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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

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]

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?

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.

Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).

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