Question

The determinant $\Delta=\begin{vmatrix}\lambda a & \lambda^2+a^2 & 1 \\ \lambda b & \lambda^2+b^2 & 1 \\ \lambda c & \lambda^2+c^2 & 1\end{vmatrix}$ 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 $C_2 \rightarrow C_2-\lambda^2 C_3$, we can write
$\Delta=\lambda\begin{vmatrix}a & a^2 & 1 \\b & b^2 & 1 \\c & c^2 & 1\end{vmatrix}$
Using $R_3 \rightarrow R_3-R_2, R_2 \rightarrow R_2-R_1$, we get
$\Delta =\lambda\begin{vmatrix} a & a^2 & 1 \\b-a & b^2-a^2 & 0 \\c-b & c^2-b^2 & 0\end{vmatrix}$
$ =\lambda(b-a)(c-b)\begin{vmatrix}1 & b+a \\1 & c+b \end{vmatrix}$
$ =\lambda(b-a)(c-b)(c-a) $
$ =\lambda(a-b)(b-c)(c-a)$