Question

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, find formulas for X, Y, and Zin 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.]

8. \[\left[ {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\{\bf{0}}&{\bf{0}}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&I\end{array}} \right]\]

Short answer

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

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 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 sides

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 \[Z = - {A^{ - 1}}B\].