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Q2.4-7Q

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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. In Exercises 5–8, find formulas for X, Y, and Z in terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint: Compute the product on the left, and set it equal to the right side.]7. $\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\{\bf{0}}&{\bf{0}}\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]$

The formulas are $$X = {A^{ - 1}}$$, $$Y = - B{A^{ - 1}}$$, and $Z = 0$.

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 of the left part of the given equation by using the row-column rule, as shown below:

$\begin{array}{c}\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{X\left( A \right) + 0\left( 0 \right) + 0\left( B \right)}&{X\left( Z \right) + 0\left( 0 \right) + 0\left( I \right)}\\{Y\left( A \right) + 0\left( 0 \right) + I\left( B \right)}&{Y\left( Z \right) + 0\left( 0 \right) + I\left( I \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{XA + 0 + 0 \cdot B}&{XZ + 0 + 0 \cdot I}\\{YA + 0 + I \cdot B}&{YZ + 0 + {I^2}}\end{array}} \right]\end{array}$

Use the matrix properties $$A \cdot I = A$$, $$A \cdot 0 = 0$$, and $${I^2} = I$$.

$\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right]$

Thus, $\left[ {\begin{array}{*{20}{c}}X&0&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&Z\\0&0\\B&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right]$.

## Step 3: Equate both the sides

Equate both the matrices as shown below:

$\left[ {\begin{array}{*{20}{c}}{XA}&{XZ}\\{YA + B}&{YZ + I}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0\\0&I\end{array}} \right]$

By comparing, the formulas become

$$\begin{array}{c}XA = I\\X\left( {{A^{ - 1}}A} \right) = {A^{ - 1}}\left( I \right)\\X \cdot I = {A^{ - 1}} \cdot I\\X = {A^{ - 1}}\end{array}$$

$\begin{array}{c}XZ = 0\\{X^{ - 1}}\left( {XZ} \right) = {X^{ - 1}}\left( 0 \right)\\Z = 0\end{array}$

and

$$\begin{array}{c}YA + B = 0\\YA = 0 - B\\YA = - B\end{array}$$

Solve further to get

$$\begin{array}{c}\left( {YA} \right){A^{ - 1}} = \left( { - B} \right){A^{ - 1}}\\Y \cdot \left( {A{A^{ - 1}}} \right) = - B{A^{ - 1}}\\Y \cdot I = - B{A^{ - 1}}\\Y = - B{A^{ - 1}}.\end{array}$$

Therefore, the formulas are $$X = {A^{ - 1}}$$, $$Y = - B{A^{ - 1}}$$, and $Z = 0$.