Question

Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)

Short answer

Theorem 2(d) is proved.

Step-by-step solution

Statement of theorem 2

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)
4. \(r\left( {AB} \right) = \left( {rA} \right)B = A\left( {rB} \right)\) (for any scalar \(r\))

Prove theorem 2(d)

The \(\left( {i,j} \right)\)- entries for \(r\left( {AB} \right),\left( {rA} \right)B,\) and \(A\left( {rB} \right)\) are all the same since

\(r\sum\limits_{k = 1}^n {{a_{ik}}{b_{kj}}} = \sum\limits_{k = 1}^n {\left( {r{a_{ik}}} \right){b_{kj}}} = \sum\limits_{k = 1}^n {{a_{ik}}\left( {r{b_{kj}}} \right)} \).

Hence, theorem 2(d) is proved.