Question

If a, b, c are real numbers, then the value of the determinant $\left|\begin{array}{ccc}1-a& a-b-c& b+c\\ 1-b& b-c-a& c+a\\ 1-c& c-a-b& a+b\end{array}\right|$ is

• A. 0
• B. $(a-b)(b-c)(c-a)$
• C. $(a+b+c{)}^2$
• D. $(a+b+c{)}^3$

Answer 1

Answer: (A)

0

We have, $△=\left[\begin{array}{ccc}1-a& a-b-c& b+c\\ 1-b& b-c-a& c+a\\ 1-c& c-a-b& a+b\end{array}\right]$

Applying $C_2\to C_2+C_3,$ we get

$△=\left[\begin{array}{ccc}1-a& a& b+c\\ 1-b& b& c+a\\ 1-c& c& a+b\end{array}\right]$

Applying $C_1\to C_1+C_2$ and $C_3\to C_3+C_2$ and taking $a+b+c$ from $C_3$ we get

$△=(a+b+c)\left[\begin{array}{ccc}1& a& 1\\ 1& b& 1\\ 1& c& 1\end{array}\right]$

$\because C_1$ and $C_3$ are identical.

So, $△=0$

Hence, A is the correct option.