Question

By using properties of determinants prove that $\left|\begin{array}{ccc}(y+z{)}^2& xy& zx\\ xy& (x+z{)}^2& yz\\ xz& yz& (x+y{)}^2\end{array}\right|$ is divisible by $(x+y+z{)}^n$. Find $n$ .

Answer 1

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

multiply $R_1,R_2,R_3$ by $x,y,z$

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

applying $C_1\to \to C_1-C_3$

and $C_2\to C_2-C_3$

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

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

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

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

Taking x,y $2$xy common $R_1,R_2$ and $R_3$

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

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

Expending along $R_3\to$

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

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

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

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