Chapter 7.2, Problem 72E

Solution

a.

To prove: The determinant $f\left(x\right)=\det \left(x{I}_2-A\right)$ is polynomial of degree 2 (the characteristic polynomial of A ).

The statement has proven.

Given Information:

The matrix is defined as,

  $A=\left[a_{ij}\right]$ and the function is $f\left(x\right)=\det \left(x{I}_2-A\right)$ .

Explanation:

Consider the given function,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Use the matrix and find the determinant.

  $\begin{array}{c}f\left(x\right)=\det \left\{x\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]\right\}\\ =\det \left[\begin{array}{cc}x-a_{11}& -a_{12}\\ -a_{21}& x-a_{22}\end{array}\right]\\ =\left(x-a_{11}\right)\left(x-a_{22}\right)-a_{12}{a}_{21}\\ =x^2-a_{22}x-a_{11}x+a_{11}{a}_{22}-a_{12}{a}_{21}\end{array}$

It can be seen that the first term of the determinant is $x^2$ .

Therefore, the polynomial $f\left(x\right)$ is degree of 2.

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,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Explanation:

Consider the given function,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

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

  $\begin{array}{c}\det A=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|\\ =a_{11}{a}_{22}-a_{12}{a}_{21}\\ =\text{constant}\end{array}$

Hence, the determinant is constant.

c.

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

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

Given Information:

The function is defined as,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Explanation:

Consider the given information,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Refer the obtained result in the previous part and simplify.

  $\begin{array}{c}f\left(x\right)=x^2-a_{22}x-a_{11}x+a_{11}{a}_{22}-a_{12}{a}_{21}\\ =x^2+\left(-a_{22}-a_{11}\right)x+a_{11}{a}_{22}-a_{12}{a}_{21}\end{array}$

And,

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

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

Therefore, the obtained coefficient is $\left(-a_{22}-a_{11}\right)$ 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,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Explanation:

Consider the given information,

  $f\left(x\right)=\det \left(x{I}_2-A\right)$

Substitute the values in the function and check.

  $\begin{array}{c}f\left(A\right)=\det \left(A{I}_2-A\right)\\ =\det A\left[I_2-1\right]\\ =\det A\left[\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)-1\right]\\ =\det A\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ =0\end{array}$

Therefore, the given statement has proved.