As all entries of square matrix $$A$$ are integers, therefore all co-factors should also be integers.

If det $$A = \pm 1\,\,$$ then $${A^{ - 1}}\,\,$$ exists. Also all entries of $${A^{ - 1}}$$ are integers.

The given equations are

$$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $$

As $$x,y,z$$ are not all zero

$$\therefore$$ The above system should not have unique (zero) solution

$$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{ { - 1} & c & b \cr c & { - 1} & a \cr b & a & { - 1} \cr } } \right| = 0$$

$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$

$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$

$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$

Let $$A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$$ $$\,\,\,$$ then $${A^2} = 1$$

$$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$$

$$ac + cd = 0\,\,\,\,bc + {d^2} = 1$$

From these four relations,

$${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$$

and $$\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$$

We can take $$a = 1,b = 0,c = 0,d = - 1$$

as one possible set of values, then

$$A = \left[ {\matrix{ 1 & 0 \cr 0 & { - 1} \cr } } \right]$$

Clearly $$A \ne I\,\,\,$$ and $$\,\,\,\,A \ne - I\,\,$$ and $$\,\,\,A = - 1$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$1$$ is true.

Also if $$A \ne I\,\,\,\,\,tr\left( A \right) = 0$$

$$\therefore$$ $$\,\,\,\,\,$$ Statement $$2$$ is false.

Given, $$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & {1 + x} & 1 \cr 1 & 1 & {1 + y} \cr } } \right|$$

Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$

and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$

$$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$

Hence, $$D$$ is divisible by both $$x$$ and $$y$$