2.4: Properties of Matrix Multiplication

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As pointed out above, it is sometimes possible to multiply matrices in one order but not in the other order. However, even if both \(AB\) and \(BA\) are defined, they may not be equal.

Example \(\PageIndex{1}\): Matrix Multiplication is Not Commutative

Compare the products \(AB\) and \(BA\), for matrices \(A = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right], B= \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right]\)

Solution

First, notice that \(A\) and \(B\) are both of size \(2 \times 2\). Therefore, both products \(AB\) and \(BA\) are defined. The first product, \(AB\) is

\[AB = \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right] \left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] = \left[ \begin{array}{rr} 2 & 1 \\ 4 & 3 \end{array} \right] \nonumber\]

The second product, \(BA\) is

\[\left[ \begin{array}{rr} 0 & 1 \\ 1 & 0 \end{array} \right] \left[ \begin{array}{rr} 1 & 2 \\ 3 & 4 \end{array} \right] = \left[ \begin{array}{rr} 3 & 4 \\ 1 & 2 \end{array} \right] \nonumber\]

Therefore, \(AB \neq BA\).

This example illustrates that you cannot assume \(AB=BA\) even when multiplication is defined in both orders. If for some matrices \(A\) and \(B\) it is true that \(AB=BA\), then we say that \(A\) and \(B\) commute. This is one important property of matrix multiplication.

The following are other important properties of matrix multiplication. Notice that these properties hold only when the size of matrices are such that the products are defined.

Proposition \(\PageIndex{1}\): Properties of Matrix Multiplication

The following hold for matrices \(A,B,\) and \(C\) and for scalars \(r\) and \(s\),

\[ \begin{align} A\left( rB+sC\right) &= r\left( AB\right) +s\left( AC\right) \label{matrixproperties1} \\[4pt] \left( B+C\right) A &=BA+CA \label{matrixproperties2} \\[4pt] A\left( BC\right) &=\left( AB\right) C \label{matrixproperties3} \end{align}\]

Proof

First we will prove \(\eqref{matrixproperties1}\). We will use Definition 2.3.1 and prove this statement using the \(ij^{th}\) entries of a matrix. Therefore,

\[\begin{aligned} \left( A\left( rB+sC\right) \right) _{ij} &=\sum_{k}a_{ik}\left( rB+sC\right) _{kj} \\[4pt] &= \sum_{k}a_{ik}\left( rb_{kj}+sc_{kj}\right) \\[4pt] &=r\sum_{k}a_{ik}b_{kj}+s\sum_{k}a_{ik}c_{kj} \\[4pt] &=r\left( AB\right) _{ij}+s\left( AC\right) _{ij} \\[4pt] &=\left( r\left( AB\right) +s\left( AC\right) \right) _{ij} \end{aligned}\]

Thus \(A\left( rB+sC\right) =r(AB)+s(AC) \) as claimed.

The proof of Equation \(\eqref{matrixproperties2}\) follows the same pattern and is left as an exercise.

Statement Equation \(\eqref{matrixproperties3}\) is the associative law of multiplication. Using Definition 2.3.1,

\[ \begin{align*}\left( A\left( BC\right) \right) _{ij} &=\sum_{k}a_{ik}\left( BC\right) _{kj} \\[4pt] &=\sum_{k}a_{ik}\sum_{l}b_{kl}c_{lj} \\[4pt] &=\sum_{l}\left( AB\right) _{il}c_{lj}=\left( \left( AB\right) C\right) _{ij}. \end{align*}\]

This proves \(\eqref{matrixproperties3}\).