Question

If $U=\left|\begin{array}{ccc}y+z& z& y\\ z& z+x& x\\ y& x& x+y\end{array}\right|$, then $U_{xyz}=?$, where $U_x$ is differentiation of $u$ w.r.t $x$

• A. $0$
• B. $1$
• C. $x+y+z$
• D. $4$

Answer 1

The correct option is D $4$

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

$u_{xyz}=4$