Chapter 2.6, Problem 3P

Explanation of Solution

Proof:

Consider a $3\times 3$ upper triangular matrix given below:

$A=\left[\begin{array}{l}{\text{a}}_{\text{11}}{\text{ a}}_{\text{12}}{\text{ a}}_{\text{13}}\\ {\text{0 a}}_{\text{22}}{\text{ a}}_{\text{23}}\\ {\text{0 0 a}}_{33}\end{array}\right]$

The third-row cofactors are as follows:

$\begin{array}{l}{A}_{33}=\left[\begin{array}{l}{\text{a}}_{\text{11}}{\text{ a}}_{\text{12}}\\ {\text{0 a}}_{\text{22}}\end{array}\right]\\ \end{array}$

Therefore, we get

$\det (A_{}$