2.1 Matrix Operations Homework Solutions

2.

A + 3B =

[2 0 -14 -5 2

]+ 3

[7 -5 11 -4 -3

]=

[23 -15 27 -17 -7

]2C − 3E cannot be calculated because the matrices do not have the same sizes.

DB =

[3 5-1 4

] [7 -5 11 -4 -3

]=

[26 -35 -12-3 -11 -13

]EC cannot be calculated because we do not have the correct dimensions to perform matrixmultiplication.

5.

a.

AB =

-1 32 45 -3

[ 4-2

]=

-10026

,

-1 32 45 -3

[ -23

]+

118

-19

So, we get

AB =

-10 110 826 -19

b.

AB =

-1 32 45 -3

[ 4 -2-2 3

]=

-10 110 826 -19

6.

a.

AB =

4 -3-3 50 1

[ 13

]=

-5123

,

4 -3-3 50 1

[ 4-2

]=

22-22-2

So, -5 22

12 -223 -2

b. 4 -3

-3 50 1

[ 1 43 -2

]=

-5 2212 -222 -2

8. B must have 5 rows. To be able to do this product, we must have that B ∈M5,n and C ∈Mn,4.

11.

AD =

1 2 32 4 53 5 6

5 0 00 3 00 0 2

=

5 6 610 12 1015 15 12

DA =

5 0 00 3 00 0 2

1 2 32 4 53 5 6

=

5 10 156 12 156 10 12

When we right multiply by D, we use the columns of D with the rows of A. each entry is thesum of 2 zeros with one non-zero from D times the entry from A in the right position. Whenwe multiply DA, we are using the rows of D with the columns of A. Because all off-diagonalentries are 0, each entry in the resulting matrices result in only one non-zero product. Butbecause we are using the same vector from D with the rows of A in AD and the columns ofA in DA, the resulting matrix for AD is the transpose of DA.A matrix that satisfies what we want here would be if A = B.

12. [3 -6-2 4

] [6 43 2

]=

[0 00 0

]15.

a. FALSEThe correct product would be the rows of A with the columns of B

b. FALSEEach column of AB is a linear combination of the columns of A with weights from thecorresponding columns of B

c. TRUE

d. TRUE

e. FALSEThe order is reversed. (AB)T = BTAT .

16.

a. TRUE

b. FALSEThere should be no ‘+’signs there - the Abi make up the columns of the product matrix.

c. TRUE

d. FALSEThe order is wrong. (ABC)T = CTBTAT

e. TRUE

19. The third column of AB will be the sum of its first two columns as well. Think of it this way:the first column of AB is Ab1 and the second column of AB is Ab2. Since the third columnof B is b3 = b1 + b2, then Ab3 = A(b1 + b2) = Ab1 + Ab2.

20. They will be equal as well. If we use the same notation as above, then the first two columnsof AB are Ab1 and Ab2. Since b1 = b2, then these columns in AB must be as well.

27.

uT v =[

-3 2 -5] a

bc

=[

-3a+2b-5c]

vTu =[

a b c] -3

2-5

=[

-3a+2b-5c]

uvT =

-32-5

[ a b c]

=

-3a -3b -3c2a 2b 2c-5a -5b -5c

vuT =

abc

[ -3 2 -5]

=

-3a 2a -5a-3b 2b -5-3c 2c -5