Question

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

Short answer

\({A^2} = \left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\0&{{I_2}}\end{array}} \right]\).

Step-by-step solution

Show the construction of a \(5 \times 5\) matrix

You have to generalize the concept of exercise 21(a) by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\), such that \({M^2} = I\).

Consider Cis a non-zero \(2 \times 3\) matrix and \(A = \left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\C&{ - {I_2}}\end{array}} \right]\).

\[\begin{array}{c}{A^2} = \left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\C&{ - {I_2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\C&{ - {I_2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{I_3} + 0}&{0 + 0}\\{C{I_3} - {I_2}C}&{0 + {{\left( { - {I_2}} \right)}^2}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\0&{{I_2}}\end{array}} \right]\end{array}\]

Thus, \({A^2} = \left[ {\begin{array}{*{20}{c}}{{I_3}}&0\\0&{{I_2}}\end{array}} \right]\).