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Math 115 Lab 4 Solutions

1. Let T : R4 R3 be a linear transformation. If T

1111

=513

, and T1102

=201

, find T5342

.Answer:

[5 3 4 2

]T = 4 [1 1 1 1]T [1 1 0 2]T .Thus,

T ([5 3 4 2

]T ) =T (4 [1 1 1 1]T [1 1 0 2]T)=4T

([1 1 1 1

]T) T ([1 1 0 2]T)=4

[5 1 3

]T [2 0 1]T=[18 4 11

]T2. Are the following transformations T : R2 R2 linear? Verify your answer.

(a) T[xy

]=[x2

y

](b) T

[xy

]=[

xy + 3

]Solution:

(a) Let ~u =[u1u2

]and ~v =

[v1v2

]. Then

T (~u+ ~v) = T([

u1 + v1u2 + v2

])=

[(u1 + v1)2

u2 + v2

]=

[u21 + 2u1v1 + v

21

u2 + v2

]But

T (~u) + T (~v) =[u21u2

]+[v21v2

]=

[u21 + v

21

u2 + v2

]6= T (~u+ ~v)

(b) Let ~u =[u1u2

]and ~v =

[v1v2

]. Then

T (~u+ ~v) = T([

u1 + v1u2 + v2

])=

[u1 + v1

u2 + v2 + 3

]But

T (~u) + T (~v) =[

u1u2 + 3

]+[

v1v2 + 3

]=

[u1 + v1

u2 + v2 + 6

]6= T (~u+ ~v)

Therefore, T is not linear.

3. In each case, find the standard matrix for the given transformation.

(a) A reflection in the y-axis followed by a rotation through pi/3.

(b) A rotation through pi/4 followed by a reflection in the x-axis.

Solution:

(a) Ry =[1 00 1

], Rpi/3 =

[cos(pi/3) sin(pi/3)sin(pi/3) cos(pi/3)

]=

[12

32

32

12

]

The standard matrix for Rpi/3 Ry is[

12

32

32

12

] [1 00 1

]=

[ 12

32

32

12

]

(b) Rpi/4 =[cos(pi/4) sin(pi/4)sin(pi/4) cos(pi/4)

]=

[12 1

212

12

]and Rx =

[1 00 1

]The standard matrix for

Rx Rpi/4 =[

12 1

212

12

] [1 00 1

]=

[12

12

12 1

2

]

4. If A is 4 4 and detA = 2, find det(15A1 6 adjA).

SolutionWe have A1 =

1detA

adjA =12adjA, so adjA = 2A1.

Using the fact that A1 is of size 4 4, this gives

det(15A1 6 adjA) = det(15A1 12A1) = det(3A1) = 34 det(A1) = 8112=

812.

5. Evaluate det

x 1 2 32 3 x 22 x 2

by first adding all other rows to the first row.Then find all values of x such that the determinant is zero.

SolutionR1 +R2 +R3

det

x 1 2 32 3 x 22 x 2

= det x 1 x 1 x 12 3 x 2

2 x 2

C2 C1, C3 C1= det

x 1 0 02 5 x 42 x+ 2 0

R2 R3= det

x 1 0 02 x+ 2 02 5 x 4

= (x 1)(x+ 2)(x 4).Hence the determinant is zero if x = 1, 2 or 4.

6. (a) Evaluate

3 5 2 61 2 1 12 4 1 53 7 5 3

using elementary row/column operations and Cofactor Expansion.(b) Use your result in (a) and Cramers Rule to solve for y in the linear system

3x+ 5y 2z + 6w = 1x+ 2y z + w = 0

2x+ 4y + z + 5w = 33x+ 7y + 5z + 3w = 8

Solution:

(a)

3 5 2 61 2 1 12 4 1 53 7 5 3

=0 1 1 31 2 1 10 0 3 30 1 8 0

= 1 1 30 3 31 8 0

= 1 1 30 3 30 9 3

= (1)3 39 3

= 18.R1 3R2 R3 +R1R3 2R2R4 3R2

(b) Evaluate the determinant of the matrix obtained by replacing column 2 of the coefficient matrixwith the vector of constants:3 1 2 61 0 1 12 3 1 53 8 5 3

=0 1 1 31 0 1 10 3 3 30 8 8 0

= 1 1 33 3 38 8 0

= 0R1 3R2R3 2R2R4 3R2

Thus, y =018 = 0.