Question

If $\left(x_0,y_0,z_0\right)$ is any solution of the system of equations $2x-y-z=1,$ $-x-y+2z=1$ and $x-2y+z=2$ , then the value of $\frac{x_0^2-y_0^2+1}{z_0}\left(where,z_0\ne 0\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

Using Cramer’s rule, we get,
$\Delta =\left|\begin{array}{ccc}2& -1& -1\\ -1& -1& 2\\ 1& -2& 1\end{array}\right|$
Using $C_1=C_1+C_2+C_3$
$\Delta =\left|\begin{array}{ccc}0& -1& -1\\ 0& -1& 2\\ 0& -2& 1\end{array}\right|=0$
${\Delta }_1=\left|\begin{array}{ccc}1& -1& -1\\ 1& -1& 2\\ 2& -2& 1\end{array}\right|=0$ (as $C_1$ and $C_2$ are identical)
${\Delta }_2=\left|\begin{array}{ccc}2& 1& -1\\ -1& 1& 2\\ 1& 2& 1\end{array}\right|$
Using $C_1=C_1-C_2+C_3$
$=\left|\begin{array}{ccc}0& 1& -1\\ 0& 1& 2\\ 0& 2& 1\end{array}\right|=0$
${\Delta }_3=\left|\begin{array}{ccc}2& -1& 1\\ -1& -1& 1\\ 1& -2& 2\end{array}\right|=0$ (as $C_2$ and $C_3$ are identical)
Therefore, infinite solutions
Now, put $z=\lambda$ , we get,
$2x-y=1+\lambda$ and
$-x-y=1-2\lambda$
Solving these $2$ equations, we get,
$x=\lambda ,y=1-\lambda$
$\frac{x_0^2-y_0^2+1}{z_0}=\frac{{\left(\lambda \right)}^2-{\left(\lambda -1\right)}^2+1}{\lambda }=2$