Chapter 11.3, Problem 62AYU

To determine

To Show : The value of the new determinant is $1$ times the value of the original determinant.

Answer to Problem 62AYU

The value of the new determinant is $-1$ times the value of the original determinant.

Explanation of Solution

Given information:

Interchange columns $1\&3$ of a 3 by 3 determinant.

Let, consider a 3 by 3 determinant.

  $=\left|\begin{array}{l}{\text{x}}_1{\text{ x}}_2{\text{ x}}_3\\ {\text{y}}_1{\text{ y}}_2{\text{ y}}_3\\ {\text{z}}_1{\text{ z}}_2{\text{ z}}_3\end{array}\right|$

Therefore, the determinant is,

  $\begin{array}{l}=x_1(y_2{z}_3-y_3{z}_2)-x_2(y_1{z}_3-y_3{z}_1)+x_3(y_1{z}_2-y_2{z}_1)\\ =x_1{y}_2{z}_3-x_1{y}_3{z}_2-x_2{y}_1{z}_3+x_2{y}_3{z}_1+x_3{y}_1{z}_2-x_3{y}_2{z}_1\\ =x_1{y}_2{z}_3+x_2{y}_3{z}_1+x_3{y}_1{z}_2-x_1{y}_3{z}_2-x_2{y}_1{z}_3-x_3{y}_2{z}_1\\ \end{array}$

Let’s Interchange column 1&3 of the determinant,

  $=\left|\begin{array}{l}{z}_1{z}_2{\text{ z}}_3\\ {\text{y}}_1{\text{y}}_2{\text{ y}}_3\\ {\text{x}}_1{\text{ x}}_2{\text{ x}}_3\end{array}\right|$

Therefore, The determinant is,

  $\begin{array}{l}={\text{z}}_1(x_3{y}_2-x_2{y}_3)-z_2(x_3{y}_1-x_1{y}_3)+z_3(x_2{y}_1-x_1{y}_2)\\ =x_3{y}_2{z}_1-x_2{y}_3{z}_1-x_3{y}_1{z}_2+x_1{y}_3{z}_2+x_2{y}_1{z}_3-x_1{y}_2{z}_3\\ =x_1{y}_3{z}_2+x_2{y}_1{z}_3+x_3{y}_2{z}_1-x_1{y}_2{z}_3-x_2{y}_3{z}_1-x_3{y}_1{z}_2\\ =-1(x_1{y}_2{z}_3+x_2{y}_3{z}_1+x_3{y}_1{z}_2-x_1{y}_3{z}_2-x_2{y}_1{z}_3-x_3{y}_2{z}_1)\end{array}$

Therefore,

  $\begin{array}{l}=-1(x_1{y}_2{z}_3+x_2{y}_3{z}_1+x_3{y}_1{z}_2-x_1{y}_3{z}_2-x_2{y}_1{z}_3-x_3{y}_2{z}_1)\\ =(x_1{y}_3{z}_2+x_2{y}_1{z}_3+x_3{y}_2{z}_1-x_1{y}_2{z}_3-x_2{y}_3{z}_1-x_3{y}_1{z}_2)\end{array}$

Hence, The value of the new determinant is $-1$ times the value of the original determinant.