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数学

Discovery

Andrius Kulikauskas

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Introduction E9F5FC

Understandable FFFFFF

Questions FFFFC0

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Book.Exercise2 istorija

Paslėpti nežymius pakeitimus - Rodyti galutinio teksto pakeitimus

2019 sausio 19 d., 19:49 atliko AndriusKulikauskas -
Pakeista 38 eilutė iš:
Where we can have any {$a$} or {$b$} such that {$ab \leq 1$}.
į:
Note that {$a$} and {$b$} can be any complex number. However, if we want a real matrix, then we must have {$a$} and {$b$} real such that {$ab \leq 1$}.
2019 sausio 19 d., 19:46 atliko AndriusKulikauskas -
Pakeista 38 eilutė iš:
Where we can have any real {$a$} or {$b$} such that {$ab \leq 1$}.
į:
Where we can have any {$a$} or {$b$} such that {$ab \leq 1$}.
2019 sausio 19 d., 19:45 atliko AndriusKulikauskas -
Pakeista 34 eilutė iš:
{$ \begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix} $}
į:
{$ \begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix} $} or
2018 lapkričio 18 d., 19:56 atliko AndriusKulikauskas -
Pakeistos 31-32 eilutės iš
But the first case is simply the special case when a or b = 0. Thus the answer is:
į:
But the first case and the second case match when a or b = 0. Thus the answer is:

{$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} $} or
{$ \begin{pmatrix}-1 & 0\\ 0 & -1\end{pmatrix} $}
2018 lapkričio 18 d., 19:54 atliko AndriusKulikauskas -
Pakeistos 17-39 eilutės iš
{$$ a_{12}a_{22} + a_{11}a_{12} = 0 $$} which means {$$ a_{22} = -a_{11} $$}.
į:
{$$ a_{12}a_{22} + a_{11}a_{12} = 0 $$}

The latter equation
means {$ a_{12}=0$} or {$ a_{22} = -a_{11} $}.

Similarly, by symmetry, {$ a_{21}=0$} or {$ a_{22} = -a_{11} $}.

Combining, we have ({$ a_{12}=0$} and {$ a_{21}=0$}) or {$ a_{22} = -a_{11} $}.

Solving further, this yields the following two possibilities:

{$ \begin{pmatrix}\pm1 & 0\\ 0 & \pm1\end{pmatrix} $} or
{$ \begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix} $} or
{$ \begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix} $}

But the first case is simply the special case when a or b = 0. Thus the answer is:

{$ \begin{pmatrix}\sqrt{1-ab} & a\\ b & -\sqrt{1-ab}\end{pmatrix} $} or
{$ \begin{pmatrix}-\sqrt{1-ab} & a\\ b & \sqrt{1-ab}\end{pmatrix} $}

Where we can have any real {$a$} or {$b$} such that {$ab \leq 1$}.

2018 lapkričio 18 d., 19:39 atliko AndriusKulikauskas -
Pakeistos 11-17 eilutės iš
{$$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$$}
į:
{$$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$$}

This yields equations of the form:

{$$ {a_{11}}^2 + a_{12}a_{21} = 1 $$}

{$$ a_{12}a_{22} + a_{11}a_{12} = 0 $$} which means {$$ a_{22} = -a_{11} $$}.
2018 lapkričio 18 d., 19:29 atliko AndriusKulikauskas -
Pridėtos 1-11 eilutės:
>>bgcolor=#E9F5FC<<
------------------------
[[Math exercises]]

'''Square root of identity matrix.'''
-------------------------
>><<

Consider the {$2x2$} identity matrix. What is its square root?

{$$ \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}\begin{pmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{pmatrix}$$}

Exercise2


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