How do you multiply #((2, 1, 0, 0), (3, 4, 2, 1), (-1, 0, 0, 1), (0, 1, 0, 0))# and #((1, 0, 0, 0), (0, 0, 1, 1), (1, 1, -1, 0), (1, 2, 3, 0))#?

1 Answer
Jun 2, 2016

#((2, 1, 0, 0),(3, 4, 2, 1), (-1, 0, 0, 1), (0, 1, 0, 0))*((1,0,0,0),(0,0,1,1),(1,1,-1,0),(1,2,3,0))=((2, 0, 1, 1),(6, 4, 5, 4),(0, 2, 3, 0), (0, 0, 1, 1))#.

Explanation:

The matrix multiplication is done raws by column. An example is better than thousand words.

Consider the first raw of the first matrix #(2,1,0,0)# and the first column of the second matrix
#((1),(0), (1), (1))#

we multiply them element by element and we sum everything

#2*1+1*0+0*1+0*1 = 2#
This is the first element of our final matrix.
Now we repeat the same for each raw of the first matrix and each column of the second matrix.

#2*0+1*0+0*1+0*2=0# (first raw, second column)
#2*0+1*1+0*(-1)+0*3=1# (first raw, third column)
#2*0+1*1+0*0+0*0=1# (first raw, fourth column)

Then the first raw of the final matrix will read #(2, 0, 1, 1)#

Second raw

#3*1 +4*0 +2*1 +1*1=6# (second raw, first column)
#3*0 +4*0 +2*1 1*2=4# (second raw, second column)
#3*0 +4*1 +2*(-1) +1*3=5# (second raw, third column)
#3*0 +4*1 +2*0 +1*0=4# (second raw, fourth column)

The second raw of the final matrix will read #(6, 4, 5, 4)#

Third raw

#-1*1+ 0*0+ 0*1+ 1*1 = 0# (third raw, first column)
#-1*0+ 0*0+ 0*1+ 1*2 = 2# (third raw, second column)
#-1*0+ 0*1+ 0*(-1)+ 1*3 = 3# (third raw, third column)
#-1*0+ 0*1+ 0*0+ 1*0 = 0# (third raw, fourth column)

The third raw of the final matrix will read #(0, 2, 3, 0)#

Fourth raw

#0*1+ 1*0+ 0*1+ 0*1 =0# (fourth raw, first column)
#0*0+ 1*0+ 0*1+ 0*2 =0# (fourth raw, second column)
#0*0+ 1*1+ 0*(-1)+ 0*3 =1# (fourth raw, third column)
#0*0+ 1*1+ 0*0+ 0*0 =1# (fourth raw, fourth column)

The fourth raw of the final matrix will read #(0, 0, 1, 1)#.

The product is then

#((2, 1, 0, 0),(3, 4, 2, 1), (-1, 0, 0, 1), (0, 1, 0, 0))*((1,0,0,0),(0,0,1,1),(1,1,-1,0),(1,2,3,0))=((2, 0, 1, 1),(6, 4, 5, 4),(0, 2, 3, 0), (0, 0, 1, 1))#.