Chapter 2.6, Problem 1P

Explanation of Solution

Verification:

Consider the following matrix,

$A=\left[\begin{array}{l}\text{1 2 3}\\ \text{4 5 6}\\ 7\text{ 8 9}\end{array}\right]$

The second-row cofactors are given below:

$\begin{array}{l}{A}_{21}=\left[\begin{array}{l}2\text{ 3}\\ \text{8 9}\end{array}\right]\\ \end{array}$

Therefore, we get

$\begin{array}{l}\det (A_{21})=18-24\\ =-6\end{array}$

Now, we have

$\begin{array}{l}{A}_{22}=\left[\begin{array}{l}\text{1 3}\\ \text{7 9}\end{array}\right]\\ \end{array}$

Therefore,

$\begin{array}{l}\det (A_{22})=9-21\\ =-12\end{array}$

We have

$\begin{array}{l}{A}_{23}=\left[\begin{array}{l}\text{1 2}\\ \text{7 8}\end{array}\right]\\ \end{array}$

Therefore,

$\begin{array}{l}\det (A_{23})=8-14\\ =-6\end{array}$

Therefore, the expansion is ,

$\begin{array}{l}\det (A)={(-1)}^{2+1}{a}_{21}\det (A_{21})+{(-1)}^{2+2}{a}_{22}\det (A_{22})+{(-1)}^{2+3}{a}_{23}\det (A_{23})\\ \end{array}$

Substituting the values, we get

$\begin{array}{l}\det (A)=-(4)(-6)+5(-12)-6(-6)\end{array}$