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Question: Exercises 9–14 require techniques from Section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\)determinants described prior to Exercises 15–18 in Section 3.1. (Note:Finding the characteristic polynomial of a \(3 \times 3\)matrix is not easy to do with just row operations, because the variableis involved.)

9.\(\left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\2&3&{ - 1}\\0&6&0\end{array}} \right)\)

Short Answer

Expert verified

Characteristic polynomial is \( - {\lambda ^3} + 4{\lambda ^2} - 9\lambda + 6\)

Step by step solution

01

Formulate the matrix \(A - \lambda I\) 

If \(A\) is an \(n \times n\) matrix, then \(det\left( {A - \lambda I} \right)\), which is a polynomial of degree \(n\), is called the characteristic polynomial of \(A\).

It is given that\(A = \left( {\begin{array}{*{20}{c}}2&7\\7&2\end{array}} \right)\)and\(I = \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\)is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{array}{c}A - \lambda I = \left( {\begin{array}{*{20}{c}}1&0&{ - 1}\\2&3&{ - 1}\\0&6&0\end{array}} \right) - \lambda \left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{1 - \lambda }&0&{ - 1}\\2&{3 - \lambda }&{ - 1}\\0&6&{ - \lambda }\end{array}} \right)\end{array}\)

02

Find the determinant of the matrix \(A - \lambda I\)

For \(n \ge 2\) the determinant of an \(n \times n\) matrix \(A = [{a_{ij}}]\) is the sum of \(n\)terms of the form \( \pm {a_{1j}}\det {A_{1j}},\) with plus and minus signs alternating, where the entries \({a_{11}},{a_{12}}, \ldots ,{a_{1n}}\) are from the first row of \(A\). In symbols,

\(\begin{gathered} {\text{det}}A = {a_{11}}\det {A_{11}} - {a_{12}}\det {A_{12}} + \ldots + {\left( { - 1} \right)^{1 + n}}{a_{1n}}\det {A_{1n}} \\ = \mathop \sum \limits_{j = 1}^n {\left( { - 1} \right)^{1 + j}}{a_{1j}}\det {A_{1j}} \\ \end{gathered} \)

With the help of above defined formula, the\(\det A\)is calculated as follows:

\(\begin{array}{c}det\left( {A - \lambda I} \right) = det\left( {\begin{array}{*{20}{c}}{1 - \lambda }&0&{ - 1}\\2&{3 - \lambda }&{ - 1}\\0&6&{ - \lambda }\end{array}} \right)\\ = \left( {1 - \lambda } \right)\left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 1}\\6&{ - \lambda }\end{array}} \right| - 0\left| {\begin{array}{*{20}{c}}2&{ - 1}\\0&{ - \lambda }\end{array}} \right| - 1\left| {\begin{array}{*{20}{c}}2&{3 - \lambda }\\0&6\end{array}} \right|\\ = \left( {1 - \lambda } \right)\left( {\left( {3 - \lambda } \right)\left( { - \lambda } \right) - \left( 6 \right)\left( { - 1} \right)} \right) - 0\left( {\left( 2 \right)\left( { - \lambda } \right) - 0\left( { - 1} \right)} \right) - 1\left( {\left( 2 \right)\left( 6 \right) - 0} \right)\\ = - 3\lambda + 3{\lambda ^2} + {\lambda ^2} - {\lambda ^3} + 6 - 6\lambda \\ = - {\lambda ^3} + 4{\lambda ^2} - 9\lambda + 6\end{array}\)

Thus, the characteristic polynomial is \( - {\lambda ^3} + 4{\lambda ^2} - 9\lambda + 6\).

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Most popular questions from this chapter

Question: Is \(\lambda = - 2\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}7&3\\3&{ - 1}\end{array}} \right)\)? Why or why not?

Assume the mapping\(T:{{\rm P}_2} \to {{\rm P}_{\bf{2}}}\)defined by \(T\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right) = 3{a_0} + \left( {5{a_0} - 2{a_1}} \right)t + \left( {4{a_1} + {a_2}} \right){t^2}\) is linear. Find the matrix representation of\(T\) relative to the bases \(B = \left\{ {1,t,{t^2}} \right\}\).

Suppose \({\bf{x}}\) is an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \).

a. Show that \(x\) is an eigenvector of \(5I - A\). What is the corresponding eigenvalue?

b. Show that \(x\) is an eigenvector of \(5I - 3A + {A^2}\). What is the corresponding eigenvalue?

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

11. \(\left( {\begin{array}{*{20}{c}}{ - 1}&4&{ - 2}\\{ - 3}&4&0\\{ - 3}&1&3\end{array}} \right)\)

(M)Use a matrix program to diagonalize

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}&0\\{14}&7&{ - 1}\\{ - 6}&{ - 3}&1\end{aligned}} \right)\)

If possible. Use the eigenvalue command to create the diagonal matrix \(D\). If the program has a command that produces eigenvectors, use it to create an invertible matrix \(P\). Then compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\). Discuss your results.

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