91Ó°ÊÓ

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

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}A&B\\0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\0&0&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{A\left( X \right) + B\left( 0 \right)}&{A\left( Y \right) + B\left( 0 \right)}&{A\left( Z \right) + B\left( I \right)}\\{0\left( X \right) + I\left( 0 \right)}&{0\left( Y \right) + I\left( 0 \right)}&{0\left( Z \right) + I\left( I \right)}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{AX + B \cdot 0}&{AY + B \cdot 0}&{AZ + B \cdot I}\\{0 \cdot X + I \cdot 0}&{0 \cdot Y + I \cdot 0}&{0 \cdot Z + I \cdot I}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{AX}&{AY}&{AZ + B}\\0&0&I\end{array}} \right]\end{array}\]

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

\[\left[ {\begin{array}{*{20}{c}}A&B\\0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\0&0&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{AX}&{AY}&{AZ + B}\\0&0&I\end{array}} \right]\]

Thus, \[\left[ {\begin{array}{*{20}{c}}A&B\\0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\0&0&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{AX}&{AY}&{AZ + B}\\0&0&I\end{array}} \right]\].

Equate both the matrices as shown below:

\[\left[ {\begin{array}{*{20}{c}}{AX}&{AY}&{AZ + B}\\0&0&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&0&0\\0&0&I\end{array}} \right]\]

By comparing, the formulas become

\[\begin{array}{c}AX = I\\\left( {{A^{ - 1}}A} \right)X = {A^{ - 1}}\left( I \right)\\I \cdot X = {A^{ - 1}} \cdot I\\X = {A^{ - 1}}\end{array}\]

\[\begin{array}{c}AY = 0\\\left( {{A^{ - 1}}A} \right)Y = {A^{ - 1}}\left( 0 \right)\\Y = 0\end{array}\]

and

\(\begin{array}{c}AZ + B = 0\\AZ = 0 - B\\AZ = - B.\end{array}\)

Solve further to get

\[\begin{array}{c}{A^{ - 1}}\left( {AZ} \right) = {A^{ - 1}}\left( { - B} \right)\\\left( {{A^{ - 1}}A} \right)Z = {A^{ - 1}}\left( { - B} \right)\\\left( {{A^{ - 1}}A} \right)Z = - {A^{ - 1}}B\\I \cdot Z = - {A^{ - 1}}B\\Z = - {A^{ - 1}}B.\end{array}\]

Therefore, the formulas are \(X = {A^{ - 1}}\), \(Y = 0\), and \[Z = - {A^{ - 1}}B\].