Question

If $p{\lambda }^4+p{\lambda }^3+p{\lambda }^2+s\lambda +t=$ $\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda +1& \lambda +3\\ \lambda +1& 2-\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|$, then value of t is

• A. $16$
• B. $18$
• C. $17$
• D. $19$

Answer 1

Answer: (B)

$18$

first we solve RHS then compare with LHS

$=>({\lambda }^2+3\lambda )\left[\right(2-\lambda \left)\right(3\lambda )-(\lambda +4\left)\right(\lambda -4\left)\right]-(\lambda +1)\left[\right(\lambda +1\left)\right(3\lambda )-(\lambda -3\left)\right(\lambda -4\left)\right]$ $+(\lambda +3)$

$\left[\right(\lambda +1\left)\right(\lambda +4-(\lambda -3)(2-\lambda )\left)\right]$

$=>[-4{\lambda }^2+6{\lambda }^3+16{\lambda }^2-12{\lambda }^3+18{\lambda }^2+48\lambda -2{\lambda }^3-10{\lambda }^2+12\lambda -2{\lambda }^2-10\lambda +12+2{\lambda }^3+10\lambda +6{\lambda }^2+30]$

$=>[-4{\lambda }^4-6{\lambda }^3+28{\lambda }^2+60\lambda +18]$

Now compare both side

$=>[p{\lambda }^4+p{\lambda }^3+p{\lambda }^2+s\lambda +t=-4{\lambda }^4-6{\lambda }^3+28{\lambda }^2+60\lambda +18]$

hence,$t=18$