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Chapter 2: Q28Q (page 93)

Exercises 27 and 28 prove special cases of the facts about elementary matrices stated in the box following Example 5. Here Ais a \({\bf{3}} \times {\bf{3}}\) matrix and \(I = {I_{\bf{3}}}\). (A general proof would require slightly more notation.)

28. Show that if row 3 of Ais replaced by \(ro{w_3}\left( A \right) - 4 \cdot ro{w_1}\left( A \right)\) the result is EA, where Eis formed from Iby replacing \(ro{w_3}\left( I \right) - 4 \cdot ro{w_{\bf{1}}}\left( I \right)\).

Short Answer

Expert verified

If row three of Ais replaced by\({\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)\), the result becomes EA.

Step by step solution

01

Apply the row operations

Suppose for\(i = 1,2,3\), matrix A is\(A = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right)}\end{aligned}} \right)\).

Use row one to obtain\({\rm{ro}}{{\rm{w}}_3}\left( A \right)\)in terms of rows three and four. Add \( - 4\) times \({\rm{ro}}{{\rm{w}}_1}\left( A \right)\) to \({\rm{ro}}{{\rm{w}}_3}\left( A \right)\).

\(\left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right)}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)}\end{aligned}} \right)\)

02

Use the equation of identity matrix

Apply the equation\({\rm{ro}}{{\rm{w}}_i}\left( A \right) = {\rm{ro}}{{\rm{w}}_i}\left( I \right) \cdot A\).

\(\begin{aligned}{c}\left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( A \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( I \right) \cdot A}\\{{\rm{ro}}{{\rm{w}}_2}\left( I \right) \cdot A}\\{{\rm{ro}}{{\rm{w}}_3}\left( I \right) \cdot A - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( I \right) \cdot A}\end{aligned}} \right)\\ = \left( {\begin{aligned}{*{20}{c}}{{\rm{ro}}{{\rm{w}}_1}\left( I \right)}\\{{\rm{ro}}{{\rm{w}}_2}\left( I \right)}\\{{\rm{ro}}{{\rm{w}}_3}\left( I \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( I \right)}\end{aligned}} \right)A\\ = EA\end{aligned}\)

Hence, proved.

Therefore, if row three of Ais replaced by\({\rm{ro}}{{\rm{w}}_3}\left( A \right) - 4 \cdot {\rm{ro}}{{\rm{w}}_1}\left( A \right)\), the result becomes EA.

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Most popular questions from this chapter

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)

Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{4}}}\\{\bf{7}}&{ - {\bf{8}}}\end{aligned}} \right)\).

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

3. \[\left[ {\begin{array}{*{20}{c}}{\bf{0}}&I\\I&{\bf{0}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}W&X\\Y&Z\end{array}} \right]\]

In Exercises 1鈥�9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1鈥�4.

4. \[\left[ {\begin{array}{*{20}{c}}I&0\\{ - X}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Assume \(A - s{I_n}\) is invertible and view (8) as a system of two matrix equations. Solve the top equation for \({\bf{x}}\) and substitute into the bottom equation. The result is an equation of the form \(W\left( s \right){\bf{u}} = {\bf{y}}\), where \(W\left( s \right)\) is a matrix that depends upon \(s\). \(W\left( s \right)\) is called the transfer function of the system because it transforms the input \({\bf{u}}\) into the output \({\bf{y}}\). Find \(W\left( s \right)\) and describe how it is related to the partitioned system matrix on the left side of (8). See Exercise 15.
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