Perform the multiplication 2 3 1 1 1 3 2 1 2 3 1 0 2 1 0 1 i

Perform the multiplication [2 3 1 -1] [1 3 2 1 -2 3] [1 0 2 -1 0 1] in each of the following orders: [2 3 1 -1] ([1 3 2 1 -2 3] [1 0 2 -1 0 1]) ([2 3 1 -1] [1 3 2 1 -2 3]) [1 0 2 -1 0 1]

Solution

Solution:

let A= 2 3

1 -1

order A=2*2

2 rows 2 columns

B= 1 3 2

1 -2 3

order =2*3

C=1 0

2 -1

0 1

order C 3*2

SolutionA:

multiply BC

order of B=2*3

order of C=3*2

since no of columnsof B=3=no of rows of C

product BC order=2*2

we can multiply BC

1 3 2 1 0

1 -2 3 2 -1

0 1

=1*1+3*2+2*0 1*0+3*-1+2*1

1*1-2*2+3*0 1*0+-2*-1+3*1

=1+6+0 0-3+2

1-4+0 0+2+3

=7 -1

-3 5

A(BC)

order=2*2*(2*2)=2*2

2 3 7 -1

1 -1 -3 5

=2*7+3*-3 2*-1+3*5

1*7+-1*-3 1*-1+-1*5

=14-9 -2+15

7 +3 -1-5

A(BC) = 5 13

10 -6

SolutionB:

(AB)C needs to be found

order A=2*2

order B=2*3

order AB=2*3

AB= 2 3 1 3 2

1 -1 1 -2 3

=2*1+3*1 2*3+3*-2 2*2+3*3

1*1+-1*1 1*3+-1*-2 1*2+-1*3

=5 0 13

0 5 -1

(AB)C=

5 0 13 1 0

0 5 -1 2 -1

0 1

=5*1+0*2+13*0 5*0+0*-1+13*1

0*1+5*2+-1*0 0*0+5*-1+-1*1

=5+0+0 0-0+13

0+10-0 0-5-1

( AB)C = 5 13

10 -6

we conclude form both that

A(BC)=(AB)C