# Let p be a non singular matrix 1+p+p^2+p^3+cdots+p^n=O (O denotes the null matrix), then p^-1 is?

Oct 25, 2017

The answer is $= - \left(I + p + \ldots \ldots \ldots {p}^{n - 1}\right)$

#### Explanation:

We know that

${p}^{-} 1 p = I$

$I + p + {p}^{2} + {p}^{3.} \ldots . {p}^{n} = O$

Multiply both sides by ${p}^{-} 1$

${p}^{-} 1 \cdot \left(1 + p + {p}^{2} + {p}^{3.} \ldots . {p}^{n}\right) = {p}^{-} 1 \cdot O$

${p}^{-} 1 \cdot 1 + {p}^{-} 1 \cdot p + {p}^{-} 1 \cdot {p}^{2} + \ldots \ldots {p}^{-} 1 \cdot {p}^{n} = O$

${p}^{-} 1 + \left({p}^{-} 1 p\right) + \left({p}^{-} 1 \cdot p \cdot p\right) + \ldots \ldots \ldots \left({p}^{-} 1 p \cdot {p}^{n - 1}\right) = O$

${p}^{-} 1 + \left(I\right) + \left(I \cdot p\right) + \ldots \ldots \ldots \left(I \cdot {p}^{n - 1}\right) = O$

Therefore,

${p}^{-} 1 = - \left(I + p + \ldots \ldots \ldots {p}^{n - 1}\right)$

Oct 25, 2017

See below.

#### Explanation:

$p \left({p}^{-} 1 + p + {p}^{2} + \cdots + {p}^{n - 1}\right) = 0$ but $p$ by hypothesis is non singular then exists ${p}^{-} 1$ so

${p}^{-} 1 p \left({p}^{-} 1 + p + {p}^{2} + \cdots + {p}^{n - 1}\right) = {p}^{-} 1 + p + {p}^{2} + \cdots + {p}^{n - 1} = 0$

and finally

${p}^{-} 1 = - {\sum}_{k = 1}^{n - 1} {p}^{k}$

Also can be solved as

${p}^{-} 1 = - p \left({\sum}_{k = 0}^{n - 2} {p}^{k}\right) = p \left({p}^{n - 1} + {p}^{n}\right) = {p}^{n} \left(1 - p\right)$