Properties of Matrix Arithmetic

Theorem

Assuming that the sizes of the matrices are such that the indicated operations can be performed, the following rules of matrix arithmetic are valid.

$$\begin{array}{rlrl}(a)& A+B=B+A& & \text{[Commutative law for matrix addition]}\\ (b)& A+(B+C)=(A+B)+C& & \text{[Associative law for matrix addition]}\\ (c)& A(BC)=(AB)C& & \text{[Associative law for matrix multiplication]}\\ (d)& A(B+C)=AB+AC& & \text{[Left distributive law]}\\ (e)& (B+C)A=BA+CA& & \text{[Right distributive law]}\\ (f)& A(B-C)=AB-AC& & \\ (g)& (B-C)A=BA-CA& & \\ (h)& a(B+C)=aB+aC& & \\ (i)& a(B-C)=aB-aC& & \\ (j)& (a+b)C=aC+bC& & \\ (k)& (a-b)C=aC-bC& & \\ (l)& a(bC)=(ab)C& & \\ (m)& a(BC)=(aB)C=B(aC)& & \end{array}$$

Remark

Unlike real numbers arithmetic where $ab=ba$, the equality of $AB$ and $BA$ can fail for 3 possible reasons:

1. $AB$ is defined, but not $BA$ (e.g., when $A$ is $2\times 3$ and $B$ is $3\times 4$)

2. $AB$ and $BA$ is defined, but have different sizes (e.g., when $A$ is $2\times 3$ and $B$ is $3\times 2$)

3. $AB$ and $BA$ is defined, have same size, but is different

   Let

   $$A=\left[\begin{array}{cc}-1& 0\\ 2& 3\end{array}\right]\text{and}B=\left[\begin{array}{cc}1& 2\\ 3& 0\end{array}\right]$$

   Multiplying gives

   $$AB=\left[\begin{array}{cc}-1& -2\\ 11& 4\end{array}\right]\text{and}BA=\left[\begin{array}{cc}3& 6\\ -3& 0\end{array}\right]$$

   Thus, $AB\ne BA$.