If $A = \begin{bmatrix} 2 & -3 \\ 4 & 1 \end{bmatrix}$,then $A + \operatorname{adj}(A)$ is:

If $K \in R_0$,then $\det(adj(KI_n))$ is equal to:

The inverse of the matrix $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ is

If $\omega$ is a complex cube root of unity and $A=\begin{bmatrix} \omega & 0 \\ 0 & \omega \end{bmatrix}$,then $A^{-1}=$

Let $I$ be a unit matrix of order $6$. Let $A = (a_{ij})$ be a square matrix of order $6$ such that $a_{ij} = \begin{cases} 1, & \text{if } i+j=7 \\ 0, & \text{if } i+j \neq 7 \end{cases}$. Then $(A(\text{adj } A) A^{-1}) A^2 = $

Which of the following matrices is invertible?
$A_{1}=\begin{bmatrix} 4 & 2 \\ 2 & 1 \end{bmatrix}$
$A_{2}=\begin{bmatrix} -1 & -2 & 3 \\ 4 & 5 & 7 \\ 2 & 4 & -6 \end{bmatrix}$
$A_{3}=\begin{bmatrix} 1 & 0 & 0 \\ 5 & 2 & 1 \\ 7 & 2 & 1 \end{bmatrix}$
$A_{4}=\begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1 \end{bmatrix}$