Question

If$\Delta = \left| \begin{matrix} x + 1 & x^{2} + 2 & x(x + 1) \\ x(x + 1) & x + 1 & x(x^{2} + 2) \\ x^{2} + 2 & x(x + 1) & x + 1 \end{matrix} ⥂ \right| = p_{0}x^{6} + p_{1}x^{5} + p_{2}x^{4} + p_{3}x^{3} + p_{4}x^{2} + p_{5}x + p_{6}$,then $(p_{5},p_{6}) =$

• A. (–3, –9)
• B. (–5, –9)
• C. (–3, –5)
• D. (3, –9)

Answer 1

Answer: (B)

(–5, –9)

Answer 2

Putting $x = 0$ in both sides, we get, $\left| \begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 2 & 0 & 1 \end{matrix} \right| = p_{6}$

$\Rightarrow$ $p_{6} = 9$ by expansion.

$p_{5}$ is the coefficient of x or constant term in the

differentiation of determinant.

Differentiate both sides,

1 & 2x & 2x + 1 \\ x(x + 1) & x + 1 & x(x^{2} + 2) \\ x^{2} + 2 & x(x + 1) & x + 1 \end{matrix} ⥂ \right| + \left| \begin{matrix} x + 1 & x^{2} + 2 & x(x + 1) \\ 2x + 1 & 1 & 3x^{2} + 2 \\ x^{2} + 2 & x(x + 1) & x + 1 \end{matrix} ⥂ \right| $$$+ \left| \begin{matrix} x + 1 & x^{2} + 2 & x(x + 1) \\ x(x + 1) & x + 1 & x(x^{2} + 2) \\ 2x & 2x + 1 & 1 \end{matrix} ⥂ \right|$ $= 6p_{0}x^{5} + 5p_{1}x^{4} + 4p_{2}x^{3} + 3p_{3}x^{2} + 2p_{4}x + p_{5}$Putting $x = 0$both sides, we get $p_{5} = - 5$; $\therefore(p_{5},p_{6}) = ( - 5,9)$.