Even if matrix A can be multiplied with matrix B and matrix B can be multiplied to matrix A, this doesn't necessarily give us that AB=BA. In other words, unlike the integers, matrices are noncommutative.

Property 1: Associative Property of Multiplication

A(BC) = (AB)C

where A,B, and C are matrices of scalar values.

Proof:

(1) Let D = AB, G = BC

(2) Let F = (AB)C = DC

(3) Let H = A(BC) = AG

(4) Using Definition 1, here, we have for each D,F,G,H:

d

_{i,j}= ∑

_{k}a

_{i,k}*b

_{k,j}

g

_{i,j}= ∑

_{k}b

_{i,k}*c

_{k,j}

f

_{i,j}= ∑

_{k}d

_{i,k}*c

_{k,j}

(5) So, expanding f

_{i,j}gives us:

f

_{i,j}= ∑

_{k}(∑

_{l}a

_{i,l}*b

_{l,j})c

_{k,j}=

(∑

_{k}∑

_{l}) a

_{i,l}*b

_{l,k}*c

_{k,j}=

= ∑

_{l}a

_{i,l}*(∑

_{k}b

_{l,k}*c

_{k,j}) =

= ∑

_{l}a

_{i,l}*g

_{l,j}= h

_{i,j}

QED

Property 2: Distributive Property of Multiplication

A(B + C) = AB + AC

(A + B)C = AC + BC

where A,B,C are matrices of scalar values.

Proof:

(1) Let D = AB such that for each:

d

_{i,j}= ∑

_{k}a

_{i,k}*b

_{k,j}

(2) Let E = AC such that for each:

e

_{i,j}= ∑

_{k}a

_{i,k}*c

_{k,j}

(3) Let F = D + E = AB + AC such that for each:

f

_{i,j}= ∑

_{k}a

_{i,k}*b

_{k,j}+a

_{i,k}*c

_{k,j}= ∑

_{k}a

_{i,k}[b

_{k,j}+ c

_{k,j}]

(4) Let G = B+C such that for each:

g

_{i,j}= b

_{i,j}+ c

_{i,j}

(5) Let H = A(B+C) = AG such that for each:

h

_{i,j}= ∑

_{k}a

_{i,k}*g

_{k,j}

(6) Then we have AB + AC = A(B+C) since for each:

h

_{i,j}= ∑

_{k}a

_{i,k}[b

_{k,j}+ c

_{k,j}]

(7) Let M = A + B such that for each:

m

_{i,j}= a

_{i,j}+ b

_{i,j}

(8) Let N = (A+B)C = MC such that:

n

_{i,j}= ∑

_{k}m

_{i,k}*c

_{k,j}=

= ∑

_{k}(a

_{i,k}+ b

_{i,k})*c

_{k,j}

(9) Let O = BC such that:

o

_{i,j}= ∑

_{k}b

_{i,k}*c

_{k,j}

(10) Let P = AC + BC = E + O such that:

p

_{i,j}= e

_{i,j}+ o

_{i,j}=

= ∑

_{k}a

_{i,k}*c

_{k,j}+ ∑

_{k}b

_{i,k}*c

_{k,j}=

= ∑

_{k}[a

_{i,k}*c

_{k,j}+ b

_{i,k}*c

_{k,j}] =

= ∑

_{k}(a

_{i,k}+ b

_{i,k})*c

_{k,j}

QED

Property 3: Scalar Multiplication

c(AB) = (cA)B = A(cB)

Proof:

(1) Let D = AB such that:

d

_{i,j}= ∑

_{k}a

_{i,k}*b

_{k,j}

(2) Let E = c(AB) = cD such that for each:

e

_{i,j}= c*d

_{i,j}= c*∑

_{k}a

_{i,k}*b

_{k,j}

(3) Let F = (cA)B such that:

f

_{i,j}= ∑

_{k}(c*a

_{i,k})*b

_{k,j}= c*∑

_{k}a

_{i,k}*b

_{k,j}

(4) Let G = A(cB) such that:

g

_{i,j}= ∑

_{k}a

_{i,k}*(c*b

_{k,j}) =

= c*∑

_{k}a

_{i,k}*b

_{k,j}

QED

Property 4: Muliplication of Matrices is not Commutative

AB does not have to = BA

Proof:

(1) Let A =

(2) Let B =

(3) AB =

(4) BA =

QED

References

- Hans Schneider, George Philip Barker, Matrices and Linear Algebra, 1989.

## 6 comments :

Hi,

I am a math teacher.

You can revise your understanding of matrices solving exercises about addition, subtraction and multiplication of matrices. Inverse,... at www.emathematics.net

Fantastic!!!

hi

i am a student taking a course in linear algebra. i just wanted to say that your tutorrial is very helpfull. so post some more on this topic.

thank you

Found it helpful. Thank you!

Hi! Thank you very much for providing this resource! I am troubled, however, by something in your proof of the associativity of matrix multiplication. (Since blogger.com will not let me use the necessary HTML tags, I will attempt to use LaTeX-like notation instead.)

In (4), you have

d_{i,j} = \sum_{k} a_{i,k}*b_{k,j}

f_{i,k} = \sum_{k} d_{i,k}*c_{k,j}

Then, in (5), you expand this last equation as

f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,j}) c_{k,j}

However, we are expanding d_{i,k} from the previous equation, not d_{i,j} (where the outer summation symbol binds the variable k). And

d_{i,k} = \sum_{l} a_{i,l}*b_{l,k}

The summation symbol that binds k becomes the outer summation symbol in the expansion. So the correct expansion of the whole equation should be:

f_{i,j} = \sum_{k} (\sum_{l} a_{i,l}*b_{l,k}) c_{k,j}

which does not lend itself to so neat a proof as the one you present. A correct proof, though with rather messier notation, can be found here:

http://linear.ups.edu/xml/0094/fcla-xml-0.94li30.xml

Another page that has a correct proof (avoiding the error noted above) can be found here:

http://www.proofwiki.org/wiki/Matrix_Multiplication_is_Associative#Proof

Hi Amittai,

Thanks very much for pointing out the mistake. I will revise the proof and post a comment on this blog when it is fixed.

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