Verify that the signals in Exercises 1 and 2 are solutions of the accompanying difference equation.

\({{\bf{3}}^k}\), \({\left( { - {\bf{3}}} \right)^k}\); \({y_{k + {\bf{2}}}} - {\bf{9}}{y_k} = {\bf{0}}\)

Short Answer

Expert verified

\({3^k}\), \({\left( { - 3} \right)^k}\) are the solution of the difference equation \({y_{k + 2}} + 2{y_{k + 1}} - 8{y_k} = 0\).

Step by step solution

01

Check the given difference equation for \({{\bf{2}}^k}\)

If \({3^k}\) is the solution,

\({y_{k + 2}} = {3^{k + 2}}\), \({y_k} = {3^k}\).

By the difference equation, you get

\(\begin{aligned} {3^{k + 2}} - 9\left( {{3^k}} \right) &= 0\\{3^k}\left( {{3^2} - 9} \right) &= 0\\{3^k}\left( {9 - 9} \right) &= 0.\end{aligned}\)

So, \({3^k}\) is the solution of the given difference equation.

02

Check the given difference equation for \({\left( { - {\bf{4}}} \right)^k}\)

If \({\left( { - 3} \right)^k}\) is the solution,

\({y_{k + 2}} = {\left( { - 3} \right)^{k + 2}}\), \({y_k} = {\left( { - 3} \right)^k}\).

By the difference equation, you get

\(\begin{aligned} {\left( { - 3} \right)^{k + 2}} - 9\left[ {{{\left( { - 3} \right)}^k}} \right] &= 0\\{\left( { - 3} \right)^k}\left( {{{\left( { - 3} \right)}^2} - 9} \right) &= 0\\{\left( { - 3} \right)^k}\left( {9 - 9} \right) &= 0.\end{aligned}\)

So, \({\left( { - 3} \right)^k}\) is the solution of the difference equation.

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Most popular questions from this chapter

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

21. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 2}&{ - 4.2}&{ - 4.8}&{ - 3.6}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

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

15. Let\(A\)be an\(m \times n\)matrix, and let\(B\)be a\(n \times p\)matrix such that\(AB = 0\). Show that\({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces\({\mathop{\rm Nul}\nolimits} A\),\({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and\({\mathop{\rm Col}\nolimits} B\)is contained in one of the other three subspaces.)

Question: Determine if the matrix pairs in Exercises 19-22 are controllable.

22. (M) \(A = \left( {\begin{array}{*{20}{c}}0&1&0&0\\0&0&1&0\\0&0&0&1\\{ - 1}&{ - 13}&{ - 12.2}&{ - 1.5}\end{array}} \right),B = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\).

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 \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

In Exercise 1, find the vector x determined by the given coordinate vector \({\left( x \right)_{\rm B}}\) and the given basis \({\rm B}\).

1. \({\rm B} = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{ - {\bf{5}}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{ - {\bf{4}}}\\{\bf{6}}\end{array}} \right)} \right\},{\left( x \right)_{\rm B}} = \left( {\begin{array}{*{20}{c}}{\bf{5}}\\{\bf{3}}\end{array}} \right)\)

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