Given A 1 2 1 0 1 0 4 6 1 B 2 9 1 3 1 1 1 2 0 and c 1 2 3

Given A = [1 2 -1 0 1 0 4 6 1], B = [2 9 1 3 -1 -1 1 2 0], and c = [1 2 3 8] Calculate A-2B and AB. Then find C^-1, the inverse of C.

Solution

Solution:

Given order of A =3*3=3 rows and 3 columns

order of B=3*3=3 rows and 3 columns

Since order of A and B are same we can find difference of 2 matrices.

To find A-2B

Find 2B

Multply each element of B to get 2B

B= 2 9 1

3 -1 -1

1 2 0

2B= 4 18 2

6 -2 -2

2 4 0

A -2B= 1 2 -1 _ 4 18 2

0 1 0 6 -2 -2

4 6 1 2 4 0

=1-4 2-18 -1-2

0-6 1-(-2) 0-(-2)

4-2 6-4 1-0

  

A-2B= -3 -16 -3

-6 3 2

2 2 1

AB=multiplying 2 matrices A and B

For multiplying 2 matrices thefollowing conditions to be satisfied

no of columns of first matrix =no of rows of second matrix

Here no of columns of A=3=no of rows of B

Henve we can multiply AB

order of AB=3*3

AB= 1 2 -1 2 9 1

0 1 0 3 -1 -1

4 6 1 1 2 0

Multiply 1 st row of A with 1 st column of B,1 st row of A with second column of B,1st row of A with third column of B to get first row elements of AB

1(2)+2(3)-1(1 ) 1(9)+2(-1)-1(2) 1(1)+2(-1)+(-1)(0)

0(2)+1(3)+0(1) 0(9)+1(-1)+0(2) 0(1)+1(-1)+0(0)

4(2)+6(3)+1(1) 4(9)+6(-1)+1(2) 4(1)+6(-1)+1(0)

7 5 -1

3 -1 -1

27 32 -2

C Inverse=

C^-1 =adjointC/DetC

Inverse of C exists if Detc not equal to zero

that is C should be singular matrix

Det(2*2) matrix=ad-bc

C= 1 2

3 8

DetC=ad-bc=1(8)-2(3)=8-6=20

Inverse of C exists.

C^-1 =1\\detC d -b

-c a

=1/2 8 -2

-3 1

=8/2 -2/2

-3/2 1/2

therefore C inverse = 4 -1

-3/2 1/2