How do you multiply #((0, 6), (5, -5))# and #((1, 1, 1), (0, 0, -2))#?

1 Answer
Feb 15, 2016

#C_color(blue)(23) =((0,6),(5,-5)) * ((1,1,1),(0,0,-2)) = ((0,0,-12),(5,5,15)) #

Explanation:

to multiply a matrix by another matrix we need the number of columns to match the number of rows.
#C_(ik)=A_(icolor(red)1)B_(color(red)1k)-A_(icolor(red)2)B_(color(red)2k)+cdots+ A_(icolor(red)N)B_(color(red)Nk) #
note the inner counter, #color(red)j# is the same that is the column of A is equal to the row of B ...
#C_(ik) =sum_(j=1)^NA_(icolor(red)j) B_(color(red)jk)#
#((c_(11),c_(12), cdots,c_(1k)), (vdots,vdots, vdots, vdots), (c_(k1),c_(k2), cdots, c_(kk)))= ((a_(11),a_(12), cdots,a_(1j)), (vdots,vdots, vdots, vdots), (a_(i1),a_(i2), cdots, a_(ij))) * ((b_(11),b_(12), cdots,b_(1k)), (vdots,vdots, vdots, vdots), (b_(j1),b_(j2), cdots, b_(jk))) #

Now that we had a quick review on Matrix multiplication in you case: #A_(2color(red)2) = ((0,6),(5,-5)); B_(color(red)(2)3)=((1,1,1),(0,0,-2)) #
So the multiplication order is:
#C_color(blue)(23) = A_(color(blue)2color(red)2)*B_(color(red)(2)color(blue)(3) #
#C_color(blue)(23) =((0,6),(5,-5)) * ((1,1,1),(0,0,-2)) = ((0,0,-12),(5,5,15)) #