Chapter 11.3, Problem 60AYU

To determine

To show: In the given mathematical expression is proved.

Explanation of Solution

Given:

  $\left|\begin{array}{l}{x}^{2}x1\\ y^{2}y1\\ z^{2}z1\end{array}\right|=\left(y-z\right)\left(x-y\right)\left(x-z\right)$

A mathematical expression is given it has to show

  $\left|\begin{array}{l}{x}^{2}x1\\ y^{2}y1\\ z^{2}z1\end{array}\right|=\left(y-z\right)\left(x-y\right)\left(x-z\right)$

L.H.S of the mathematical expression is

  $\left|\begin{array}{l}{x}^{2}x1\\ y^{2}y1\\ z^{2}z1\end{array}\right|$

Now for showing this the determinant of a $3\times 3$ matrix is defined as shown.

  $\left|\begin{array}{l}{x}^{2}x1\\ y^{2}y1\\ z^{2}z1\end{array}\right|=x^2\left(y-z\right)-x\left(y^2-z^2\right)+1\left(z{y}^2-y{z}^2\right)$

R.H.S of the mathematical expression is

  $\begin{array}{l}\left(y-z\right)\left(x-y\right)\left(x-z\right)\\ =\left(y-z\right)\left(x^2-xz-xy+yz\right)\\ =x^{2}y-xyz-x{y}^2+y^{2}z-x^{2}z+z^{2}x+xyz-z^{2}y\\ =x^{2}y-x{y}^2+y^{2}z-x^{2}z+x{z}^2-y{z}^2\end{array}$

It’s not written in the same order, but the terms do match so they are equivalent.

Hence, $\left|\begin{array}{l}{x}^{2}x1\\ y^{2}y1\\ z^{2}z1\end{array}\right|=\left(y-z\right)\left(x-y\right)\left(x-z\right)$ is proved.