Let $A$ be a symmetric matrix such that $A^{5} = 0$ and $B = I + A + A^{2} + A^{3} + A^{4}$ , then $B$ is

symmetric

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singular

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non-singular

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skew symmetric

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Solution

The correct options are
A symmetric
C non-singular
Given $A$ be a symmetric matrix such that $A^{5} = 0$ and $B = I + A + A^{2} + A^{3} + A^{4}$
$B^{- 1} = (I + A + A^{2} + A^{3} + A^{4})^{- 1} = I^{'} + A^{- 1} + (A^{- 1})^{2} + (A^{- 1})^{3} + (A^{- 1})^{4} = I + A + A^{2} + A^{3} + A^{4} = B$
$∴ B$ is symmetric.
$B A = (I + A + A^{2} + A^{3} + A^{4}) A = A + A^{2} + A^{3} + A^{4} + A^{5} = B - I$
$\Rightarrow B A = B - I$
$\Rightarrow B (I - A) = I \Rightarrow B^{- 1} = I - A$
$B^{- 1}$ exists.
$∴ B$ is non-singualr.
Hence, options A and C.

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Let A be a symmetric matrix such that A5=0 and B=I+A+A2+A3+A4, then B is

Let $A$ be a symmetric matrix such that $A^{5} = 0$ and $B = I + A + A^{2} + A^{3} + A^{4}$ , then $B$ is