We prove each property by examining the entries of the matrices.
Property 1: Associative Property of Multiplication
Let
\(A = [a_{ij}]\) be an
\(m \times n\) matrix,
\(B = [b_{jk}]\) be an
\(n \times p\) matrix, and
\(C = [c_{k\ell}]\) be a
\(p \times q\) matrix.
For the left side \(A(BC)\text{,}\) we first compute \(BC = [d_{j\ell}]\) where
\begin{equation*}
d_{j\ell} = \sum_{k=1}^{p} b_{jk}c_{k\ell}
\end{equation*}
Then the \((i,\ell)\)-entry of \(A(BC)\) is:
\begin{align*}
[A(BC)]_{i\ell} \amp= \sum_{j=1}^{n} a_{ij}d_{j\ell}\\
\amp= \sum_{j=1}^{n} a_{ij}\left(\sum_{k=1}^{p} b_{jk}c_{k\ell}\right)\\
\amp= \sum_{j=1}^{n} \sum_{k=1}^{p} a_{ij}b_{jk}c_{k\ell}
\end{align*}
For the right side \((AB)C\text{,}\) we first compute \(AB = [e_{ik}]\) where
\begin{equation*}
e_{ik} = \sum_{j=1}^{n} a_{ij}b_{jk}
\end{equation*}
Then the \((i,\ell)\)-entry of \((AB)C\) is:
\begin{align*}
[(AB)C]_{i\ell} \amp= \sum_{k=1}^{p} e_{ik}c_{k\ell}\\
\amp= \sum_{k=1}^{p} \left(\sum_{j=1}^{n} a_{ij}b_{jk}\right)c_{k\ell}\\
\amp= \sum_{k=1}^{p} \sum_{j=1}^{n} a_{ij}b_{jk}c_{k\ell}
\end{align*}
By
Lemma 2.3.4, we can exchange the order of summation:
\begin{equation*}
\sum_{j=1}^{n} \sum_{k=1}^{p} a_{ij}b_{jk}c_{k\ell} = \sum_{k=1}^{p} \sum_{j=1}^{n} a_{ij}b_{jk}c_{k\ell}
\end{equation*}
Therefore, \([A(BC)]_{i\ell} = [(AB)C]_{i\ell}\) for all \(i\) and \(\ell\text{,}\) which proves \(A(BC) = (AB)C\text{.}\)
Property 2: Left Distributive Property
Let
\(A = [a_{ij}]\) be an
\(m \times n\) matrix, and let
\(B = [b_{jk}]\) and
\(C = [c_{jk}]\) be
\(n \times p\) matrices.
The \((i,k)\)-entry of \(A(B+C)\) is:
\begin{align*}
[A(B+C)]_{ik} \amp= \sum_{j=1}^{n} a_{ij}(B+C)_{jk}\\
\amp= \sum_{j=1}^{n} a_{ij}(b_{jk} + c_{jk})\\
\amp= \sum_{j=1}^{n} (a_{ij}b_{jk} + a_{ij}c_{jk})\\
\amp= \sum_{j=1}^{n} a_{ij}b_{jk} + \sum_{j=1}^{n} a_{ij}c_{jk}\\
\amp= [AB]_{ik} + [AC]_{ik}\\
\amp= [AB + AC]_{ik}
\end{align*}
Therefore, \(A(B+C) = AB + AC\text{.}\)
Property 3: Right Distributive Property
Let
\(A = [a_{ik}]\) and
\(B = [b_{ik}]\) be
\(m \times n\) matrices, and let
\(C = [c_{k\ell}]\) be an
\(n \times p\) matrix.
The \((i,\ell)\)-entry of \((A+B)C\) is:
\begin{align*}
[(A+B)C]_{i\ell} \amp= \sum_{k=1}^{n} (A+B)_{ik}c_{k\ell}\\
\amp= \sum_{k=1}^{n} (a_{ik} + b_{ik})c_{k\ell}\\
\amp= \sum_{k=1}^{n} (a_{ik}c_{k\ell} + b_{ik}c_{k\ell})\\
\amp= \sum_{k=1}^{n} a_{ik}c_{k\ell} + \sum_{k=1}^{n} b_{ik}c_{k\ell}\\
\amp= [AC]_{i\ell} + [BC]_{i\ell}\\
\amp= [AC + BC]_{i\ell}
\end{align*}
Therefore, \((A+B)C = AC + BC\text{.}\)
Property 4: Scalar Multiplication Compatibility
Let
\(A = [a_{ij}]\) be an
\(m \times n\) matrix,
\(B = [b_{jk}]\) be an
\(n \times p\) matrix, and let
\(c\) be a scalar.
For \(c(AB)\text{,}\) the \((i,k)\)-entry is:
\begin{align*}
[c(AB)]_{ik} \amp= c \cdot [AB]_{ik}\\
\amp= c \cdot \sum_{j=1}^{n} a_{ij}b_{jk}\\
\amp= \sum_{j=1}^{n} c(a_{ij}b_{jk})\\
\amp= \sum_{j=1}^{n} (ca_{ij})b_{jk}\\
\amp= [(cA)B]_{ik}
\end{align*}
This shows that \(c(AB) = (cA)B\text{.}\)
Similarly, we can also write:
\begin{align*}
\sum_{j=1}^{n} c(a_{ij}b_{jk}) \amp= \sum_{j=1}^{n} a_{ij}(cb_{jk})\\
\amp= [A(cB)]_{ik}
\end{align*}
This shows that \(c(AB) = A(cB)\text{.}\)
Combining these results, we have
\(c(AB) = (cA)B = A(cB)\text{.}\)
This completes the proof of all four properties.
