14

4 4 4 441 4 70 6 6 18 O o O o

You can use this blank page.

6

Ii tia O

bIsd 3

y s 3 y sty

3. Let A =

1

2

�1

0

!. Find a basis and the dimension for the following subspace of

2⇥ 2 matrices:

M = {B : AB = BA} .

7

B C C 811 fa c b d

ka as

a e atab abt C o

b de a atb d o

2a c 2d La c 22 0

2b c 2bxc o

I off p

2 0 I 2 02 0 I 20

is 4 4 10 2 1 o o O O O O O

You can use this blank page.

8

i iti il

Ac a d pa E B b E

B E ie E aG Plap

aBasis vectors

M Span I 9 f dim m 2

can be chosen as

a

4. Let S = {A2⇥2: A = At} and K = {B2⇥2

: B = �Bt}. (S and K are called the space

of symmetric and skew-symmetric matrices.)

(a) Find a basis and the dimension for S.

(b) Find a basis and the dimension for K.

(c) If you collect the basis vectors for S and K together, do they form a basis for

the space of all 2 ⇥ 2 matrices? If so, write a general 2 ⇥ 2 matrix as a linear

combination of these basis vectors.

9

A et cha a

b c a d freeA 9 a af 4 8 4

s adim S 3 Basis vectors

b 13 61 C sa a a so

b c c bB f if

de de do aBasis vector

dim4 I

You can use this blank page.

10

I c 8 e a1G X G w Caf 8Catch y E YECa Ca Z G YEHence

EY 9 04 1 0189YET 8

Yes the basis vectors for Stk togetherfam a basis for the spaceof 2 2 matrices

They can span and they have 4reetoisintotal Since dim of 2 2 matrices is 4 Hence

theyre automatically linear independent

5. Consider the following linear transformation from R4to R3

:

T (X) =

0

B@1

1

1

1

0

2

2

1

3

�3

1

�7

1

CAX.

(a) Is T onto?

If yes, for a general Y =

0

B@a

b

c

1

CA, find X =

0

BBB@

x

y

z

w

1

CCCAsuch that T (X) = Y .

If no, find an explicit condition on Y such that there is an X such that T (X) = Y .

Find also an example of Y such that there is no X so that T (X) = Y .

(b) Is T one-to-one?

If no, find an Y and X1 6= X2 such that Y = T (X1) and Y = T (X2).

12

I fo i s p0 1 1 4 c a

Zero row

fxkk.getenet is not onto

We need a b za in order for TIX Y to besolvable

eg Y 8 then there is not X s t

THI Y

You can use this blank page.

13

b T is not one to one since there is a

free variable

gSet 8

1 4 iii itO O O O O 9 qX a p

EY WAy 2 48HI P O X 1

Y 1 2 1 w o

2 0

pal Xi 1 y 4 2 0 4 1

Then it I k

fo 71 1I 2 3 7 o

x x

I