Question

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

Answer 1

Applying $R_1\to x{R}_1,R_2,R_3\to z{R}_3$ to $\Delta 4$and dividing by $xyz$ we get

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

Taking common factor $x,y,z$ from $C_1,C_2$ $C_3$ respectively, we get

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

Applying $C_2\to C_2,C-3\to C_3-C_1$ we have

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

Taking common factor $(x+y+z)$ from $C_2$ and $C_3$ we have

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

Applying $R_1\to R_1-(R_2+R_3)$

we have

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

Applying $C_2\to C_2+\frac{1}{y}{C}_1$ and $C_3\to C_3+\frac{1}{2}{C}_1$, we get

$\Delta ={(x+y+z)}^2\left[\begin{array}{ccc}2yz& 0& 0\\ y^2& x+z& \frac{y^2}{z}\\ z^2& \frac{z^2}{y}& x+y\end{array}\right]$

Finally expanding along $R_1$, we have

$\Delta =(x+y+z{)}^2(2yz\left)\right[(x+z)(x+y)-yz]=(x+y+z{)}^2\left(2yz\right)(x^2+xy+xz)=2xyz(x+y+z{)}^3$

Comparing L.H.S and R.H.S,

$\Rightarrow m=3$