Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

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

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

\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}I&0&0\\X&I&0\\Y&0&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{{A_{21}}}&{{A_{22}}}\\{{A_{31}}}&{{A_{32}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\0&{{B_{22}}}\\0&{{B_{32}}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{I\left( {{A_{11}}} \right) + 0\left( {{A_{21}}} \right) + 0\left( {{A_{31}}} \right)}&{I\left( {{A_{12}}} \right) + 0\left( {{A_{22}}} \right) + 0\left( {{A_{32}}} \right)}\\{X\left( {{A_{11}}} \right) + I\left( {{A_{21}}} \right) + 0\left( {{A_{31}}} \right)}&{X\left( {{A_{12}}} \right) + I\left( {{A_{22}}} \right) + 0\left( {{A_{32}}} \right)}\\{Y\left( {{A_{11}}} \right) + 0\left( {{A_{21}}} \right) + I\left( {{A_{31}}} \right)}&{Y\left( {{A_{12}}} \right) + 0\left( {{A_{22}}} \right) + I\left( {{A_{32}}} \right)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\0&{{B_{22}}}\\0&{{B_{32}}}\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{X{A_{11}} + {A_{21}}}&{X{A_{12}} + {A_{22}}}\\{Y{A_{11}} + {A_{31}}}&{Y{A_{12}} + {A_{32}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\0&{{B_{22}}}\\0&{{B_{32}}}\end{array}} \right]\end{array}\]

Thus, \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{X{A_{11}} + {A_{21}}}&{X{A_{12}} + {A_{22}}}\\{Y{A_{11}} + {A_{31}}}&{Y{A_{12}} + {A_{32}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\0&{{B_{22}}}\\0&{{B_{32}}}\end{array}} \right]\].

Equate both the matrices, as shown below:

\[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{12}}}\\{X{A_{11}} + {A_{21}}}&{X{A_{12}} + {A_{22}}}\\{Y{A_{11}} + {A_{31}}}&{Y{A_{12}} + {A_{32}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\0&{{B_{22}}}\\0&{{B_{32}}}\end{array}} \right]\]

By comparing, the formulas obtained are shown below:

\[\begin{array}{c}X{A_{11}} + {A_{21}} = 0\\X{A_{11}} = - {A_{21}}\\X\left( {{A_{11}}^{ - 1}{A_{11}}} \right) = - {A_{11}}^{ - 1}{A_{21}}\\X = - {A_{11}}^{ - 1}{A_{21}}\end{array}\]

And,

\[\begin{array}{c}Y{A_{11}} + {A_{31}} = 0\\Y{A_{11}} = - {A_{31}}\\Y\left( {{A_{11}}^{ - 1}{A_{11}}} \right) = - {A_{11}}^{ - 1}{A_{31}}\\Y = - {A_{11}}^{ - 1}{A_{31}}\end{array}\]

Therefore, the formulas are \[X = - {A_{11}}^{ - 1}{A_{21}}\] and \[Y = - {A_{11}}^{ - 1}{A_{31}}\].

By equating the entries, it is also observed that \[X{A_{12}} + {A_{22}} = {B_{22}}\].

\[\begin{array}{l}{B_{22}} = X{A_{12}} + {A_{22}}\\{B_{22}} = - {A_{11}}^{ - 1}{A_{21}}{A_{12}} + {A_{22}}\\{B_{22}} = {A_{22}} - {A_{21}}{A_{11}}^{ - 1}{A_{12}}\end{array}\]

Thus, \[{B_{22}} = {A_{22}} - {A_{21}}{A_{11}}^{ - 1}{A_{12}}\].