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Q2.4-3Q

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Found in: Page 93

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.3. $\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]$

The product is $\left[ {\begin{array}{*{20}{c}}Y&Z\\W&X\end{array}} \right]$.

See the step by step solution

## Step 1: 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:

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

## Step 2: Obtain the product

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

$$\begin{array}{c}\left[ {\begin{array}{*{20}{c}}0&I\\I&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{0\left( W \right) + I\left( Y \right)}&{0\left( X \right) + I\left( Z \right)}\\{I\left( W \right) + 0\left( Y \right)}&{I\left( X \right) + 0\left( Z \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{IY}&{IZ}\\{IW}&{IX}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}Y&Z\\W&X\end{array}} \right]\end{array}$$

Thus, $\left[ {\begin{array}{*{20}{c}}0&I\\I&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}Y&Z\\W&X\end{array}} \right]$.