If $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$, then $A^2 - 5A + 14I = 0$. Which of the following is equivalent to $A^2$?

The equation $\left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & -(2+x^2) \\ 2x+1 & 3x & 1-5x \\ x+1 & 2x & 2-3x \end{array} \right| + \left| \begin{array}{ccc} (1+x)^2 & 2x+1 & x+1 \\ (1-x)^2 & 3x & 2x \\ 1-2x & 3x-2 & 2x-3 \end{array} \right| = 0$

Let $X = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$. Define $f: X \rightarrow \mathbb{R}$ by $f(A) = \operatorname{det}(A), \forall A \in X$. Then, $f$ is

Which of the following determinant$(s)$ vanish(es)?

Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$,$\alpha \in R$ such that $A^{32} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Then a value of $\alpha$ is

If $z = \begin{bmatrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{bmatrix}$, then which of the following is true? (where $i = \sqrt{-1}$)