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Chapter 4: Q2E (page 191)

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

[M] Let \(A = \left[ {\begin{array}{*{20}{c}}7&{ - 9}&{ - 4}&5&3&{ - 3}&{ - 7}\\{ - 4}&6&7&{ - 2}&{ - 6}&{ - 5}&5\\5&{ - 7}&{ - 6}&5&{ - 6}&2&8\\{ - 3}&5&8&{ - 1}&{ - 7}&{ - 4}&8\\6&{ - 8}&{ - 5}&4&4&9&3\end{array}} \right]\).

  1. Construct matrices \(C\) and \(N\) whose columns are bases for \({\mathop{\rm Col}\nolimits} A\) and \({\mathop{\rm Nul}\nolimits} A\), respectively, and construct a matrix \(R\) whose rows form a basis for Row\(A\).
  2. Construct a matrix \(M\) whose columns form a basis for \({\mathop{\rm Nul}\nolimits} {A^T}\), form the matrices \(S = \left[ {\begin{array}{*{20}{c}}{{R^T}}&N\end{array}} \right]\) and \(T = \left[ {\begin{array}{*{20}{c}}C&M\end{array}} \right]\), and explain why \(S\) and \(T\) should be square. Verify that both \(S\) and \(T\) are invertible.

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

Is it possible for a nonhomogeneous system of seven equations in six unknowns to have a unique solution for some right-hand side of constants? Is it possible for such a system to have a unique solution for every right-hand side? Explain.

Consider the polynomials \({{\bf{p}}_{\bf{1}}}\left( t \right) = {\bf{1}} + t\), \({{\bf{p}}_{\bf{2}}}\left( t \right) = {\bf{1}} - t\), \({{\bf{p}}_{\bf{3}}}\left( t \right) = {\bf{4}}\), \({{\bf{p}}_{\bf{4}}}\left( t \right) = {\bf{1}} + {t^{\bf{2}}}\), and \({{\bf{p}}_{\bf{5}}}\left( t \right) = {\bf{1}} + {\bf{2}}t + {t^{\bf{2}}}\), and let H be the subspace of \({P_{\bf{5}}}\) spanned by the set \(S = \left\{ {{{\bf{p}}_{\bf{1}}},\,{{\bf{p}}_{\bf{2}}},\;{{\bf{p}}_{\bf{3}}},\,{{\bf{p}}_{\bf{4}}},\,{{\bf{p}}_{\bf{5}}}} \right\}\). Use the method described in the proof of the Spanning Set Theorem (Section 4.3) to produce a basis for H. (Explain how to select appropriate members of S.)

In Exercises 27-30, use coordinate vectors to test the linear independence of the sets of polynomials. Explain your work.
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