Given A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right] $\therefore$ $A \times A$ = ${A^2}$ = \left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right] and ${A^3}$ = ${A^2} \times A$ = \left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right] So we can say ${A^n}$ = \left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right] Now $nA - \left( {n - 1} \right){\rm I}$ = \left[ {\matrix{ n & 0 \cr n & n \cr } } \right] - \left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right] = ${A^n}$ $\therefore$ ${A^n} = nA - \left( {n - 1} \right){\rm I}$