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Question: Let p be a non-singular matrix, and \(I+p+{{p}^{2}}+.....+{{p}^{n}}=0\), then find \({{p}^{-1}}\). ...

Let p be a non-singular matrix, and I+p+p2+.....+pn=0I+p+{{p}^{2}}+.....+{{p}^{n}}=0, then find p−1{{p}^{-1}}.
A. II
B. pn+1{{p}^{n+1}}
C. pn{{p}^{n}}
D. (pn+1−I)(p−I)\left( {{p}^{n+1}}-I \right)\left( p-I \right)

Explanation

Solution

Since p is a non –singular matrix i.e. p is invertible. Multiply the LHS and RHS of the given equation by p−1{{p}^{-1}}and simplify the obtained equation by using I×p=p×I=pI\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{{A}^{-1}} such that they satisfy A×A−1=IA\times {{A}^{-1}}=I. Where “I” is the identity matrix.
Let us assume the inverse of the given matrix ‘p’ as p−1{{p}^{-1}}.
We have,
I+p+p2+.....+pn=0I+p+{{p}^{2}}+.....+{{p}^{n}}=0……………..(1)
Multiplying both sides of equation by p−1{{p}^{-1}}, we will get,
⇒p−1(I+p+p2+.....+pn)=0.p−1 ⇒Ip−1+p−1p+p−1p2+.....p−1pn=0.p−1 \begin{aligned} & \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{aligned}
As, we know I.A=AI.A=A, so Ip−1=p−1I{{p}^{-1}}={{p}^{-1}} and for any matrix A, A.A−1=IA.{{A}^{-1}}=I if A−1{{A}^{-1}} is defined.
⇒p−1p=I\Rightarrow {{p}^{-1}}p=I
We can write the above equation as following by using above properties:
⇒p−1+I+(p−1.p)p+....(p−1.p)pn−1=0 ⇒p−1+I+Ip+....Ipn−1=0 \begin{aligned} & \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{aligned}
Taking ‘I’ common, we will get,
⇒p−1+I(I+p+p2+.....+pn−1)=0 ⇒p−1=−I(I+p+p2+.....+pn−1) \begin{aligned} & \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{aligned}
According to equation (1),
I+p+p2+.....+pn=0 ⇒I+p+p2+p3+.....+pn−1=−pn \begin{aligned} & I+p+{{p}^{2}}+.....+{{p}^{n}}=0 \\\ & \Rightarrow I+p+{{p}^{2}}+{{p}^{3}}+.....+{{p}^{n-1}}=-{{p}^{n}} \\\ \end{aligned}
Replacing I+p+p2+p3+.....+pn−1=−pnI+p+{{p}^{2}}+{{p}^{3}}+.....+{{p}^{n-1}}=-{{p}^{n}}in above equation of p−1{{p}^{-1}}, we will get,

& {{p}^{-1}}=-I\left( -{{p}^{n}} \right) \\\ & \Rightarrow {{p}^{-1}}=I{{p}^{n}} \\\ \end{aligned}$$ As for any matrix $IA=A,I{{p}^{n}}={{p}^{n}}$ $\Rightarrow {{p}^{-1}}={{p}^{n}}$ **Hence, the required value of ${{p}^{-1}}={{p}^{n}}$ and option (C) is the correct answer.** **Note:** When we got ${{p}^{-1}}=I+p+{{p}^{2}}+.....+{{p}^{n}}$, students can do mistake by applying GP here but we can’t apply formula of GP here as common ratio = p and ‘p’ is not a number, it is a matrix, calculation won’t be possible. If we apply formula for sum of GP here, we will get, $$\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 $p – 1.$