Question

Using properties of determinants, prove that

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

Answer 1

LHS :

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

Applying $R_1\to z{R}_1,R_2\to x{R}_2,R_3\to y{R}_3$, we get,

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

Applying $C_1\to \frac{1}{z}{C}_1,C_2\to \frac{1}{x}{C}_2,C_3\to \frac{1}{y}{C}_3$, we get,

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

Applying $C_1\to C_1-C_3,C_2\to C_2-C_3$, we get,

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

Applying $R_3\to R_3-(R_1+R_2)$, we get,

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

Applying $C_1\to C_1+\frac{1}{z}{C}_3,C_2\to C_2+\frac{1}{x}{C}_3$, we get,

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

$=(x+y+z{)}^2[2xz(xz+xy+yz+y^2-xz)]$

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

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