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Question: P is a nonsingular matrix and \(I+p+{{p}^{2}}+........{{p}^{n}}=0\) then \({{p}^{-1}}\) is (a) \(...

P is a nonsingular matrix and I+p+p2+........pn=0I+p+{{p}^{2}}+........{{p}^{n}}=0 then p−1{{p}^{-1}} is
(a) pn{{p}^{n}}
(b) pn−1{{p}^{n-1}}
(c) pn−1{{p}^{n-1}}
(d) −pn−1-{{p}^{n-1}}

Explanation

Solution

Hint: Here, we can apply the property of a matrix that a matrix multiplied by its inverse is given as an identity matrix. Multiplying the equation with p−1{{p}^{-1}}, it gives an identity matrix and further simplification we get the required answer.

Complete step-by-step answer:
Since, we have been given that p is a non-singular matrix. An n×nn\times n matrix A is called non singular if the determinant of the matrix is not equal to 0. Non- singularity is the first condition of a matrix to be invertible and the next condition for a matrix to be invertible is that there exists an n×nn\times n matrix B such that AB=BA=II. So, a matrix B such that AB=BA=II is called an inverse of matrix A. There can only be one inverse for a given matrix.
Since, it is given that:
I+p+p2+........pn=0...........(1)I+p+{{p}^{2}}+........{{p}^{n}}=0...........\left( 1 \right)
Since the matrix p is non-singular, its inverse exists, that is there is a matrix p−1{{p}^{-1}} such that p×p−1=1p\times {{p}^{-1}}=1.
On multiplying p−1{{p}^{-1}} on both sides of the equation, we get:
p−1(I+p+p2+........pn)=p−10 ⇒p−1I+p−1.p+p−2.p+......+p−1pn=0 ⇒p−1+I+p+p2+p3+.......+pn−1=0..........(2) \begin{aligned} & {{p}^{-1}}\left( I+p+{{p}^{2}}+........{{p}^{n}} \right)={{p}^{-1}}0 \\\ & \Rightarrow {{p}^{-1}}I+{{p}^{-1}}.p+{{p}^{-2}}.p+......+{{p}^{-1}}{{p}^{n}}=0 \\\ & \Rightarrow {{p}^{-1}}+I+p+{{p}^{2}}+{{p}^{3}}+.......+{{p}^{n-1}}=0..........\left( 2 \right) \\\ \end{aligned}
Now, from equation (1), we can write:
I+p+p2+........pn−1=−pnI+p+{{p}^{2}}+........{{p}^{n-1}}=-{{p}^{n}}
Using this in equation (2), we can write:
p−1−pn=0 ⇒p−1=pn \begin{aligned} & {{p}^{-1}}-{{p}^{n}}=0 \\\ & \Rightarrow {{p}^{-1}}={{p}^{n}} \\\ \end{aligned}

So, the value of p−1{{p}^{-1}} comes out to be pn{{p}^{n}}.
Hence, option (a) is the correct answer.

Note: Students can make a mistake of considering the given expression as a sum of G.P with first term as I and common ratio as p. It is wrong because geometric progressions exist for numbers and not for matrices. Also, in the formula for sum of G.P, in the denominator we have r - 1, where r is the common ratio. If we consider the given expression as a G.P, then the common ratio becomes the matrix p, and the subtraction between a matrix and a scalar does not exist. So, such mistakes should be avoided.