Let #A = ((0,1,0 ), (2,-1,1 ), (0,2,−1 ))# and #B = ((−1,2,0 ), (4,6,0), (1,0,1 )) # then the product of #A*B = C# is given by:
#cij = a_(i1)b_(1j) + a_(i2)b_(2j) + cdots + a_(1n)b_(1n)#
#cij = sum_(k=1)^na_(ik)b_(kj)#
Now we have n = 3
#cij = sum_(k=color(red)(1))^color(red)3a_(ik)b_(kj)#
#c_(color(red)(1)color(blue)(1)) = a_(color(red)(1)1)b_(1color(blue)(1) ##+a_(color(red)(1)2)b_(2color(blue)(1) + ##a_(color(red)(1)3)b_(3color(blue)(1) #
#c_(color(red)(1)color(blue)(2)) = a_(color(red)(1)1)b_(1color(blue)(2) ##+a_(color(red)(1)2)b_(2color(blue)(2) + ##a_(color(red)(1)3)b_(3color(blue)(2)#
#cdots#
Note: each entry is made by three term obtained by multiplying row-line by column- line. That is:
row line = # ((color(red)0, color(blue)1, color(green)0))# #((color(red)(-1)), (color(blue)(2)), (color(green)(0)) )#
#color(red)("red"xx"red")# + #color(blue)("blue"xx"blue")# + #color(green)("green"xx"green")#
Using this approach we find:
#C = ((4,6,0 ), (-5,-2,1 ), (7,12,-1 ))#
