Question

Using the properties of determinants, show that:

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

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

Answer 1

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

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

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

Taking $a+b+c$ common from $C_1,C_2$

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

$R_2\to R_2+R_1$

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

Expanding along first column,

$={(a+b+c)}^2[-1(-c+a+b-2a-2b\left)\right]$

$={(a+b+c)}^3$

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

$R_1\to R_1-R_2,R_2\to R_2-R_3$

$=\left|\begin{array}{ccc}x+y+z& -(x+y+z)& 0\\ 0& x+y+z& -(x+y+z)\\ z& x& z+x+2y\end{array}\right|$

Taking $(x+y+z)$ common from $R_1,R_2$

$={(x+y+z)}^2\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ z& x& z+x+2y\end{array}\right|$

$C_2\to C_2+C_3$

$={(x+y+z)}^2\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& -1\\ z& x+z& z+x+2y\end{array}\right|$

Expanding along first row,

$={(x+y+z)}^2\left|\begin{array}{cc}1& -1\\ x+z& x+z+2y\end{array}\right|$

$={(x+y+z)}^2(x+z+2y+x+z)$

$=2(x+y+{z)}^3$