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Chapter 4: Q29E (page 191)

Write the difference equations in Exercises 29 and 30 as first order systems, \({x_{k + {\bf{1}}}} = A{x_k}\), for all \(k\).

\({y_{k + {\bf{4}}}} - {\bf{6}}{y_{k + {\bf{3}}}} + {\bf{8}}{y_{k + {\bf{2}}}} + {\bf{6}}{y_{k + {\bf{1}}}} - {\bf{9}}{y_k} = {\bf{0}}\)

Short Answer

Expert verified

\(A = \left[ {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\9&{ - 6}&{ - 8}&6\end{array}} \right]\)

Step by step solution

01

Write vectors \({x_k}\)and \({x_{k + {\bf{1}}}}\)

Vectors\({x_k}\)and\({x_{k + 1}}\)can be expressed as

\({x_k} = \left[ {\begin{array}{*{20}{c}}{{y_k}}\\{{y_{k + 1}}}\\{{y_{k + 2}}}\\{{y_{k + 3}}}\end{array}} \right]\) and \({x_{k + 1}} = \left[ {\begin{array}{*{20}{c}}{{y_{k + 1}}}\\{{y_{k + 2}}}\\{{y_{k + 3}}}\\{{y_{k + 4}}}\end{array}} \right]\).

02

Write \({x_{k + {\bf{1}}}}\) in the matrix form

Thematrix formfor \({x_{k + 1}}\) is

\(\begin{aligned} {x_{k + 1}} &= \left[ {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\9&{ - 6}&{ - 8}&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{y_k}}\\{{y_{k + 1}}}\\{{y_{k + 2}}}\\{{y_{k + 3}}}\end{array}} \right]\\ &= A{x_k}\end{aligned}\).

So, matrix \(A = \left[ {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\9&{ - 6}&{ - 8}&6\end{array}} \right]\).

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Most popular questions from this chapter

Prove theorem 3 as follows: Given an \(m \times n\) matrix A, an element in \({\mathop{\rm Col}\nolimits} A\) has the form \(Ax\) for some x in \({\mathbb{R}^n}\). Let \(Ax\) and \(A{\mathop{\rm w}\nolimits} \) represent any two vectors in \({\mathop{\rm Col}\nolimits} A\).

  1. Explain why the zero vector is in \({\mathop{\rm Col}\nolimits} A\).
  2. Show that the vector \(A{\mathop{\rm x}\nolimits} + A{\mathop{\rm w}\nolimits} \) is in \({\mathop{\rm Col}\nolimits} A\).
  3. Given a scalar \(c\), show that \(c\left( {A{\mathop{\rm x}\nolimits} } \right)\) is in \({\mathop{\rm Col}\nolimits} A\).

Let S be a maximal linearly independent subset of a vector space V. In other words, S has the property that if a vector not in S is adjoined to S, the new set will no longer be linearly independent. Prove that S must be a basis of V. [Hint: What if S were linearly independent but not a basis of V?]

Use Exercise 28 to explain why the equation\(Ax = b\)has a solution for all\({\rm{b}}\)in\({\mathbb{R}^m}\)if and only if the equation\({A^T}x = 0\)has only the trivial solution.

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

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

Let be a basis of\({\mathbb{R}^n}\). .Produce a description of an \(B = \left\{ {{{\bf{b}}_{\bf{1}}},....,{{\bf{b}}_n}\,} \right\}\)matrix A that implements the coordinate mapping \({\bf{x}} \mapsto {\left( {\bf{x}} \right)_B}\). Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)). (See Exercise 21.)
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