Find the value(s) of p for the pair of equati | vedclass

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(C) The given pair of linear equations is:$-x + py - 1 = 0 \quad ...(i)$$px - y - 1 = 0 \quad ...(ii)$Comparing these with the standard form $ax + by

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$1$

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(C) The given pair of linear equations is:$-x + py - 1 = 0 \quad ...(i)$$px - y - 1 = 0 \quad ...(ii)$Comparing these with the standard form $ax + by + c = 0$,we get:$a_1 = -1, b_1 = p, c_1 = -1$ (from Eq. $i$)$a_2 = p, b_2 = -1, c_2 = -1$ (from Eq. $ii$)For a pair of linear equations to have no solution,the lines must be parallel,which implies:$\frac{a_1}{a_2} = \frac{b_1}{b_2} 
eq \frac{c_1}{c_2}$Substituting the values:$\frac{-1}{p} = \frac{p}{-1} 
eq \frac{-1}{-1}$Taking the first two parts:$\frac{-1}{p} = \frac{p}{-1} \Rightarrow p^2 = 1 \Rightarrow p = \pm 1$Taking the condition $\frac{p}{-1} 
eq \frac{-1}{-1}$:$\frac{p}{-1} 
eq 1 \Rightarrow p 
eq -1$Since $p = \pm 1$ and $p 
eq -1$,the only valid value is $p = 1$.

Find the value$(s)$ of $p$ for the pair of equations $-x + py = 1$ and $px - y = 1$ if the system has no solution.

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