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Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).

Short Answer

Expert verified

The values are\({S^2} = \left( {\begin{aligned}{*{20}{c}}0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\),\({S^3} = \left( {\begin{aligned}{*{20}{c}}0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\),\({S^4} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\),\({S^5} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\), and\({S^6} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

Step by step solution

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01

Create the matrix

Consider the matrix\(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\).

Write the above matrix in the MATLAB command.

\( > > {\rm{S}} = \left( {{\rm{0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 0; 0 0 0 0 1; 0 0 0 0 0}}} \right){\rm{;}}\)

02

Obtain \({S^k}\) using \(k = {\bf{2}}\)

Put\(k = 2\)to obtain the matrix\({S^2}\)by using the MATLAB command shown below:

\( > > S*S\)

The output matrix is shown below:

\({S^2} = \left( {\begin{aligned}{*{20}{c}}0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus, \({S^2} = \left( {\begin{aligned}{*{20}{c}}0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

03

Obtain \({S^k}\) using \(k = {\bf{3}}\)

Put\(k = 3\)to obtain the matrix\({S^3}\)by using the MATLAB command shown below:

\( > > S\^{\bf{3}}\)

The output matrix is shown below:

\({S^3} = \left( {\begin{aligned}{*{20}{c}}0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus, \({S^3} = \left( {\begin{aligned}{*{20}{c}}0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

04

Obtain \({S^k}\) using \(k = {\bf{4}}\)

Put\(k = 4\)to obtain the matrix\({S^4}\)by using the MATLAB command shown below:

\( > > S\^4\)

The output matrix is shown below:

\({S^4} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus, \({S^4} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&1\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

05

Obtain \({S^k}\) by using \(k = 5\)

Put\(k = 5\)to obtain the matrix\({S^5}\)by using the MATLAB command shown below:

\( > > S\^5\)

The output matrix is shown below:

\({S^5} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus, \({S^5} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

06

Obtain \({S^k}\) using \(k = {\bf{6}}\)

Put\(k = 6\)to obtain the matrix\({S^6}\)by using the MATLAB command shown below:

\( > > S\^{\bf{6}}\)

The output matrix is shown below:

\({S^6} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\)

Thus, \({S^6} = \left( {\begin{aligned}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{aligned}} \right)\).

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Most popular questions from this chapter

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

In Exercise 9 mark each statement True or False. Justify each answer.

9. a. In order for a matrix B to be the inverse of A, both equations \(AB = I\) and \(BA = I\) must be true.

b. If A and B are \(n \times n\) and invertible, then \({A^{ - {\bf{1}}}}{B^{ - {\bf{1}}}}\) is the inverse of \(AB\).

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ab - cd \ne {\bf{0}}\), then A is invertible.

d. If A is an invertible \(n \times n\) matrix, then the equation \(Ax = b\) is consistent for each b in \({\mathbb{R}^{\bf{n}}}\).

e. Each elementary matrix is invertible.

Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.

In Exercises 1โ€“9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5โ€“8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

8. \[\left[ {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}X&Y&Z\\{\bf{0}}&{\bf{0}}&I\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&I\end{array}} \right]\]

In Exercises 1โ€“9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1โ€“4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

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