46 CHAPTER 2. LINEAR TRANSFORMATIONS

2.1.3 Properties of Matrix Multiplication

As pointed out above, sometimes it is possible to multiply matrices in one order but notin the other order. What if it makes sense to multiply them in either order? Will they beequal then?

Example 2.1.14 Compare

(1 2

3 4

)(0 1

1 0

)and

(0 1

1 0

)(1 2

3 4

).

The first product is (1 2

3 4

)(0 1

1 0

)=

(2 1

4 3

),

the second product is (0 1

1 0

)(1 2

3 4

)=

(3 4

1 2

),

and you see these are not equal. Therefore, you cannot conclude that AB = BA for matrixmultiplication. However, there are some properties which do hold.

Proposition 2.1.15 If all multiplications and additions make sense, the following hold formatrices, A,B,C and a, b scalars.

A (aB + bC) = a (AB) + b (AC) (2.13)

(B + C)A = BA+ CA (2.14)

A (BC) = (AB)C (2.15)

Proof: Using the above definition of matrix multiplication,

(A (aB + bC))ij =∑k

Aik (aB + bC)kj

=∑k

Aik (aBkj + bCkj)

= a∑k

AikBkj + b∑k

AikCkj

= a (AB)ij + b (AC)ij= (a (AB) + b (AC))ij

showing that A (B + C) = AB +AC as claimed. Formula 2.14 is entirely similar.Consider 2.15, the associative law of multiplication. Before reading this, review the

definition of matrix multiplication in terms of entries of the matrices.

(A (BC))ij =∑k

Aik (BC)kj

=∑k

Aik

∑l

BklClj

=∑l

(AB)il Clj

= ((AB)C)ij .■

46 CHAPTER 2. LINEAR TRANSFORMATIONS2.1.3 Properties of Matrix MultiplicationAs pointed out above, sometimes it is possible to multiply matrices in one order but notin the other order. What if it makes sense to multiply them in either order? Will they beequal then?1 2 1 1 1 2Example 2.1.14 Compare 0 and 0 .3.4 1 0 1 0 3.4The first product is1 2 01) (213 4 10) \4 3)?01 12\) (341 0 34) \i1 2]?and you see these are not equal. Therefore, you cannot conclude that AB = BA for matrixmultiplication. However, there are some properties which do hold.the second product isProposition 2.1.15 Jf all multiplications and additions make sense, the following hold formatrices, A,B,C and a,b scalars.A(aB + 6C) =a(AB) +b(AC) (2.13)(B+C)A=BA+CA (2.14)A(BC) = (AB)C (2.15)Proof: Using the above definition of matrix multiplication,(A(aB+bC)),, = So Aix (aB+bC),;k= So Aix (aBrj + bCx;)k= a> Ain Brj +0 > AinCrk k= a(AB),; +b(AC);,= (a(AB) +b(AC)),,showing that A(B+C) = AB+ AC as claimed. Formula 2.14 is entirely similar.Consider 2.15, the associative law of multiplication. Before reading this, review thedefinition of matrix multiplication in terms of entries of the matrices.(A (BC));; = > Aix (BC);kS> Ain S> Brijk 1S> (AB), CjI((AB)C),;