Chapter 7.2, Problem 73E

Solution

a.

To prove: The determinant $f\left(x\right)=\det \left(x{I}_3-A\right)$ is polynomial of degree 3 (the characteristic polynomial of A ).

The statement has prove.

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right]$ and the function is $f\left(x\right)=\det \left(x{I}_3-A\right)$ .

Explanation:

Consider the given function,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Use the matrix and find the determinant.

  $f\left(x\right)=\det \left\{\left[\begin{array}{ccc}x& 0& 0\\ 0& x& 0\\ 0& 0& x\end{array}\right]-\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]\right\}$

  $=\left[\begin{array}{ccc}x-a_{11}& -a_{12}& -a_{13}\\ -a_{21}& x-a_{22}& -a_{23}\\ -a_{31}& -a_{32}& x-a_{33}\end{array}\right]$

  $=\left(x-a_{11}\right)\left[\left(x-a_{22}\right)\left(x-a_{33}\right)-a_{23}{a}_{32}\right]+a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]$

  $=\left\{\begin{array}{c}\left[\left(x-a_{11}\right)\left(x^2-x{a}_{33}-a_{22}x+a_{22}{a}_{22}\right)-\left(x-a_{11}\right)a_{23}{a}_{32}\right]\\ +a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]\end{array}\right\}$

  $=\left\{\begin{array}{c}\left[x^3-x^2{a}_{33}-a_{22}{x}^2+x{a}_{22}{a}_{22}-a_{11}\left(x^2-x{a}_{33}-a_{22}x+a_{22}{a}_{22}\right)-\left(x-a_{11}\right)a_{23}{a}_{32}\right]\\ +a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]\end{array}\right\}$

It can be seen that the first term of the determinant is $x^3$ .

Therefore, the polynomial $f\left(x\right)$ is degree of 3.

b.

To determine: The reason of that the constant tem of $f\left(x\right)$ related to det A.

The constant can make negative determinant.

Given Information:

The function is defined as,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Explanation:

Consider the given function,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Take the given x as constant. Now, find the determinant.

  $\begin{array}{c}\det A=a_{11}\left(a_{22}{a}_{33}-a_{23}{a}_{32}\right)-a_{12}\left(a_{21}{a}_{33}-a_{23}{a}_{31}\right)+a_{13}\left(a_{21}{a}_{32}-a_{22}{a}_{31}\right)\\ =-\left[-a_{11}\left(a_{22}{a}_{33}-a_{23}{a}_{32}\right)+a_{12}\left(a_{21}{a}_{33}-a_{23}{a}_{31}\right)-a_{13}\left(a_{21}{a}_{32}-a_{22}{a}_{31}\right)\right]\\ =-\det A\end{array}$

Hence, the result express above.

c.

To determine: The constant term of $x^2$ from the function.

The obtained coefficient is $\left(-a_{33}-a_{22}-a_{11}\right)$ and this is trace of A .

Given Information:

The function is defined as,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Explanation:

Consider the given information,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Refer the obtained result in the previous part and simplify.

  $f\left(x\right)=\left[\begin{array}{l}{x}^3-x^2{a}_{33}-a_{22}{x}^2+x{a}_{22}{a}_{22}+\left(-a_{11}{x}^2+a_{11}x{a}_{33}+a_{11}{a}_{22}x-a_{11}{a}_{22}{a}_{22}\right)\hfill \\ -\left(x-a_{11}\right)a_{23}{a}_{32}+a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]\hfill \end{array}\right]$

  $=\left[\begin{array}{l}{x}^3-x^2{a}_{33}-a_{22}{x}^2-a_{11}{x}^2+x{a}_{22}{a}_{22}+a_{11}x{a}_{33}+a_{11}{a}_{22}x-a_{11}{a}_{22}{a}_{22}\hfill \\ -\left(x-a_{11}\right)a_{23}{a}_{32}+a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]\hfill \end{array}\right]$

  $=\left[\begin{array}{c}{x}^3+\left(-a_{33}-a_{22}-a_{11}\right)x^2+x{a}_{22}{a}_{22}+a_{11}x{a}_{33}+a_{11}{a}_{22}x-a_{11}{a}_{22}{a}_{22}\\ -\left(x-a_{11}\right)a_{23}{a}_{32}+a_{12}\left[-a_{21}\left(x-a_{33}\right)-a_{23}{a}_{31}\right]-a_{13}\left[a_{21}{a}_{32}+a_{31}\left(x-a_{22}\right)\right]\end{array}\right]$

And,

  $\begin{array}{c}\left(-a_{33}-a_{22}-a_{11}\right)=-\left(a_{11}+a_{22}+a_{33}\right)\\ =-\text{Trace }A\end{array}$

The coefficient of $x^2$ is representing the trace of $A$ required explanation given above.

Therefore, the obtained coefficient is $\left(-a_{33}-a_{22}-a_{11}\right)$ and this is trace of A .

d.

To prove: The given express is true.

  $f\left(A\right)=0$

The given statement has proved.

Given Information:

The function is defined as,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Explanation:

Consider the given information,

  $f\left(x\right)=\det \left(x{I}_3-A\right)$

Substitute the values in the function and check.

  $\begin{array}{c}f\left(A\right)=\det \left(A{I}_3-A\right)\\ =\det A\left[I_3-1\right]\end{array}$

  $=\det A\left[\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)-1\right]$

  $=\det A\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$

  $=0$

Therefore, the given statement has proved.