Prove Theorem 2(b) and 2(c). Use the row-column rule. The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\) or \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} \).

If the product AB is defined, the entry in row \(i\) and column \(j\) of ABis the sum of the products of corresponding entries from the row \(i\)of Aand column \(j\) of B. If \({\left( {AB} \right)_{ij}}\) denotes the \(\left( {i,j} \right)\)- entry in AB, and if Ais a \(m \times n\) matrix, then

\({\left( {AB} \right)_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + ... + {a_{in}}{b_{nj}}\).

Theorem 2states that Abe a \(m \times n\) matrix let Band Chave sizes for which the indicated sums and products are defined.

1. \(A\left( {BC} \right) = \left( {AB} \right)C\) (associative law of multiplication)
2. \(A\left( {B + C} \right) = AB + AC\) (left distributive law)
3. \(\left( {B + C} \right)A = BA + CA\) (right distributive law)

The \(\left( {i,j} \right)\)- entry in \(A\left( {B + C} \right)\) can be written as \({a_{i1}}\left( {{b_{1j}} + {c_{1j}}} \right) + ... + {a_{in}}\left( {{b_{nj}} + {c_{nj}}} \right)\).

The \(\left( {i,j} \right)\)- entry of \(A\left( {B + C} \right)\) equals to the \(\left( {i,j} \right)\)- entry of \(AB + AC\) since \(\sum\limits_{k = 1}^n {{a_{ik}}\left( {{b_{kj}} + {c_{kj}}} \right)} = \sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} + \sum\limits_{k = 1}^n {{a_{ik}}{c_{kj}}} \).

The \(\left( {i,j} \right)\)- entry of \(\left( {B + C} \right)A\) equals to the \(\left( {i,j} \right)\)- entry of \(BA + CA\) since \(\sum\limits_{k = 1}^n {\left( {{b_{ik}} + {c_{ik}}} \right){a_{kj}}} = \sum\limits_{k = 1}^n {{b_{ik}}{a_{kj}}} + \sum\limits_{k = 1}^n {{c_{ik}}{a_{kj}}} \).

Hence, the theorems 2(b) and 2(c) are proved.