Question

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

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

Answer 1

Answer: (C)

$18$

$\left|\begin{array}{ccc}{\lambda }^2+3\lambda & \lambda -1& \lambda +3\\ \lambda +1& 2-\lambda & \lambda -4\\ \lambda -3& \lambda +4& 3\lambda \end{array}\right|$

Expanding along $R_1$

$=\left({\lambda }^2+3\lambda \right)\left[3\lambda \left(2-\lambda \right)-\left({\lambda }^2-16\right)\right]-\left(\lambda -1\right)\left[3{\lambda }^2+3\lambda -\left(\lambda -4\right)\left(\lambda -3\right)\right]+\left(\lambda +3\right)\left[\left(\lambda +1\right)\left(\lambda +4\right)-\left(2-\lambda \right)\left(\lambda -3\right)\right]$

$=\left({\lambda }^2+3\lambda \right)\left[-4{\lambda }^2+6\lambda +16\right]-\left(\lambda -1\right)\left[2{\lambda }^2+10\lambda -12\right]+\left(\lambda +3\right)\left[2{\lambda }^2+10\right]$

$=-4{\lambda }^4-6{\lambda }^3+32{\lambda }^2+80\lambda +18$

$oncomparingitwith$p{\lambda ^4} + q{\lambda ^3} + r{\lambda ^2} + s{\lambda ^4} + t$

we get, $t=18$