If $A$ is a square matrix,such that $A^2=A$,then $(I+A)^3$ is equal to

If $a > 0$ and the discriminant of $ax^2 + 2bx + c$ is negative,then $\left| \begin{array}{ccc} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0 \end{array} \right|$ is

Let $\alpha$ be a root of the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$,where $a, b, c$ are distinct real numbers such that the matrix $\begin{bmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{bmatrix}$ is singular. Then the value of $\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$ is:

Let $A = \begin{bmatrix} 2 & b & 1 \\ b & b^2+1 & b \\ 1 & b & 2 \end{bmatrix}$ where $b > 0$. Then the minimum value of $\frac{\det(A)}{b}$ is

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Then the maximum value of $\operatorname{det}(A)$ is:

Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & -\sin t & \cos t \end{bmatrix}$. Let $\lambda_{1}, \lambda_{2}, \lambda_{3}$ be the roots of $\det(A - \lambda I_{3}) = 0$, where $I_{3}$ denotes the identity matrix. If $\lambda_{1} + \lambda_{2} + \lambda_{3} = \sqrt{2} + 1$, then the set of possible values of $t$ for $-\pi \leq t < \pi$ is: