2.1 matrix operations homework...
TRANSCRIPT
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