solutions sample final exam

7/26/2019 Solutions Sample Final Exam

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Solutions to Sample Final Exam


Vector Geometry and Linear Algebra MATH 1300 Solutions to Sample Final Exam 1


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1. OB OA+AB=OA+OC= =a + c

+ a = a - c

c = a a + a b + b a + b b 0 since

CA = CO + OA= OC +OA =- c

2. c = a + b c c = (a + b) (a + b)

c but = b = b a a ba

c c = a a + b b

2 2 2c = a + b ,etc.

2c since c = c

1) (3)(4) 1) 12 2 2 8 = = =v

3. a. u (3, 1, 2) (4, 2,= ( 1)(2) (2)(+ +

b.2 2 23 ( 1) 2 9 1 4 14 + = + + == +u

( ), , (1 4), ( 3 8), (6 4) ( 3,11,10)2 1 4 1 4 2

= = + =

v1 2 3 2 3 1

c. u

d.8 8u v 8

=14 21 2 7 3 7 7 6

= = =u v

cos

e.8 31 16 8

proj (4, 2, 1) , ,21 21 21 21

= =

v

u vu = v

v v

4. Normal to plane Point on plane Q ((2,1, 2)= n 3,1,0)=

QP

(5,4,6) (3,1,0) (2,3,6)= =

distance =QP (2,3,6) (2,1, 2) 4 3 12 5 5

3 3 3

+ = =

n

4 1 4= =

+ +n

2


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5. a. PQ (5, 4, 1) (2,3,1) (3,1, 2)= = = v

( , , ) (2,3,1) (3,1, 2)t x y z t + = + x = p v

parametric equations are: x= 2 + 3t, y= 3 + t, z= 1 2t.

b. 2( 2 3 ) (3 ) 2(1 2 ) 15 3 9 15 3 6 2t t t t t t + + + + = + = = =

2 3(2) 8, 3 2 5, 1 2(2) 3x y z= + = = + = = = point of intersection: or (8,5,3)

6. a.

1 1 1 4

2 1 1 2

3 1 2 5

1 1 41

1 1 1 4 1 1 1 4

b. 2 1 1 2 0 3 3 6 0 1 1 2

3 1 2 5 0 4 7 0 4 5 7

13 332 2 1

3 3 1

R - RR R -2R

R R -3R

1 1 0 7 5 2 10 7 3

B 3 2 1 3 2 1 31 21 9

4 2 1 4 2 1 38 26 11

= =

4

4

5 3

5

1 0 0 2 1 0 2

0 1 2 0 1 3 M

0 0 1 1 0 0 1 1

2 2 31 1 2

3 3 2

R R +RR R -R

R R +4Ratrix in RREF

0

01

c. Solution: x= 2, y= 3, z = 1.

7. a. A

b.T

2 2 0 7 3 4 9 5

2A+B 6 4 2 5 2 2 11 6

8 4 2 2 1 1 10

= + =

1 2 0

1 0

0 0 1

c. row 1 of B = row 1 of A + 2 row 2 of A E 0

=

.



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d. C = 2B A =

14 10 4 1 1 0 13 9 4

6 4 2 3 2 1 3 2 1

8 4 2 4 2 1 4 2 1

=

1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0

8. 3 2 1 0 1 0 0 1 1 3 1 0 0 1 1 3 1 0

4 2 1 0 0 1 0 2 1 4 0 1 0 2 1 4 0 1

1 0 1 2 1 0 1 0 1

0 1 1 3 1 0 0

0 0 1 2 2 1

2 2 1 2 2

3 3 1

3 31 1 2

3 3 2

R R -3R R -R

R R -4R

R -RR R -R

R R +2R

3 -1

2 1 0

1 1 3 1 0

0 0 1 2 2 1

1 0 0 0 1 1 0 1 1

0 1 0 1 1 1 A 1 1 10 0 1 2 2 1 2 2 1

=

1 1

2 2 3

R R -R

R R +R

Verifying:-1

1 1 0 0 1 1 1 0 0

AA 3 2 1 1 1 1 0 1 0

4 2 1 2 2 1 0 0 1

= =

9. a. 1 1 1 2 1 31 4 3 4 3 1+ + += + + = + =A ( 1) (2) ( 1) (3) ( 1) (1) 2(6) 3(2) 1( 4) 21 2 1 2 1 1

b.3 1

31

3 1C ( 1) 12 1 11

1 4

+ = =a= = 3 2322 1

( 1) (8 3) 53 4

b C += = = =

10. a. det(AB) = det(A) det(B) = 35 = 15

b. det(2A) = 2 d = 83

et )(A 3 = 24

c.-1 -1 1 3AB )=det(A)det(B ) 3

5 5

det( = =

n n-1A adj(A det(A)det(adj(A))=[det(A)] det (adj(A))=[det(A)]

d. )=det(A)I

3 1 2det(adj(A)) 3 3 9 = = =

4


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11.

1 5 1

2 7 1

3 4 1 273

1 1 1 9

2 1 1

3 2 1

y

= =

=

1 1 2 2 1 2 1 2a b a b a a b b+ +

1 1 1 1a b ka kbk

b a kb ka

=

1 1 2 2 1 2 1 2

1 2 1 21 1 2

a b b

c c c c

+ + + = +

12. det and so vectors form a linearly dependent set.

1 2 3

4 5 6 0

7 8 9

=

13. a. so we have closure under addition1 1 2 2 1 2 1 2( )b a b a b b a a+ = + +

1 1 1 1

a b a b a

so we have closure under scalar mult. Hence a subspace.

b. not closed under addition so not a subspace.

14. a. basis for row space is set of nonzero rows of RREF {(1,2,0,0), (0,0,1,2)}

b. basis for column space is set of columns from A corresponding to the columns in RREFthat contain leading 1's. {(1,2,2,1), (1,1,2,2)}

c. dimension of column space of A = rank of A = number of nonzero rows in RREF = 2

d. Ax= 0 x= (-2s, s, -2t, t) = s(2, 1, 0, 0) + t(0, 0, 2, 1)

Since (2, 1, 0, 0) and (0, 0, 2, 1) span the solution space and are linearly independent,they form a basis for the solution space.

e. rank A + nullity A = number of columns of A

2 + nullity A = 4nullity A = 2

solutions sample final exam

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