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Finding Square Root of a Number

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Question

If there are three square matrix A, B, C of same order satisfying the equation $A^{2} = A^{- 1}$ and let $B = A^{2^{n}}$ and $C = A^{2^{(n - 2)}}$ , then $(B - C) = ?, n \in N$

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Solution

$B = A^{2^{n}} = A^{2. 2^{n - 1}}$
$= (A^{2})^{2^{n - 1}}$ $= (A^{- 1})^{2^{n - 1}}$ $= (A^{2^{n - 1}})^{- 1}$
$= {(A^{2. 2^{n - 2}})}^{- 1}$ $= {((A^{2})^{2^{n - 2}})}^{- 1}$
$= {((A^{- 1})^{- 1})}^{2^{n - 2}} = A^{2^{(n - 2)}}$
So, $B = C \Rightarrow (B - C) = O \Rightarrow$ dt. $(B - C) = 0$

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A, B, C

be three square matrices of same order satisfying the equations

A^{2} = A^{- 1},

B = A^{2^{n}}

C = A^{2^{(n - 2)}} .

Then which of the following statements is(are) TRUE ?
(Here,

O

is a null matrix of the same order.)

a_{r} > 0; \forall r, n \in N

a_{1}, a_{2}, a_{3}, . . . . . a_{2 n}

are in A.P, then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}} =

A, B, C

are three square matrices of order

n

A = B + C

C

is a nilpotent matrix with index

2

B

C

A^{10}

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a, a_{1}, a_{2}, a_{3}, a_{4}, . . . . . ., a_{2 n}, b

a, g_{1}, g_{2}, g_{3}, g_{4}, . . . . . . . ., g_{2 n}, b

h

a

b

\frac{a_{1} + a_{2 n}}{g_{1} g_{2 n}} + \frac{a_{2} + a_{2 n - 1}}{g_{2} g_{2 n - 1}} + . . . . + \frac{a_{n} + a_{n + 1}}{g_{n} g_{n + 1}}

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