Let $a = \lim_{x \to 1} \left( \frac{x}{\ln x} - \frac{1}{x \ln x} \right)$,$b = \lim_{x \to 0} \frac{x^3 - 16x}{4x + x^2}$,$c = \lim_{x \to 0} \frac{\ln(1 + \sin x)}{x}$,and $d = \lim_{x \to -1} \frac{(x + 1)^3}{3(\sin(x + 1) - (x + 1))}$. Then the matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is:

Let $A$ be a symmetric matrix and $B$ be a skew-symmetric matrix,such that $A - B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$. Then $|A|$ is:

If the minimum and the maximum values of the function $f : \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$,defined by $f(\theta) = \left|\begin{array}{ccc} -\sin^2 \theta & -1-\sin^2 \theta & 1 \\ -\cos^2 \theta & -1-\cos^2 \theta & 1 \\ 12 & 10 & -2 \end{array}\right|$ are $m$ and $M$ respectively,then the ordered pair $(m, M)$ is equal to:

Let $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$ and $A$ be a $2 \times 2$ matrix satisfying $\left(A^T\right)^{-1}=A$. If $X=A B A^T$,then $A^T X^{2021} A=$

If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$ and $\det(A^n - I) = 1 - \lambda^n$ for $n \in N$,then $\lambda$ is:

The maximum value of $f(x) = \left|\begin{array}{ccc} \sin^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x & \cos^{2} x & \sin 2x \end{array}\right|, x \in R$ is: