Question

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

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

Answer 1

Answer: (C)

$18$

To get $t$,

Put $\lambda =0$

We get, value of determinant as,

$\left[0\right(0+16)+1(0-12)+3(4+6\left)\right]=t$

=> $t=18$