How do you multiply #(4, 5)(3, 4)# by #(3, 0)(0, 3)#?

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Answer 1 · 2015-07-04T17:07:58

#((4, 5),(3, 4)) ((3, 0),(0, 3)) = ((12,15),("9","12"))#

#((4, 5),(3, 4)) ((3, 0),(0, 3))#

Find the first row of the product

Take the first row of #((4, 5),(3, 4))#, and make it vertical in front of #((3, 0),(0, 3))#. (We'll do the same for the second row in a minute.)

It looks like:

#{: (4),(5) :}((3, 0),(0, 3))#

Now multiply times the first column and add to get the first number in the first row of the answer:
#4 xx 3 + 5 xx 0 = 12 + 0 = 12#

Next multiply times the second column and add to get the second number in the first row of the answer:
#4 xx 0 + 5 xx 3 = 0 + 15 = 15#

The first row of the product is: #((12,15))#

A this point we know that the product looks like:

#((4, 5),(3, 4)) ((3, 0),(0, 3)) = ((12,15),("-","-"))#

Find the second row of the product
Find the second row of the product by the same process using the second row of #((4, 5),(3, 4))#

#{: (3),(4) :} ((3, 0),(0, 3)) # to get: #9# and #12#

The second row of the product is: #((9,12))#

Write the answer

#((4, 5),(3, 4)) ((3, 0),(0, 3)) = ((12,15),("9","12"))#

Answer 2 · 2015-07-04T20:40:10

The second matrix is just a scalar multiple (#3#) of the identity matrix, so the result is just the same scalar multiple of the first matrix, i.e.

#3((4,5),(3,4))=((12,15),(9,12))#

Jim's explanation is really helpful if you are multiplying any two square matrices, but in this particular example there's a short cut.

#((3,0),(0,3)) = 3((1,0),(0,1))#

So:

#((4,5),(3,4))((3,0),(0,3))#

#=3((4,5),(3,4))((1,0),(0,1))#

#=3((4,5),(3,4))#

#=((12,15),(9,12))#