Question

If $p, q , r$ are the roots of the equation $\begin{vmatrix}x&1&2\\ 1&x&2\\ 1&2&x\end{vmatrix} = 0$ , then $\frac{p^{4} +q^{4}+r^{4}}{p^{2}+q^{2} +r^{2}}$ is equal to

• A. 7
• B. 6
• C. 1/6
• D. 1/7

Answer 1

Answer: (A)

$p, q , r$ the roots of equation ,
$\begin{vmatrix}x&1&2\\ 1&x&2\\ 1&2&x\end{vmatrix} = 0$
Operating $C_1 \to C_1 + C_2 + C_3$, we get
$\begin{vmatrix}x+3&1&2\\ x+3&x&2\\ x+3&2&x\end{vmatrix}=0$
$\Rightarrow \left(x+3\right) \begin{vmatrix}1&1&2\\ 1&x&2\\ 1&2&x\end{vmatrix} = 0 $
Operating $R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 $
$\Rightarrow \left(x+3\right) \begin{vmatrix}1&1&2\\ 0&x-1&0\\ 0&1&x-2\end{vmatrix} $
Expanding along $C_1$, we get $(x + 3) [(x - l)(x - 2) - 0]= 0 x = 1, 2, - 3$ i.e., roots of equation
Let $ p = 1, q = 2, r = - 3$.
$\therefore \frac{p^{4} +q^{4}+r^{4}}{p^{2}+q^{2}+r^{2}} = \frac{1+16+81}{1+4+9} =\frac{98}{14}=7$