if a= [0 1 2] and b= [3 2 1], then show that: (ab)'=b'a'

Question

a is vertical mtx and b is horizontal mtx how to slve please explain

Answer 1

I think what you really mean is show that (ab)^t = b^ta^t

 

For (ab),

 

Row 1 = (0*3  0*2  0*1) = (0 0 0)

Row 2 = (1*3  1*2  1*1) = (3 2 1)

Row 3 = (2*3  2*2  2*1) = (6 4 2)

 

For (ab)^t,

 

Row 1 = (0 3 6)

Row 2 = (0 2 4)

Row 3 = (0 1 2)

 

For b^ta^t,

 

Row 1 = (3*0  3*1  3*2) = (0 3 6)

Row 2 = (2*0  2*1  2*2) = (0 2 4)

Row 3 = (1*0  1*1  1*2) = (0 1 2)

 

Now compare (ab)^t to b^ta^t and see that they are the same.

Answer 2

Hi Rahul.  I am in Virginia also!!!

 

The " (ab)' " represents the transpose of the matrix obtained by multiplying matrix a with matrix b.

The "  b' a'  "  represents the matrix obtained by first transposing a and b, them multiplying these transposed matrices.

 

So, start by multiplying matrix a  with matrix b.   Matrix a is 3x1  and matrix b is 1x3 so resulting multiplied matrix will be 3x3.

 

        [0]                     [ 0 0 0 ]

ab = [1] x [3 2 1] =    [ 3 2 1 ] 

        [2]                     [ 6 4 2 ]

 

(ab)' is the transpose matrix, so make each row a column:  

 

(ab)' =   [ 0 3 6 ]

             [ 0 2 4 ]

             [ 0 1 2 ]

 

 

Now start by transposing matrix b and matrix a.

 

        [ 3 ]

b' =  [ 2 ]    and   a' = [ 0 1 2 ]

        [ 1 ] 

 

Now multiply these two transposed matrices:

 

 

           [ 0 3 6 ]

b'a' =   [ 0 2 4 ]

           [ 0 1 2 ]

 

 

Notice that  (ab)' is the same matrix as b'a'.  You have just proved it!   

 

Hope that helped.  :-)

 

Vicki