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 $\frac{x_{0}^{2} - y_{0}^{2} + 1}{z_{0}}\left(where , z_{0} \neq 0\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

Using Cramer’s rule, we get,
$\Delta =\begin{vmatrix} 2 & -1 & -1 \\ -1 & -1 & 2 \\ 1 & -2 & 1 \end{vmatrix}$
Using $C_{1}=C_{1}+C_{2}+C_{3}$
$\Delta =\begin{vmatrix} 0 & -1 & -1 \\ 0 & -1 & 2 \\ 0 & -2 & 1 \end{vmatrix}=0$
$\Delta _{1}=\begin{vmatrix} 1 & -1 & -1 \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix}=0$ (as $C_{1}$ and $C_{2}$ are identical)
$\Delta _{2}=\begin{vmatrix} 2 & 1 & -1 \\ -1 & 1 & 2 \\ 1 & 2 & 1 \end{vmatrix}$
Using $C_{1}=C_{1}-C_{2}+C_{3}$
$=\begin{vmatrix} 0 & 1 & -1 \\ 0 & 1 & 2 \\ 0 & 2 & 1 \end{vmatrix}=0$
$\Delta _{3}=\begin{vmatrix} 2 & -1 & 1 \\ -1 & -1 & 1 \\ 1 & -2 & 2 \end{vmatrix}=0$ (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 $
$\frac{x_{0}^{2} - y_{0}^{2} + 1}{z_{0}}=\frac{\left(\lambda \right)^{2} - \left(\lambda - 1\right)^{2} + 1}{\lambda }=2$