Let $$\left| A \right| = 8$$  and $$A$$ is a square matrix of order 3.
We know that $$\left| {adj\,A} \right| = {\left| A \right|^{n - 1}}.I$$    where $$'n'$$ is the order of the matrix $$A.$$
$$\therefore \left| {adj\,A} \right| = {8^{3 - 1}} = {8^2} = 64$$

For the rank to be \[3,\left| {\begin{array}{*{20}{c}} 1&2&3 \\ \lambda &2&4 \\ 2&{ - 3}&1 \end{array}} \right| \ne 0.\]

Let,
\[A = \left[ {\begin{array}{*{20}{c}} { - 1}&2&5\\ 2&{ - 4}&{a - 4}\\ 1&{ - 2}&{a + 1} \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} { - 1}&2&5\\ 0&0&{a + 6}\\ 0&0&{a + 6} \end{array}} \right]\left[ {{R_2} \to {R_2} + 2{R_1},{R_3} \to {R_3} + {R_1}} \right]\]
Clearly rank of $$A$$ is $$1$$ if $$a = - 6$$
Also, for $$a = 1,$$  \[\left| A \right| = \left| {\begin{array}{*{20}{c}} { - 1}&2&5\\ 2&{ - 4}&{ - 3}\\ 1&{ - 2}&2 \end{array}} \right| = 0\]
and \[\left| {\begin{array}{*{20}{c}} 2&5\\ { - 4}&{ - 3} \end{array}} \right| = - 6 + 20 = 14 \ne 0\]
∴ rank of $$A$$ is $$2$$ if $$a = 1$$

Since, adjoint of the square matrix $$A$$ is $$B$$
$$\eqalign{ & \Rightarrow \,\frac{B}{{\left| A \right|}} = {A^{ - 1}} \cr & \Rightarrow \,\frac{{AB}}{{\left| A \right|}} = A{A^{ - 1}} = I \cr & \Rightarrow \,AB = \left| A \right|I \cr & \Rightarrow \,AB = \alpha I \cr} $$

We have, \[\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \gamma &\delta \end{array}} \right]\left[ {\begin{array}{*{20}{c}} \alpha &\beta \\ \gamma &\delta \end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}} {{\alpha ^2} + \beta \gamma }&{\alpha \beta + \beta \delta }\\ {\alpha \gamma + \delta \gamma }&{\beta \gamma + {\delta ^2}} \end{array}} \right]\]
$$\eqalign{ & \Rightarrow {\alpha ^2} + \beta \gamma = 1,\beta \left( {\alpha + \delta } \right) = 0,\gamma \left( {\alpha + \delta } \right) = 0,\beta \gamma + {\delta ^2} = 1 \cr & \Rightarrow \beta = 0 = \gamma ,\alpha \ne - \delta {\text{ and }}{\alpha ^2} = {\delta ^2} \cr & \Rightarrow \delta = \alpha . \cr} $$

\[\left| {\begin{array}{*{20}{c}} 1&{bc}&{a\left( {b + c} \right)}\\ 1&{ac}&{b\left( {c + a} \right)}\\ 1&{ab}&{c\left( {a + b} \right)} \end{array}} \right|\]
Applying $${C_3} \to {C_2} + {C_3}$$
\[\begin{array}{l} \left| {\begin{array}{*{20}{c}} 1&{bc}&{ab + bc + ac}\\ 1&{ac}&{ab + bc + ac}\\ 1&{ab}&{ab + bc + ac} \end{array}} \right|\\ = \,\left( {ab + bc + ac} \right)\left| {\begin{array}{*{20}{c}} 1&{bc}&1\\ 1&{ac}&1\\ 1&{ab}&1 \end{array}} \right| \end{array}\]
$$ = \,\left( {ab + bc + ac} \right) \times 0 = 0$$

\[\begin{array}{l} \therefore \,\left| {\begin{array}{*{20}{c}} {1 + x}&1&1\\ 1&{1 + x}&1\\ 1&1&{1 + x} \end{array}} \right|\\ = \left( {1 + x} \right)\left[ {{{\left( {1 + x} \right)}^2} - 1} \right] - 1\left[ {\left( {1 + x} \right) - 1} \right] + 1\left( {1 - \left( {1 + x} \right)} \right)\\ = \left( {1 + x} \right)\left[ {1 + 2x + {x^2} - 1} \right] - 1\left[ x \right] + 1\left( x \right)\\ = \left[ {1 + x + 2x + 2{x^2} + {x^2} + {x^3} - 1 - x} \right]\\ = {x^3} + 3{x^2}\\ = {x^2}\left( {x + 3} \right) \end{array}\]

\[{A^2} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ a&b&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ a&b&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\]
$$\eqalign{ & {A^2} = {A^4} = {A^6} = {I_3} \cr & \Rightarrow {A^2} + 2{A^4} + 4{A^6} \cr} $$
\[\begin{array}{l} = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 2&0&0\\ 0&2&0\\ 0&0&2 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 4&0&0\\ 0&4&0\\ 0&0&4 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 7&0&0\\ 0&7&0\\ 0&0&7 \end{array}} \right] = 7{I_3} = 7{A^8} \end{array}\]

Using $${R_2} \to {R_2} - {R_3},$$
\[\vartriangle = \left| {\begin{array}{*{20}{c}} 1&1&1 \\ {2 \cdot {2^x} \cdot 2 \cdot {2^{ - x}}}&{2 \cdot {3^x} \cdot 2 \cdot {3^{ - x}}}&{2 \cdot {5^x} \cdot 2 \cdot {5^{ - x}}} \\ {{{\left( {{2^x} - {2^{ - x}}} \right)}^2}}&{{{\left( {{3^x} - {3^{ - x}}} \right)}^2}}&{{{\left( {{5^x} - {5^{ - x}}} \right)}^2}} \end{array}} \right|\]
\[\vartriangle = 4\left| {\begin{array}{*{20}{c}} 1&1&1 \\ 1&1&1 \\ {{{\left( {{2^x} - {2^{ - x}}} \right)}^2}}&{{{\left( {{3^x} - {3^{ - x}}} \right)}^2}}&{{{\left( {{5^x} - {5^{ - x}}} \right)}^2}} \end{array}} \right| = 0.\]

\[{A^2} = \left[ \begin{array}{l} \alpha \,\,\,\,\,\,\beta \\ \beta \,\,\,\,\,\,\alpha \end{array} \right] = \left[ \begin{array}{l} a\,\,\,\,\,\,b\\ b\,\,\,\,\,\,a \end{array} \right]\left[ \begin{array}{l} a\,\,\,\,\,\,\,b\\ b\,\,\,\,\,\,\,a \end{array} \right]\]
\[\,\,\,\,\,\,\,\,\, = \left[ \begin{array}{l} {a^2} + {b^2}\,\,\,\,\,\,\,\,2ab\\ \,\,\,2ab\,\,\,\,\,\,\,\,\,\,\,{a^2} + {b^2} \end{array} \right]\]
$$\alpha = {a^2} + {b^2};\beta = 2ab$$