Let $$A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$$ where $$a,b,c,d$$ $$ \ne 0$$

$${A^2} = \left( {\matrix{ a & b \cr c & d \cr } } \right)\left( {\matrix{ a & b \cr c & d \cr } } \right)$$

$$ \Rightarrow {A^2} = \left( {\matrix{ {{a^2} + bc} & {ab + bd} \cr {ac + cd} & {bc + {d^2}} \cr } } \right)$$

$$ \Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$$

$$ab + bd = ac + cd = 0$$

$$c \ne 0\,\,\,\,\,b \ne 0$$

$$ \Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$$

$$\left| A \right| = ad - bc = - {a^2} - bc = - 1$$

$$\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {b + 1} & {c - 1} \cr {a - 1} & {b - 1} & {c + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$$

$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$$

(Taking transpose of second determinant)

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$$

$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$$

$$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$

$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$$

$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$

$${R_1} + {R_3}$$

$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$$

$$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$$

$$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$$

$$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$$ $$\,\,\,\,\,$$ as $$\,\,\,\,\,b\left( {a + c} \right) \ne 0$$

$$ \Rightarrow n$$ should be an odd integer.

We know that $$\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$$b

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|$$

$$\therefore$$ $$\,\,\,\,\,\,$$ Both the statements are true and statement $$-2$$ is a correct explanation for statement - $$1.$$

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.