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Matrix A such that ${{A}^{2}}=2A-I$, where I is the identity matrix. Then for $n\ge 2$,${{A}^{n}}$ is equal to
(a) $nA - (n - 1)I$
(b) $nA – I$
(c) ${{2}^{n-1}}A-\left( n-1 \right)I$
(d) ${{2}^{n-1}}A-I$

Answer
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Hint:Find ${{A}^{3}}$ by multiplying the given equation ${{A}^{2}}=2A-I\to A$, further calculate ${{A}^{4}}$by multiplying the equation of ${{A}^{3}}$, similarly, get ${{A}^{5}},{{A}^{6}},{{A}^{7}}......$ by multiplying the previous equation formed by A. and use the equation ${{A}^{2}}=2A-I$ to make all the equations of ${{A}^{3}},{{A}^{4}},{{A}^{5}}......$ only in terms of A and I. Now, observe all the equations and get the values of ${{A}^{n}}$by observing the equations only.

Complete step-by-step answer:
Equation with matrix is given as
${{A}^{2}}=2A-I$……………..(i)
Where, I is the identity matrix.
And with the help of equation (i), we need to calculate the value of ${{A}^{n}}$, for $n\ge 2$.
As, we know multiplication of any matrix M to M will give resultant as ${{M}^{2}}$ and similarly, if we multiply ${{M}^{2}}$by M, we will get ${{M}^{3}}$ and so on.
 So, let us multiply the equation (i) by ‘A’ to both sides. Hence, we get
$\begin{align}
  & A.{{A}^{2}}=\left( 2A-I \right)A \\
 & {{A}^{3}}=2A.A-IA \\
 & {{A}^{3}}=2{{A}^{2}}-A \\
\end{align}$
Where, we know multiplication of an identity matrix with any other matrix will give the same matrix.
So, IA = A.
Hence, we get equation after multiplying the equation (i) by ‘A’ as
${{A}^{3}}=2{{A}^{2}}-A$ …………..(ii)
Now, replace ${{A}^{2}}\to $ 2A – I from the equation (i) to the equation (ii).
Hence, we get
$\begin{align}
  & {{A}^{3}}=2\left( 2A-I \right)-A \\
 & {{A}^{3}}=4A-A-2I \\
\end{align}$
${{A}^{3}}=3A-2I$……………..(iii)
Now, let us multiply the above equation by A to both sides. So, we get
$\begin{align}
  & A.{{A}^{3}}=\left( 3A-2I \right)A \\
 & {{A}^{4}}=3{{A}^{2}}-2AI,{{A}^{4}}=3{{A}^{2}}-2A \\
\end{align}$
Now, we can replace ${{A}^{2}}$by 2A – I from the equation (i) to the above equation. So, we get
$\begin{align}
  & {{A}^{4}}=3\left( 2A-I \right)-2A \\
 & {{A}^{4}}=6A-3I-2A \\
 & {{A}^{4}}=6A-2A-3I \\
\end{align}$
${{A}^{4}}=4A-3I$……………….(iv)
Now, similarly, multiply the equation (iv) by A to both sides as well, so, we get’
\[\begin{align}
  & A.{{A}^{4}}=\left( 4A-3I \right)A \\
 & {{A}^{5}}=4{{A}^{2}}-3IA,{{A}^{5}}=4{{A}^{2}}-3A \\
\end{align}\]
Now, replace\[{{A}^{2}}\to 2A-I\]in the above equation from the equation (i). so, we get
$\begin{align}
  & {{A}^{5}}=4\left( 2A-I \right)-3A \\
 & {{A}^{5}}=8A-4I-3A \\
\end{align}$
${{A}^{5}}=5A-4I$……………(v)
Now, we can observe the equation (i), (ii), (iv) and (v) and hence get that the coefficient of A is following the sequence 2, 3, 4, 5…… and coefficient of I in all the equations is following the sequence (-1, -2, -3, -4……..)
It means, coefficient of A for ${{A}^{2}}=2$
Coefficient of A for ${{A}^{3}}=3$
Coefficient of A for \[{{A}^{4}}=4\]
Coefficient of A for ${{A}^{5}}=5$
Similarly, the coefficient of A for ${{A}^{n}}$ should be = n.
And coefficient of I for ${{A}^{2}}=-1=-\left( 2-1 \right)$
Coefficient of I for${{A}^{3}}=-2=-\left( 3-1 \right)$
Coefficient of I for${{A}^{4}}=-3=\left( 4-1 \right)$
Coefficient of I for${{A}^{5}}=-4=-\left( 5-1 \right)$
Hence, coefficient of I for ${{A}^{n}}$ id given as – (n - 1).
Hence, we get the value of ${{A}^{n}}$, with the help of above observations as
${{A}^{n}}=nA-\left( n-1 \right)I$
So, option (a) is the correct answer.

Note: One may go wrong if he/she applies the following approach as
$\begin{align}
  & {{A}^{2}}=2A-I \\
 & {{A}^{2}}-2A+I=0 \\
 & {{\left( A-I \right)}^{2}}=0 \\
\end{align}$
Now, one may take A = I and hence, will get ${{A}^{n}}\to I$, which is wrong. So, take care with this approach.
Observing the pattern of ${{A}^{2}},{{A}^{3}},{{A}^{4}}.........{{A}^{n}}$ was the key point of the solution. One may always use this approach for these kind of questions.