/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q10E In Exercises 7-10, let \(B = \le... [FREE SOLUTION] | 91影视

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Chapter 4: Q10E (page 191)

In Exercises 7-10, let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},{{\mathop{\rm b}\nolimits} _2}} \right\}\) and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},{{\mathop{\rm c}\nolimits} _2}} \right\}\) be bases for \({\mathbb{R}^2}\). In each exercise, find the change-of-coordinates matrix from \(B\) to \(C\) and the change-of-coordinates matrix from \(C\) to \(B\).

10. \({{\mathop{\rm b}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}7\\{ - 2}\end{array}} \right),{{\mathop{\rm b}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\end{array}} \right),{{\mathop{\rm c}\nolimits} _1} = \left( {\begin{array}{*{20}{c}}4\\1\end{array}} \right),{{\mathop{\rm c}\nolimits} _2} = \left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right)\).

Short Answer

Expert verified

The change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}8&3\\{ - 5}&{ - 2}\end{array}} \right)\). The change-of-coordinates matrix from \(C\) to \(B\) is \(\mathop P\limits_{B \leftarrow C} = \left( {\begin{array}{*{20}{c}}2&3\\{ - 5}&{ - 8}\end{array}} \right)\).

Step by step solution

01

Change-of-coordinate matrix

Let \(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\)and \(C = \left\{ {{{\mathop{\rm c}\nolimits} _1},...,{{\mathop{\rm c}\nolimits} _n}} \right\}\) be bases of a vector space \(V\). Then according toTheorem 15,there is a unique \(n \times n\) matrix \(\mathop P\limits_{C \leftarrow B} \) such that

\({\left( {\mathop{\rm x}\nolimits} \right)_C} = \mathop P\limits_{C \leftarrow B} {\left( {\mathop{\rm x}\nolimits} \right)_B}\). 鈥�(1)

The columns of \(\mathop P\limits_{C \leftarrow B} \) are the \(C - \)coordinate vectors of the vectors in the basis \(B\). That is, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}{{{\left( {{{\mathop{\rm b}\nolimits} _1}} \right)}_C}}&{{{\left( {{{\mathop{\rm b}\nolimits} _2}} \right)}_C}}& \cdots &{{{\left( {{{\mathop{\rm b}\nolimits} _n}} \right)}_C}}\end{array}} \right)\).

02

Determine the change-of-coordinates from \(B\) to \(C\)\(\frac{1}{4}\)

Write the augmented matrix as shown below:

Perform an elementary row operation to produce a row-reduced echelon form of the matrix.

At row 1, multiply row 1 by .

At row 2, subtract row 1 from row 2.

\( \sim \left( {\begin{array}{*{20}{c}}1&{1.25}&{1.75}&{0.5}\\0&{ - 3}&{12}&{ - 9}\end{array}} \right)\)

\( \sim \left( {\begin{array}{*{20}{c}}1&{1.25}&{1.75}&{0.5}\\0&{0.75}&{ - 3.75}&{ - 1.5}\end{array}} \right)\)

At row 2, multiply row 2 by \(1.333\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{1.25}&{1.75}&{0.5}\\0&1&{ - 5}&{ - 2}\end{array}} \right)\)

At row 1, multiply row 2 by \(1.25\) and subtract it from row 1.

\( \sim \left( {\begin{array}{*{20}{c}}1&0&8&3\\0&1&{ - 5}&{ - 2}\end{array}} \right)\)

Therefore, \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}8&3\\{ - 5}&{ - 2}\end{array}} \right)\).

Thus, the change-of-coordinates matrix from \(B\) to \(C\) is \(\mathop P\limits_{C \leftarrow B} = \left( {\begin{array}{*{20}{c}}8&3\\{ - 5}&{ - 2}\end{array}} \right)\).

03

Determine the change-of-coordinates from \(C\) to \(B\)

It is known that \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\) is the matrix that converts \(C - \)coordinates into \(B - \)coordinates. That is, \({\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}} = \mathop P\limits_{B \leftarrow C} \).

