The Multiplication can occur in two ways #(1) :[AB, and, (2) : BA#.
In General, #ABneBA#.
Let #A_(3xx3)=[a_(ij)]# and #B_(3xx3)=[b_(jk)]#
For #AB# to be defined, we must have,
No. of Columns in #A#=no. of Rows in #B#.
We find that this cond. is satisfied in our case, so, #AB# is defined.
Let #AB=C=[c_(ik)]_(3xx3),# where,
#c_(ik)=sum_(j=1)^(j=3) (a_(ij)*b_(jk)), i,k=1,2,3#.
#:. c_11=sum_(j=1)^(j=3) (a_(1j)*b_(j1))=a_11*b_11+a_12*b_21+a_13*b_31#
#:. c_11=1*1+0*0+2*3=7#. Similarly,
#c_12=1*1+0*2+2(-1)=-1#
#c_13=1*0+0*1+2*0=0#
#c_21=3*1+(-1)0+0*3=3#
#c_22=3*1+(-1)2+0*(-1)=1#
#c_23=3*0+(-1)1+0*0=-1#
#c_31=0*1+5*0+1*3=3#
#c_32=0*1+5*2+1*(-1)=9#
#c_33=0*0+5*1+1*0=5#
#:. [C]=[AB]=[(7,-1,0),(3,1,-1),(3,9,5)]#.
Similarly, we can work out, #[BA]=[(4,-1,2),(6,3,1),(0,1,6)]#.
Enjoy Maths.!
