Question

$\left[\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right]=$

• A. $0$
• B. $1$
• C. $abc$
• D. $(a-b)(b-c)(c-a)$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Compute the determinant of the given matrix.

A matrix $\left[\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2-bc& b^2-ca& c^2-ab\end{array}\right]$ is given.

Find the determinant $△$ as follows:

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

Apply $C_2\to C_2-C_1,C_3\to C_3-C_1$.

$$\begin{gathered} △=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2-bc& b^2-a^2-ca+bc& c^2-a^2-ab+bc\end{array}\right| \\ \Rightarrow △=\left|\begin{array}{ccc}1& 0& 0\\ a& b-a& c-a\\ a^2-bc& (b-a)\left(a+b+c\right)& (c-a)(a+b+c)\end{array}\right| \\ \Rightarrow △=\left(b-a\right)\left(c-a\right)\left|\begin{array}{ccc}1& 0& 0\\ a& 1& 1\\ a^2-bc& \left(a+b+c\right)& (a+b+c)\end{array}\right| \\ \Rightarrow △=\left(b-a\right)\left(c-a\right)\left[1\left|\begin{array}{cc}1& 1\\ a+b+c& a+b+c\end{array}\right|+0+0\right] \\ \Rightarrow △=\left(b-a\right)\left(c-a\right)\left[a+b+c-(a+b+c)\right] \\ \Rightarrow △=0 \end{gathered}$$

Therefore, the determinant of the given matrix is zero.

Hence, option $A$ is the correct answer.