Question

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

4. \[\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Short answer

The product is \(\left[ {\begin{array}{*{20}{c}}A&B\\{ - XA + C}&{ - XB + D}\end{array}} \right]\).

Step-by-step solution

State the row-column rule

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:

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

Obtain the product

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

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{I\left( A \right) + 0\left( C \right)}&{I\left( B \right) + 0\left( D \right)}\\{ - X\left( A \right) + I\left( C \right)}&{ - X\left( B \right) + I\left( D \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{IA}&{IB}\\{ - XA + IC}&{ - XB + ID}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}A&B\\{ - XA + C}&{ - XB + D}\end{array}} \right]\end{array}\)

Thus, \(\left[ {\begin{array}{*{20}{c}}0&I\\I&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}A&B\\{ - XA + C}&{ - XB + D}\end{array}} \right]\).