Question

Show that

$\left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& b+a& c\end{array}\right|=(a+b+c)(a^2+b^2+c^2)$

Answer 1

Let $A=\left|\begin{array}{ccc}a& b-c& c+b\\ a+c& b& c-a\\ a-b& b+a& c\end{array}\right|$

On multiplying and dividing first column by a, we get,

$A=\frac{1}{a}\left|\begin{array}{ccc}{a}^2& b-c& c+b\\ a^2+ac& b& c-a\\ a^2-ab& b+a& c\end{array}\right|$

Applying $C_1\to C_1+b{C}_2+c{C}_3$, we get,

$A=\frac{1}{a}\left|\begin{array}{ccc}{a}^2+b^2+c^2& b-c& c+b\\ a^2+b^2+c^2& b& c-a\\ a^2+b^2+c^2& b+a& c\end{array}\right|$

On taking $a^2+b^2+c^2$ common from $C_1$, we get,

$A=\frac{1}{a}(a^2+b^2+c^2)\left|\begin{array}{ccc}1& b-c& c+b\\ 1& b& c-a\\ 1& b+a& c\end{array}\right|$

Applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, we get,

$A=\frac{1}{a}(a^2+b^2+c^2)\left|\begin{array}{ccc}1& b-c& c+b\\ 0& c& -a-b\\ 0& a+c& -b\end{array}\right|$

Now, expanding along $C_1$, we get,

$A=\frac{1}{a}(a^2+b^2+c^2)\times 1\times \left|\begin{array}{cc}c& -a-b\\ a+c& -b\end{array}\right|$

Therefore, $A=\frac{1}{a}(a^2+b^2+c^2)(a^2+ac+ba)$

$A=\frac{1}{a}(a^2+b^2+c^2)\times a(a+b+c)$

$A=(a+b+c)(a^2+b^2+c^2)$

Hence, proved.