find the matrix multification of A and
B A= [1/2 -2/3 ] B = [ 1/2 2/3 3/1 ]
Answers
Answer:
This is a matrix productA and B are two matrix and AB represents their product
(i)HereA=[
a
−b
b
a
]
B=[
a
b
−b
a
]
Here we multiply first row of A with first column of B and that will make first element for the first row of product matrix.
i.e First element of AB will be a∗a+b∗b=a
2
+b
2
We will get second element of first row for product matrix by multiplying first row of A with second column of Bi.e −a∗b+b∗a=0
We will get first element of second row for product matrix by multiplying second row of A with first column of B
i.e −b∗a+a∗b=0
We will get second element of second row for product matrix by multiplying second row of A with second column of B
i.e −b∗−b+a∗a=−b
2
+a
2
So final product matrix AB will be
AB=[
a
2
+b
2
0
0
a
2
+b
2
]
(ii)HereA=
⎣
⎢
⎢
⎡
1
2
3
⎦
⎥
⎥
⎤
B=[
2
3
4
]
Here we multiply first row of A with first column of B and that will make first element for the first row of product matrix.
i.e First element of AB will be 1∗2=2
We will get second element of first row for product matrix by multiplying first row of A with second column of Bi.e 1∗3=3
We will get third element of first row for product matrix by multiplying first row of A with third column of B
i.e 1∗4=4
We will get first element of second row for product matrix by multiplying second row of A with first column of B
i.e 2∗2=4
We will get second element of second row for product matrix by multiplying second row of A with second column of B
i.e 2∗3=6
We will get third element of second row for product matrix by multiplying second row of A with third column of B
i.e 2∗4=8
We will get first element of third row for product matrix by multiplying third row of A with first column of B
i.e 3∗2=6
We will get second element of third row for product matrix by multiplying third row of A with second column of B
i.e 3∗3=9
We will get third element of third row for product matrix by multiplying third row of A with third column of B
i.e 3∗4=12
So final product matrix AB will be
AB=
⎣
⎢
⎢
⎡
2
4
6
3
6
9
4
8
12
⎦
⎥
⎥
⎤
(iii)A=[
1
2
−2
3
]
B=[
1
2
2
3
3
1
]
AB= [
1
2
−2
3
]x[
1
2
2
3
3
1
]
⇒ AB=[
1∗1+(−2∗2)
2∗1+3∗2
1∗2+(−2∗3)
2∗2+2∗3
1∗3+(−2∗1)
2∗3+3∗1
]
⇒AB=[
−3
8
−4
13
1
9
]
(iv)A=
⎣
⎢
⎢
⎡
2
3
4
3
4
5
4
5
6
⎦
⎥
⎥
⎤
B=
⎣
⎢
⎢
⎡
1
0
3
−3
2
0
5
4
5
⎦
⎥
⎥
⎤
AB=
⎣
⎢
⎢
⎡
2
3
4
3
4
5
4
5
6
⎦
⎥
⎥
⎤
x
⎣
⎢
⎢
⎡
1
0
3
−3
2
0
5
4
5
⎦
⎥
⎥
⎤
⇒ AB=
⎣
⎢
⎢
⎡
2∗1+3∗0+4∗3
3∗1+4∗0+5∗3
4∗1+5∗0+6∗3
(2∗−3)+3∗2+4∗0
(3∗−3)+4∗2+5∗0
(4∗−3)+5∗2+6∗0
2∗5+3∗4+4∗5
3∗5+4∗4+5∗5
4∗5+5∗4+6∗5
⎦
⎥
⎥
⎤
⇒AB=
⎣
⎢
⎢
⎡
15
18
22
0
−1
2
42
56
70
⎦
⎥
⎥
⎤
(v)A=
⎣
⎢
⎢
⎡
2
3
−1
1
2
1
⎦
⎥
⎥
⎤
B=[
1
−1
0
2
1
1
]
⇒AB=
⎣
⎢
⎢
⎡
2
3
−1
1
2
1
⎦
⎥
⎥
⎤
x[
1
−1
0
2
1
1
]
⇒ AB=
⎣
⎢
⎢
⎡
2∗1+(1∗−1)
3∗1+(2∗−1)
(−1∗1)+(1∗−1)
2∗0+1∗2
3∗0+2∗2
(−1∗0)+1∗2
2∗1+1∗1
3∗1+2∗1
(−1∗1)+1∗1
⎦
⎥
⎥
⎤
⇒AB=
⎣
⎢
⎢
⎡
1
1
−2
2
4
2
3
5
0
⎦
⎥
⎥
⎤
(vi)A=[
3
−1
−1
0
3
2
]
B=
⎣
⎢
⎢
⎡
2
1
3
−3
0
1
⎦
⎥
⎥
⎤
⇒AB=[
3
−1
−1
0
3
2
]x
⎣
⎢
⎢
⎡
2
1
3
−3
0
1
⎦
⎥
⎥
⎤
⇒ AB=[
3∗2+(−1∗1)+3∗3
(−1∗2)+0∗1+2∗3
(3∗−3)+−1∗0+3∗1
(−1∗−3)+(0∗0)+2∗1
]
⇒AB=[
14
4
0
5
]