Question

The determinant $\Delta =\left|\begin{array}{ccc}\lambda a& {\lambda }^2+a^2& 1\\ \lambda b& {\lambda }^2+b^2& 1\\ \lambda c& {\lambda }^2+c^2& 1\end{array}\right|$ equals

• A. $\lambda (a-b)(b-c)(c-a)$
• B. $\lambda \left(a^2+b^2+c^2\right)$
• C. $\lambda (a+b+c)$
• D. ${\lambda }^2(a-b)(b-c)(c-a)$

Answer 1

Answer: (A)

Using $C_2\to C_2-{\lambda }^2{C}_3$, we can write
$\Delta =\lambda \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|$
Using $R_3\to R_3-R_2,R_2\to R_2-R_1$, we get
$\Delta =\lambda \left|\begin{array}{ccc}a& a^2& 1\\ b-a& b^2-a^2& 0\\ c-b& c^2-b^2& 0\end{array}\right|$
$=\lambda (b-a)(c-b)\left|\begin{array}{cc}1& b+a\\ 1& c+b\end{array}\right|$
$=\lambda (b-a)(c-b)(c-a)$
$=\lambda (a-b)(b-c)(c-a)$