Let-A-a-b-c-d-Find-a-condition-on-a-b-c-d-so-that-A-n-1-A-n-nA-n-N-A-n-1-A-n-nA-A-n-A-n-1-n-1-A-A-n-1-A-n-2-n-2-A-A-n-2-A-n-3-n-3-A-A-4-A-3-3A-A-3-A-2-2A

Question Number 3566 by Yozzii last updated on 15/Dec/15

L e t A = (\begin{matrix} a & b \\ c & d \end{matrix}) . F i n d a c o n d i t i o n o n

a, b, c, d s o t h a t A^{n + 1} - A^{n} = n A, n \in N .

A^{n + 1} - A^{n} = n A

A^{n} - A^{n - 1} = (n - 1) A

A^{n - 1} - A^{n - 2} = (n - 2) A

A^{n - 2} - A^{n - 3} = (n - 3) A

\dots

A^{4} - A^{3} = 3 A

A^{3} - A^{2} = 2 A

A^{2} - A = A

S o, A^{n + 1} - A = (n + n - 1 + n - 2 + \dots + 3 + 2 + 1) A

A^{n + 1} = A + \frac{n (n + 1)}{2} A

A^{n + 1} = \frac{n^{2} + n + 2}{2} A

A^{n} = \frac{{(n - 1)}^{2} + n + 1}{2} A

A^{n} = \frac{n^{2} - n + 2}{2} A (n ⩾ 1), A = (\begin{matrix} a & b \\ c & d \end{matrix}) .

Commented by prakash jain last updated on 15/Dec/15

A^{2} = (\begin{matrix} a^{2} + b c & a b + b d \\ a c + c d & b c + d^{2} \end{matrix}) = (\begin{matrix} 2 a & 2 b \\ 2 c & 2 d \end{matrix})

a^{2} + b c = 2 a (1)

a b + b d = 2 b (2)

a c + c d = 2 c (3)

b c + d^{2} = 2 d (4)

4 v a r i a b l e s a n d 4 e q u a t i o n

f r o m (1) a n d (4)

a^{2} - d^{2} = 2 (a - d) \Rightarrow a = d o r a + d = 2

c a s e 1 a = d

f r o m (2) d b + b d = 2 b \Rightarrow b = 0 o r d = 1

1 a . c a s e b = 0

f r o m (4) d^{2} = 2 d \Rightarrow d = 0 o r d = 2

1 a a \cdot case d = 0

a = b = c = d = 0 \dots . (A)

1 a b . c s s e d = 2

a = d = 2

a c + d c = 2 c \Rightarrow 2 \sqrt{2} c = 2 c \Rightarrow c = 0

a = d = 2, b = c = 0 (B)

1 b . c a s e d = 1

a = d = 1, b c = 1 \dots (C)

c a s e 2 : a + d = 2

b, c c a n t a k e a n y v a l e (D)

To satiafy A^{2} = 2 A solution are

(S ol A) a = b = c = d = 0 t h is satisifies genral condtion

(Sol B) a = d = 2, b = c = 0

t o b e c h e c k e d f o r h i g h e r p o w e r s .

(Sol C) a = d = 1, c b = 1

t o b e c h e c k e d f o r h i g h e r p o w e r s .

(Sol D) b, c a, d = 1 - a

t o b e c h e c k e d f o r h i g h e r p o w e r s .

Continue

Answered by prakash jain last updated on 15/Dec/15

A = (\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}) A^{2} = (\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}) A^{n} = (\begin{matrix} 2^{n} & 0 \\ 0 & 2^{n} \end{matrix})

d o e s n o t s a t i s f y g e n e r a l c o m d i t i o n .

s o

a = b = c = d = 0 i s s o l u t i o n

a = d = 2, b = d = 0 i s n o t s o l u t i o n .

S ol (C) and (D) to be checked .

Answered by prakash jain last updated on 15/Dec/15

Actually there is an easier way .

A^{2} = 2 A

A^{3} - A^{2} = 2 A

A^{3} = 4 A

A^{4} - A^{3} = 3 A

A^{3} \cdot A - 4 A = 3 A

4 A^{2} - 4 A = 3 A

8 A - 4 A = 3 A

\Rightarrow a = b = c = d = 0