Question

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

• A. 2xyz
• B. 1
• C. xyz
• D. $x^2{y}^2{z}^2$

Answer 1

The correct option is B 1

$\left|\begin{array}{ccc}y+z& x& y\\ z+x& z& x\\ x+y& y& z\end{array}\right|=(x+y+z)\left|\begin{array}{ccc}2& 1& 1\\ z+x& z& x\\ x+y& y& z\end{array}\right|by{R}_1\to R_1+R_2+R_3=(x+y+z)\left|\begin{array}{ccc}1& 1& 1\\ x& z& x\\ x& y& z\end{array}\right|;by{C}_1\to C_1-C_2=(x+y+z).\left\{\right(z^2-xy)-(xz-x^2)+(xy-xz\left)\right\}=(x+y+z)(x-z{)}^2\Rightarrow k=1$.
Trick : Put x=1, y=2, z=3, then
$\left|\begin{array}{ccc}5& 1& 2\\ 4& 3& 1\\ 3& 2& 3\end{array}\right|=5\left(7\right)-1(12-3)+2(8-9)=35-9-2=24$ and $(x+y+z)(x-z{)}^2=(6\left)\right(-2{)}^2=24$
$\therefore k=\frac{24}{24}=1$.