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A = $ \begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix} $ . I wrote X = $ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $. So $X^2$=$ \begin{pmatrix} a^2+bc & ab+bd \\ ac+bc & bc+d^2 \\ \end{pmatrix} $ = $ \begin{pmatrix} 1 & 1 \\ -1 & 1 \\ \end{pmatrix} $

Unfortunatelly I don't know how to continue from here and maybe someone can help be find the matrix X

MAKA PAKA
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  • See https://math.stackexchange.com/questions/59384/find-the-square-root-of-a-matrix – user1551 Nov 19 '17 at 14:22
  • You can use this formula https://en.wikipedia.org/wiki/Square_root_of_a_2_by_2_matrix#One_formula or other methods here https://en.wikipedia.org/wiki/Square_root_of_a_matrix – Raffaele Nov 19 '17 at 17:36

2 Answers2

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Hints:

From top-right and bottom-left elements we get: $$(a+b)d = (a+b)c = -1 \Rightarrow c=-d$$

The top-left and bottom-right elements give us: $$a^2 + bc = bc + d^2 = 1 \Rightarrow a^2 = d^2 \Rightarrow a = \pm d$$

Can you take it from here?

naslundx
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You get:

$a^2 +bc$ = 1

$ac + bc$ = -1

$ab + bd$ = 1

$bc + d^2$ = 1

Hence use substitution to find a, b, c and d.

43zombiegit
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