Question

Find the value of the following determinants.
$\left|\begin{matrix}a&b&c\\ a-b&b-c&c-a\\ b+c&c+a&a+b\end{matrix}\right|$

• A. $\left(a+b+c\right)^{2}$
• B. $a^{3}+b^{3}+c^{3}-3abc$
• C. $\left(a+b+c\right)^{3}$
• D. $a^{3}+b^{3}+c^{3}$

Answer 1

Answer: (B)

Let $\Delta=\left|\begin{matrix}a&b&c\\ a-b&b-c&c-a\\ b+c&a+c&a+b\end{matrix}\right|$

Applying $C_{1}\rightarrow C_{1}+C_{2}+C_{3}$, we get

$\Delta=\left|\begin{matrix}a+b+c&b&c\\ 0&b-c&c-a\\ 2\left(a+b+c\right)&a+c&a+b\end{matrix}\right|$

Taking $\left(a + b + c\right)$ common from $C_{1}$, we get

$\Delta=\left(a+b+c\right)\left|\begin{matrix}1&b&c\\ 0&b-c&c-a\\ 2&a+c&a+b\end{matrix}\right|$

Applying $R_{3}\rightarrow R_{3}-2R_{1}$, we get

$\Delta=\left(a+b+c\right)\left|\begin{matrix}1&b&c\\ 0&b-c&c-a\\ 0&a+c-2b&a+b-2c\end{matrix}\right|$

Expanding along $C_{1}$, we get

$\Delta=\left(a+b+c\right)\left[\left(b-c\right)\left(a+b-2c\right)-\left(c-a\right)\left(a+c-2b\right)\right]$

$=\left(a+b+c\right)\left[a^{2}+b^{2}+c^{2}-ab-bc-ac\right]$

$=a^{3}+b^{3}+c^{3}-3abc$