Question

If $\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-z^2\\ xz-y^2& xy-z^2& yz-x^2\\ xy-z^2& yz-x^2& zx-y^2\end{array}\right|=\left|\begin{array}{ccc}{r}^2& u^2& u^2\\ u^2& r^2& u^2\\ u^2& u^2& r^2\end{array}\right|$, then

• A. $r^2=x+y+z$
• B. $r^2=x^2+y^2+z^2$
• C. $u^2=yz+zx+xy$
• D. $u^2=xyz$

Answer 1

Answer: (B), (C)

$r^2=x^2+y^2+z^2$
$u^2=yz+zx+xy$

Given,$\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-z^2\\ xz-y^2& xy-z^2& yz-x^2\\ xy-z^2& yz-x^2& zx-y^2\end{array}\right|=\left|\begin{array}{ccc}{r}^2& u^2& u^2\\ u^2& r^2& u^2\\ u^2& u^2& r^2\end{array}\right|$
$RHS=\left|\begin{array}{ccc}{r}^2& u^2& u^2\\ u^2& r^2& u^2\\ u^2& u^2& r^2\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}{r}^2-u^2& -(r^2-u^2)& 0\\ 0& r^2-u^2& -(r^2-u^2)\\ u^2& u^2& r^2\end{array}\right|$
$={(r^2-u^2)}^2\left|\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ u^2& u^2& r^2\end{array}\right|$
$\Rightarrow RHS={(r^2-u^2)}^2(r^2+2u^2)$
Now, $LHS=\left|\begin{array}{ccc}yz-x^2& zx-y^2& xy-z^2\\ xz-y^2& xy-z^2& yz-x^2\\ xy-z^2& yz-x^2& zx-y^2\end{array}\right|$
$C_1\to C_1{+C}_2+C_3$
$=(xy+yz+xz-x^2-y^2-z^2)\left|\begin{array}{ccc}1& zx-y^2& xy-z^2\\ 1& xy-z^2& yz-x^2\\ 1& yz-x^2& zx-y^2\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=(xy+yz+xz-x^2-y^2-z^2)\left|\begin{array}{ccc}0& (z-y)(x+y+z)& (x-z)(x+y+z)\\ 0& (x-z)(x+y+z)& (y-x)(x+y+z)\\ 1& yz-x^2& zx-y^2\end{array}\right|$
$LHS=(xy+yz+xz-x^2-y^2-z^2{)}^2{(x+y+z)}^2$
So, on comparing $r^2-u^2=xy+yz+xz-x^2-y^2-z^2$ ....(i)
and $r^2+2u^2={(x+y+z)}^2$ .....(ii)
Subtracting (i) from (ii), we get
$\Rightarrow u^2=xy+yz+zx$
Also, $r^2=x^2+y^2+z^2$
Hence, options 'B' and 'C' are correct.