Question
Matrix $$A$$ such that $$A^2 = 2A – I,$$ where $$I$$ is the identity matrix, then for $$n \geqslant 2,{A^n}$$ is equal to
A.
$${2^{n - 1}} A - \left( {n - 1} \right)I$$
B.
$$2^{n - 1} A - I$$
C.
$$ nA - \left( {n - 1} \right)I$$
D.
$$nA - I$$
Answer :
$$ nA - \left( {n - 1} \right)I$$
Solution :
$$\eqalign{
& {\text{Given,}}\,{A^2} = 2A - I \cr
& {\text{Now,}}\,{A^3} = A\left( {{A^2}} \right) = A\left( {2A = I} \right) \cr
& = \,2{A^2} - A = 2\left( {2A - I} \right) - A = 3A - 2I \cr
& {A^4} = A\left( {{A^3}} \right) = A\left( {3A - 2I} \right) \cr
& = \,3{A^2} - 2A = 3\left( {2A - I} \right) - 2A = 4A - 3I \cr} $$
Following this, we can say $${A^n} = nA - \left( {n - 1} \right)I$$