For $A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 3 \\ 1 & 1 & 0 \end{bmatrix}$,if $A^3 - 2A^2 + kA - 4I_3 = 0$,then $k = $ . . . . . . .

If $A=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}$,prove that $A^{n}=\begin{bmatrix} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{bmatrix}$ for all $n \in N$.

Let $A$ be a $3 \times 3$ invertible matrix. If $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$,then $|A^2|$ is equal to

If $\frac{x^2+7}{(x^2+1)(x-2)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+1}$, then the determinant of the matrix $\begin{bmatrix} A & B \\ C & \frac{2}{5} \end{bmatrix}$ is

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

Let $a = \text{Minimum} \{x^2 + 2x + 3, x \in R\}$ and $b = \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2}$. The value of $\sum_{r = 0}^n a^r \cdot b^{n - r}$ is