Q. 60 - Precalculus Enhanced with Graphing Utilities

Question

Show that $\left|\begin{array}{lll}{x}^2& x& 1\\ y^2& y& 1\\ z^2& z& 1\end{array}\right|=(y-z)(x-y)(x-z)$

Step-by-Step Solution

Answer

The given expression is proved.

Step 1. Given Information

We are given a determinant and we have to prove,

$\left|\begin{array}{lll}{x}^2& x& 1\\ y^2& y& 1\\ z^2& z& 1\end{array}\right|=(y-z)(x-y)(x-z)$

Step 2. Proving the expression

Applying the row transformation, we get

$$\begin{gathered} R_2\to R_2-R_1 \\ {R}_3\to R_3-R_1 \end{gathered}$$

$$\begin{gathered} =\left|\begin{array}{ccc}{x}^2& x& 1\\ y^2-x^2& y-x& 0\\ z^2-x^2& z-x& 0\end{array}\right| \\ =(y-x)(z-x)\left|\begin{array}{ccc}{x}^2& x& 1\\ y+x& 1& 0\\ z+x& 1& 0\end{array}\right| \end{gathered}$$

Expanding the determinant, we get

$$\begin{gathered} =(y-x)(z-x)·\left(\right(y+x)·1-(z+x)·1) \\ =(x-y)(y-z)(z-x) \end{gathered}$$

Hence Proved.