Let p be a non-singular matrix, and $I+p+{{p}^{2}}+.....+{{p}^{n}}=0$, then find ${{p}^{-1}}$.
A. \[I\]
B. ${{p}^{n+1}}$
C. \[{{p}^{n}}\]
D. \[\left( {{p}^{n+1}}-I \right)\left( p-I \right)\]

Hint: Since p is a non –singular matrix i.e. p is invertible. Multiply the LHS and RHS of the given equation by ${{p}^{-1}}$and simplify the obtained equation by using $I\times p=p\times I=p$. Where ‘I’ is an identity matrix and ‘p’ is only a matrix.

Complete step-by-step solution:
Inverse of a matrix A is defined as ${{A}^{-1}}$ such that they satisfy $A\times {{A}^{-1}}=I$. Where “I” is the identity matrix.
Let us assume the inverse of the given matrix ‘p’ as ${{p}^{-1}}$.
We have,
$I+p+{{p}^{2}}+.....+{{p}^{n}}=0$……………..(1)
Multiplying both sides of equation by ${{p}^{-1}}$, we will get,
$\begin{align}
  & \Rightarrow {{p}^{-1}}\left( I+p+{{p}^{2}}+.....+{{p}^{n}} \right)=0.{{p}^{-1}} \\
 & \Rightarrow I{{p}^{-1}}+{{p}^{-1}}p+{{p}^{-1}}{{p}^{2}}+.....{{p}^{-1}}{{p}^{n}}=0.{{p}^{-1}} \\
\end{align}$
As, we know $I.A=A$, so $I{{p}^{-1}}={{p}^{-1}}$ and for any matrix A, $A.{{A}^{-1}}=I$ if ${{A}^{-1}}$ is defined.
$\Rightarrow {{p}^{-1}}p=I$
We can write the above equation as following by using above properties:
$\begin{align}
  & \Rightarrow {{p}^{-1}}+I+\left( {{p}^{-1}}.p \right)p+....\left( {{p}^{-1}}.p \right){{p}^{n-1}}=0 \\
 & \Rightarrow {{p}^{-1}}+I+Ip+....I{{p}^{n-1}}=0 \\
\end{align}$
Taking ‘I’ common, we will get,
$\begin{align}
  & \Rightarrow {{p}^{-1}}+I\left( I+p+{{p}^{2}}+.....+{{p}^{n-1}} \right)=0 \\
 & \Rightarrow {{p}^{-1}}=-I\left( I+p+{{p}^{2}}+.....+{{p}^{n-1}} \right) \\
\end{align}$
According to equation (1),
$\begin{align}
  & I+p+{{p}^{2}}+.....+{{p}^{n}}=0 \\
 & \Rightarrow I+p+{{p}^{2}}+{{p}^{3}}+.....+{{p}^{n-1}}=-{{p}^{n}} \\
\end{align}$
Replacing $I+p+{{p}^{2}}+{{p}^{3}}+.....+{{p}^{n-1}}=-{{p}^{n}}$in above equation of ${{p}^{-1}}$, we will get,
\[\begin{align}
  & {{p}^{-1}}=-I\left( -{{p}^{n}} \right) \\
 & \Rightarrow {{p}^{-1}}=I{{p}^{n}} \\
\end{align}\]
As for any matrix $IA=A,I{{p}^{n}}={{p}^{n}}$
$\Rightarrow {{p}^{-1}}={{p}^{n}}$
Hence, the required value of ${{p}^{-1}}={{p}^{n}}$ and option (C) is the correct answer.

Note: When we got ${{p}^{-1}}=I+p+{{p}^{2}}+.....+{{p}^{n}}$, students can do mistake by applying GP here but we can’t apply formula of GP here as common ratio = p and ‘p’ is not a number, it is a matrix, calculation won’t be possible. If we apply formula for sum of GP here, we will get,
\[\text{sum of n terms =}\dfrac{a\times {{r}^{n-1}}}{r-1}\]
Here a = I, r = p
So, we will get $I+p+{{p}^{2}}+.....+{{p}^{n}}=\dfrac{I\times {{p}^{n-1}}}{p-1}$.
In the denominator, we are getting $p – 1$, ‘p’ is a matrix and ‘1’ is an integer and so we can’t calculate $p – 1.$