Let $A = \begin{bmatrix} 4 & -2 \\ \alpha & \beta \end{bmatrix}$. If $A^2 + \gamma A + 18I = O$,then $\operatorname{det}(A)$ is equal to

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$

The number of elements in the set $\{A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I-A)^3 = I-A^3 \}$,where $I$ is the $2 \times 2$ identity matrix,is:

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$,then:

In a square matrix $A$ of order $3$,$a_{ii} = i + m_i$,where $i = 1, 2, 3$ and $m_i$ are the slopes (in increasing order of their absolute value) of the $3$ normals concurrent at the point $(9, -6)$ to the parabola $y^2 = 4x$. All other entries of the matrix are $1$. The value of $\det(A)$ is equal to

If ${a^2} + {b^2} + {c^2} = -2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$,then $f(x)$ is a polynomial of degree: