Question

if $\Delta$ = $\left|\begin{array}{ccc}x& 2y-z& -z\\ y& 2x-z& -z\\ y& 2y-z& 2x-2y-z\end{array}\right|$ , then

• A. (x-y)is a factor of$\Delta$
• B. $(x-y{)}^2$is a factor of$\Delta$
• C. $(x-y{)}^3$is a factor of$\Delta$
• D. $\Delta$ is independent of z

Answer 1

Answer: (A), (B)

The correct options are
A (x-y)is a factor of$\Delta$
B $(x-y{)}^2$is a factor of$\Delta$

$\Delta =\left|\begin{array}{ccc}x& 2y-z& -z\\ y& 2x-z& -z\\ y& 2y-z& 2x-2y-z\end{array}\right|$
so,by column transformation,
$C_2\to C_2-C_3$
so,$\Delta =\left|\begin{array}{ccc}x& 2y& -z\\ y& 2x& -z\\ y& 4y-2x& 2x-2y-z\end{array}\right|$
$\Delta =\frac{1}{2}\left|\begin{array}{ccc}2x& 2y& -z\\ 2y& 2x& -z\\ 2y& 4y-2x& 2x-2y-z\end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}2(x-y)& 2y& -z\\ 2(y-x)& 2x& -z\\ 2(x-y)& 4y-2x& 2x-2y-z\end{array}\right|$
$=\frac{1}{2}(x-y)\left|\begin{array}{ccc}z& 2y& -z\\ -z& 2x& -z\\ z& 4y-2x& 2x-2y-z\end{array}\right|$
$=(x-y{)}^2\left|\begin{array}{ccc}0& (y+x)& 0\\ -z& 2x& -z\\ z& 4y-2x& 2x-2y-z\end{array}\right|$