Question

If

${\Delta }_1$, $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$, ${\Delta }_2$ = $\left|\begin{array}{ccc}1& bc& a\\ 1& ca& b\\ 1& ab& c\end{array}\right|$ then

• A. ${\Delta }_1+{\Delta }_2$ = 0
• B. ${\Delta }_1+2{\Delta }_2$ = 0
• C. ${\Delta }_1={\Delta }_2$
• D. ${\Delta }_1$ = 2 ${\Delta }_2$

Answer 1

The correct option is A

${\Delta }_1+{\Delta }_2$ = 0

Given that,

${\Delta }_1$ = $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$ ${\Delta }_2$ = $\left|\begin{array}{ccc}1& bc& a\\ 1& ca& b\\ 1& ab& c\end{array}\right|$

Here we are going to take the first term, use some properties of the determinant to see if we are able to reach the second term. We would use 3 properties of determinants in this. First is the scalar multiplication. By this, something multiplied to the whole determinant will be equal to the same thing multiplied to any one row or one column of the determinant. Second is the property which says the determinant of a matrix and its transpose are the same. And by third one, we can see that if we interchange any two rows or columns of a matrix its determinant will change its sign.

${\Delta }_1$ = $\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$

(by multiplying and dividing the RHS with )

=$\frac{1}{abc}$ $\left|\begin{array}{ccc}abc& abc& abc\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$

(taking out from first, second and third columns respectively)

= abc . $\frac{1}{abc}$ $\left|\begin{array}{ccc}bc& ac& ab\\ 1& 1& 1\\ a& b& c\end{array}\right|$

= $\left|\begin{array}{ccc}bc& 1& a\\ ac& 1& b\\ ab& 1& c\end{array}\right|$ = - $\left|\begin{array}{ccc}1& bc& a\\ 1& ac& b\\ 1& ab& c\end{array}\right|$

= - ${\Delta }_2$

$\therefore$ ${\Delta }_1+{\Delta }_2$ = 0