Question

$\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=(a+b+c{)}^3$

$\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=2(x+y+z{)}^3$

Answer 1

$LHS=\left|\begin{array}{ccc}a-b-c& 2a& 2a\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|=\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$ (using $R_1\to R_1+R_2+R_3$)

Take out(a+b+c) common from $R_1$, we get
$=(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ 2b& b-c-a& 2b\\ 2c& 2c& c-a-b\end{array}\right|$

$=(a+b+c)\left|\begin{array}{ccc}1& 0& 0\\ 2b& -b-c-a& 0\\ 2c& 0& -c-a-b\end{array}\right|$ (using $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$)

Expanding along $R_1$. we get
$=(a+b+c)\left[1\right(-b-c-a\left)\right(-c-a-b\left)\right]=(a+b+c)[-(b+c+a)\times (-\left)\right(c+a+b\left)\right]=(a+b+c)(a+b+c)(a+b+c)=(a+b+c{)}^3=RHS$ Hence proved.

$LHS=\left|\begin{array}{ccc}x+y+2z& x& y\\ z& y+z+2x& y\\ z& x& z+x+2y\end{array}\right|=\left|\begin{array}{ccc}2(x+y+z)& x& y\\ 2(x+y+z)& y+z+2x& y\\ 2(x+y+z)& x& z+x+2y\end{array}\right|$
(using $C_1\to C_1+C_2+C_3$)

Take out $2(x+y+z)$ common from $C_1$, we get
$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 1& y+z+2x& y\\ 1& x& z+x+2y\end{array}\right|$

$=2(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& y+z+x& 0\\ 0& 0& z+x+y\end{array}\right|$ (using $R_2\to R_2-R_1,R_3\to R_3-R_1$)

Take out $(x+y+z)$ common from $R_2$ and $R_3$, we get
$=2(x+y+z)(x+y+z)(x+y+z)\left|\begin{array}{ccc}1& x& y\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|$

Expanding along $R_3$, we get
$=2(x+y+z{)}^3[\left(1\right)(1-0)]=2(x+y+z{)}^3=RHS$.