Chapter 8, Problem 10PS

To determine

TO-DETERMINE: Verify the given relation.

Answer to Problem 10PS

It is verified that LHS equalsRHS

Explanation of Solution

Given:

The matrix relation

  $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left(a–b\right)\left(b–c\right)\left(c–a\right)$

To verify

  $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left(a–b\right)\left(b–c\right)\left(c–a\right)$

Now applying

  $\begin{array}{l}{C}_1\to C_1-C_2\\ C_2\to C_2-C_3\end{array}$

  $\begin{array}{l}\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ a^2-b^2& b^2-c^2& c^2\end{array}\right|\\ =\left|\begin{array}{ccc}0& 0& 1\\ a-b& b-c& c\\ \left(a-b\right)\left(a+b\right)& \left(b+c\right)\left(b-c\right)& c^2\end{array}\right|\\ =\left(a-b\right)\left(b-c\right)\left|\begin{array}{ccc}0& 0& 1\\ 1& 1& c\\ \left(a+b\right)& \left(b+c\right)& c^2\end{array}\right|\end{array}$

Expanding along the first row gives:

  $\begin{array}{l}\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left(a-b\right)\left(b-c\right)\left(1\left(1\left(b+c\right)-1\left(a+b\right)\right)\right)\\ =\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\\ =\left(a-b\right)\left(b-c\right)\left(c-a\right)\end{array}$

Hence,

  $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left(a-b\right)\left(b-c\right)\left(c-a\right)$