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

Let V be a vector space that contains a linearly independent set \(\left\{ {{u_{\bf{1}}},{u_{\bf{2}}},{u_{\bf{3}}},{u_{\bf{4}}}} \right\}\). Describe how to construct a set of vectors \(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\) in V such that \(\left\{ {{v_{\bf{1}}},{v_{\bf{3}}}} \right\}\) is a basis for Span\(\left\{ {{v_{\bf{1}}},{v_{\bf{2}}},{v_{\bf{3}}},{v_{\bf{4}}}} \right\}\).

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