If $B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}$ is the inverse of a $3 \times 3$ matrix $A$,then the sum of all values of $\alpha$ for which $\det(A) + 1 = 0$ is:

Let $A$ be a $2 \times 2$ symmetric matrix such that $A \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$ and the determinant of $A$ is $1$. If $A^{-1} = \alpha A + \beta I$,where $I$ is an identity matrix of order $2 \times 2$,then $\alpha + \beta$ equals:

Let $A$ be any $3 \times 3$ invertible matrix. Then which one of the following is not always true?

If $A = \begin{bmatrix} 1 & 2 \\ 3 & -5 \end{bmatrix}$,then $A^{-1} = $

If $A = \begin{bmatrix} 2 & -3 \\ -4 & 1 \end{bmatrix}$,then $\text{adj}(3A^2 + 12A)$ is equal to

If matrix $A = \begin{bmatrix} -1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1 \end{bmatrix}$, then the sum of all elements in $adj(A^{-1})$ is equal to...