JEE Main 2019 (Online) 10th January Morning ShiftQuestion 1
Let $d\in R$, and $A=\begin{bmatrix}-2&4+d&(\sin\theta)-2\\1&(\sin\theta)+2&d\\5&(2\sin\theta)-d&(-\sin\theta)+2+2d\end{bmatrix}$, $\theta\in[0,2\pi]$. If the minimum value of $\det(A)$ is $8$, then a value of $d$ is :
Determinant
Solution
$\det A=(d+2-\sin\theta)(d+2+\sin\theta)=(d+2)^2-\sin^2\theta$, minimum $(d+2)^2-1=8$, so $d+2=\pm3$, $d=1$ or $-5$. So $d=-5$.