\(\begin{aligned} \mathop P\limits_{B \leftarrow C} &= {\left( {\mathop P\limits_{C \leftarrow B} } \right)^{ - 1}}\\ &= {\left( {\begin{array}{*{20}{c}}8&3\\{ - 5}&{ - 2}\end{array}} \right)^{ - 1}}\\ &= \frac{1}{{ - 1}}\left( {\begin{array}{*{20}{c}}{ - 2}&{ - 3}\\5&8\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}2&3\\{ - 5}&{ - 8}\end{array}} \right)\end{aligned}\)

Thus, the change-of-coordinates matrix from \(C\) to \(B\) is \(\mathop P\limits_{B \leftarrow C} = \left( {\begin{array}{*{20}{c}}2&3\\{ - 5}&{ - 8}\end{array}} \right)\).

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Most popular questions from this chapter

(M) Show that \(\left\{ {t,sin\,t,cos\,{\bf{2}}t,sin\,t\,cos\,t} \right\}\) is a linearly independent set of functions defined on \(\mathbb{R}\). Start by assuming that

\({c_{\bf{1}}} \cdot t + {c_{\bf{2}}} \cdot sin\,t + {c_{\bf{3}}} \cdot cos\,{\bf{2}}t + {c_{\bf{4}}} \cdot sin\,t\,cos\,t = {\bf{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\bf{5}} \right)\)

Equation (5) must hold for all real t, so choose several specific values of t (say, \(t = {\bf{0}},\,.{\bf{1}},\,.{\bf{2}}\)) until you get a system of enough equations to determine that the \({c_j}\) must be zero.

Suppose a nonhomogeneous system of six linear equations in eight unknowns has a solution, with two free variables. Is it possible to change some constants on the equations鈥� right sides to make the new system inconsistent? Explain.

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

17. A submatrix of a matrix A is any matrix that results from deleting some (or no) rows and/or columns of A. It can be shown that A has rank \(r\) if and only if A contains an invertible \(r \times r\) submatrix and no longer square submatrix is invertible. Demonstrate part of this statement by explaining (a) why an \(m \times n\) matrix A of rank \(r\) has an \(m \times r\) submatrix \({A_1}\) of rank \(r\), and (b) why \({A_1}\) has an invertible \(r \times r\) submatrix \({A_2}\).

The concept of rank plays an important role in the design of engineering control systems, such as the space shuttle system mentioned in this chapter鈥檚 introductory example. A state-space model of a control system includes a difference equation of the form

\({{\mathop{\rm x}\nolimits} _{k + 1}} = A{{\mathop{\rm x}\nolimits} _k} + B{{\mathop{\rm u}\nolimits} _k}\)for \(k = 0,1,....\) (1)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(\left\{ {{{\mathop{\rm x}\nolimits} _k}} \right\}\) is a sequence of 鈥渟tate vectors鈥� in \({\mathbb{R}^n}\) that describe the state of the system at discrete times, and \(\left\{ {{{\mathop{\rm u}\nolimits} _k}} \right\}\) is a control, or input, sequence. The pair \(\left( {A,B} \right)\) is said to be controllable if

\({\mathop{\rm rank}\nolimits} \left( {\begin{array}{*{20}{c}}B&{AB}&{{A^2}B}& \cdots &{{A^{n - 1}}B}\end{array}} \right) = n\) (2)

The matrix that appears in (2) is called the controllability matrix for the system. If \(\left( {A,B} \right)\) is controllable, then the system can be controlled, or driven from the state 0 to any specified state \({\mathop{\rm v}\nolimits} \) (in \({\mathbb{R}^n}\)) in at most \(n\) steps, simply by choosing an appropriate control sequence in \({\mathbb{R}^m}\). This fact is illustrated in Exercise 18 for \(n = 4\) and \(m = 2\). For a further discussion of controllability, see this text鈥檚 website (Case study for Chapter 4).

If A is a \({\bf{6}} \times {\bf{4}}\) matrix, what is the smallest possible dimension of Null A?

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).
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