$$ax + y + z = \alpha - 1$$

$$x + \alpha \,y + z = \alpha - 1;$$

$$x + y + z\alpha = \alpha - 1$$

$$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & 1 & \alpha \cr } } \right|$$

$$ = \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)$$

$$ = \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)$$

For infinite solutions, $$\Delta = 0$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0$$

$$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0$$

$$ \Rightarrow \alpha = - 2,1;$$

But $$\alpha \ne 1.\,\,\,$$ $$\therefore$$ $$\,\,\alpha = - 2$$

Given $${A^2} - A + I = 0$$

$${A^{ - 1}}{A^2} - {A^{ - 1}}A + {A^{ - 1}}.I = {A^{ - 1}}.0$$

(Multiplying $$\,\,\,{A^{ - 1}}$$ on both sides)

$$ \Rightarrow A - 1 + {A^{ - 1}} = 0$$

or $$\,\,\,\,\,\,{A^{ - 1}} = 1 - A$$

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$

$$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \cr } } \right|$$

$$ = \left| {\matrix{ {\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \cr {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \cr {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \cr } } \right|$$

$$ = 0\left[ \, \right.$$ Apply $$\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]$$

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

clearly $$\,\,\,A \ne 0.\,$$ Also $$\,\,\left| A \right| = - 1 \ne 0$$

$$\therefore$$ $${A^{ - 1}}\,\,$$ exists, further

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

Also $${A^2} = \left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]\left[ {\matrix{ 0 & 0 & { - 1} \cr 0 & { - 1} & 0 \cr { - 1} & 0 & 0 \cr } } \right]$$

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