Determinants Matrices Question 177
Question: For what value of p, is the system of equations : $ p^{3}x+{{(p+1)}^{3}}y={{(p+2)}^{3}} $
$ px+(p+1)y=p+2 $
$ x+y=1 $ consistent-
Options:
A) $ p=0 $
B) $ p=1 $
C) $ p=-1 $
D) For all $ p>1 $
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] The given system of equations are: $ p^{3}x+{{(p+1)}^{3}}y={{(p+2)}^{3}} $
…. (1) $ px+(p+1)y=(p+2) $
…. (2) $ x+y=1 $
…. (3) This system is consistent, if values of x and y from first two equation satisfy the third equation. which
$ \Rightarrow \begin{vmatrix} p^{3} & {{(p+1)}^{3}} & {{(p+2)}^{3}} \\ p & (p+1) & (p+2) \\ 1 & 1 & 1 \\ \end{vmatrix}=0 $
$ \Rightarrow \begin{vmatrix} p^{3} & {{(p+1)}^{3}}-p^{3} & {{(p+2)}^{3}}-p^{3} \\ p & 1 & 2 \\ 1 & 0 & 0 \\ \end{vmatrix}=0 $
$ \Rightarrow 2{{(p+1)}^{3}}-2p^{3}-{{(p+2)}^{3}}+p^{3}=0 $
$ \Rightarrow 2(p^{3}+1+3p^{2}+3p)-2p^{3}-(p^{3}+8+12p+6p^{2})+p^{3}=0 $
$ \Rightarrow 2p^{3}+2+6p^{2}+6p-2p^{3}-p^{3}-8-12p $
$ -6p^{2}+p^{3}=0 $
$ \Rightarrow -6-6p=0\Rightarrow p=-1 $
BETA
