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

We have
L.H.S =$\left|\begin{array}{ccc}(x+y{)}^2& zx& zy\\ zx& (z+y{)}^2& xy\\ zy& xy& (z+x{)}^2\end{array}\right|$

Operating $R_1\to z{R}_1,R_2\to x{R}_2$ and $R_3\to y{R}_3$

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

Taking $z,x$ and $y$ common form $C_1,C_2$ and $C_3$ respectively.

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

Operating $C_1\to C_1-C_3$ and $C_2\to C_2-C_3,$ we have

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

Taking $(x+y+z)$ common form $C_1$ and $C_2$

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

Operating $R_3\to R_3-R_1-R_2$, we have

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

Applying, $C_{1}z$ and $C_{2}x$, we have

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

Adding $C_1\to C_1+C_3$ and $C_2\to C_2+C_3$, we have

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

Expanding along $R_3$, we have

$=\frac{(x+y+z{)}^2}{zx}\times 2zx\left[\right(xz+yz\left)\right(xy+xz)-(x^2{z}^2\left)\right]$

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

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

$=2\left(xyz\right)(x+y+z{)}^3=R.H.S.$