In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

If the sum of the products of matching entries from row\(i\)of matrix A and column\(j\)of matrix B equals the item in row\(i\)and column\(j\)of AB, then it can be said that product AB is defined.

The product is shown below:

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

Compute the product by using the row-column rule, as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}E&0\\0&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{E\left( A \right) + 0\left( C \right)}&{E\left( B \right) + 0\left( D \right)}\\{0\left( A \right) + F\left( C \right)}&{0\left( B \right) + F\left( D \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{EA}&{EB}\\{FC}&{FD}\end{array}} \right]\end{array}\)

Thus, \(\left[ {\begin{array}{*{20}{c}}E&0\\0&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{EA}&{EB}\\{FC}&{FD}\end{array}} \right]\).