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## Book.Exercise2 istorija

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

Naujausi pakeitimai

 Puslapis paskutinį kartą pakeistas 2019 sausio 19 d., 19:49