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Chapter 2: Q2.4-22Q (page 93)

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

Expert verified

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

Step by step solution

01

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

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}\\{\bf{5}}&{{\bf{12}}}\end{aligned}} \right),{b_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}\\{\bf{3}}\end{aligned}} \right),{b_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{ - {\bf{5}}}\end{aligned}} \right),{b_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{6}}\end{aligned}} \right),\) and \({b_{\bf{4}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{5}}\end{aligned}} \right)\).

  1. Find \({A^{ - {\bf{1}}}}\), and use it to solve the four equations \(Ax = {b_{\bf{1}}},\)\(Ax = {b_2},\)\(Ax = {b_{\bf{3}}},\)\(Ax = {b_{\bf{4}}}\)\(\)
  2. The four equations in part (a) can be solved by the same set of row operations, since the coefficient matrix is the same in each case. Solve the four equations in part (a) by row reducing the augmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{b_{\bf{1}}}}&{{b_{\bf{2}}}}&{{b_{\bf{3}}}}&{{b_{\bf{4}}}}\end{aligned}} \right)\).

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

Give a formula for \({\left( {ABx} \right)^T}\), where \({\bf{x}}\) is a vector and \(A\) and \(B\) are matrices of appropriate sizes.

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

When a deep space probe launched, corrections may be necessary to place the probe on a precisely calculated trajectory. Radio elementary provides a stream of vectors, \({{\bf{x}}_{\bf{1}}},....,{{\bf{x}}_k}\), giving information at different times about how the probe鈥檚 position compares with its planned trajectory. Let \({X_k}\) be the matrix \(\left[ {{x_{\bf{1}}}.....{x_k}} \right]\). The matrix \({G_k} = {X_k}X_k^T\) is computed as the radar data are analyzed. When \({x_{k + {\bf{1}}}}\) arrives, a new \({G_{k + {\bf{1}}}}\) must be computed. Since the data vector arrive at high speed, the computational burden could be serve. But partitioned matrix multiplication helps tremendously. Compute the column-row expansions of \({G_k}\) and \({G_{k + {\bf{1}}}}\) and describe what must be computed in order to update \({G_k}\) to \({G_{k + {\bf{1}}}}\).
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