MatrixSolutions - Revision history

Ccn admin: 6 revisions

2014-01-16T03:43:42Z

6 revisions

← Older revision	Revision as of 03:43, 16 January 2014
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Mscohen at 01:15, 13 January 2009

2009-01-13T01:15:08Z

← Older revision		Revision as of 01:15, 13 January 2009
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	Of course, you can use Cramer's rule if you remember it, or some other solving algorithm. I think you can get a matrix solution app for your iPhone, as well.		Of course, you can use Cramer's rule if you remember it, or some other solving algorithm. I think you can get a matrix solution app for your iPhone, as well.

			We will use this general concept a lot in ''Principles of Neuroimaging.'' In particular, when we imagine the image result of an experiment to be the sum of a variety of influences (experimental and otherwise), we will use matrix form to evaluate the strength of each of these influences in creating our experimental result.

Mscohen at 01:11, 13 January 2009

2009-01-13T01:11:13Z

← Older revision		Revision as of 01:11, 13 January 2009
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	<math>\mathbf{X}=\left[\begin{array}{cc}		<math>\mathbf{X}=\left[\begin{array}{cc}
	1 & 4\\		1 & 4\\
	1 & 2\end{array}\right]^{-1}</math>		-1 & 2\end{array}\right]^{-1}</math>

	I think that the easiest solution is to solve as a set of simultaneous equations. If <math>\mathbf{X} = \left[\begin{array}{cc}		I think that the easiest solution is to solve as a set of simultaneous equations. If <math>\mathbf{X} = \left[\begin{array}{cc}
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	<math>\left[\begin{array}{cc}		<math>\left[\begin{array}{cc}
	1 & 4\\		1 & 4\\
	1 & 2\end{array}\right] \left[\begin{array}{cc}		-1 & 2\end{array}\right] \left[\begin{array}{cc}
	a & b\\		a & b\\
	c & d\end{array}\right] = \left[\begin{array}{cc}		c & d\end{array}\right] = \left[\begin{array}{cc}

Mscohen at 01:04, 13 January 2009

2009-01-13T01:04:18Z

← Older revision		Revision as of 01:04, 13 January 2009
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	For the problem posed in the quiz:		For the problem posed in the quiz:

	find <math>\mathbf{X}=\left[\begin{array}{cc}		find

			<math>\mathbf{X}=\left[\begin{array}{cc}
	1 & 4\\		1 & 4\\
	1 & 2\end{array}\right]^{-1}</math>		1 & 2\end{array}\right]^{-1}</math>
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	I think that the easiest solution is to solve as a set of simultaneous equations. If <math>\mathbf{X} = \left[\begin{array}{cc}		I think that the easiest solution is to solve as a set of simultaneous equations. If <math>\mathbf{X} = \left[\begin{array}{cc}
	a & b\\		a & b\\
	c & d\end{array}\right]</math> ~~is the inverse~~,		c & d\end{array}\right]</math>,

	<math>\left[\begin{array}{cc}		<math>\left[\begin{array}{cc}

Mscohen at 01:03, 13 January 2009

2009-01-13T01:03:38Z

← Older revision		Revision as of 01:03, 13 January 2009
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	The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf{X}^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I'''. '''I''', in turn, is the matrix which, when multiplied by any matrix '''B''' just gives ~~bavk~~ '''B'''.		The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf{X}^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I'''. '''I''', in turn, is the matrix which, when multiplied by any matrix '''B''' just gives back '''B'''.

	For the problem posed in the quiz:		For the problem posed in the quiz:

	find <math>\left[\begin{array}{cc}		find <math>\mathbf{X}=\left[\begin{array}{cc}
	1 & 4\\		1 & 4\\
	1 & 2\end{array}\right]^{-1}</math>		1 & 2\end{array}\right]^{-1}</math>

Mscohen at 01:02, 13 January 2009

2009-01-13T01:02:32Z

← Older revision		Revision as of 01:02, 13 January 2009
Line 1:		Line 1:
	The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I'''. '''I''', in turn, is the matrix which, when multiplied by any matrix '''B''' just gives bavk '''B'''.		The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf{X}^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I'''. '''I''', in turn, is the matrix which, when multiplied by any matrix '''B''' just gives bavk '''B'''.

	For the problem posed in the quiz:		For the problem posed in the quiz:
Line 13:		Line 13:
	<math>\left[\begin{array}{cc}		<math>\left[\begin{array}{cc}
	1 & 4\\		1 & 4\\
	1 & 2\end{array}\right] * \~~mathbf~~{X} = \left[\begin{array}{cc}		1 & 2\end{array}\right] \left[\begin{array}{cc}
			a & b\\
			c & d\end{array}\right] = \left[\begin{array}{cc}
	1 & 0\\		1 & 0\\
	0 & 1\end{array}\right]</math>		0 & 1\end{array}\right]</math>

	Thus, the rules of matrix multiplication give us:		Thus, the rules of matrix multiplication give us this set of four equations:
	<math>a+4c=1</math><br>
	<math>b+4d=0-</math><br>		:<math>a+4c=1</math><br>
	<math>-a+2c=0</math><br>		:<math>b+4d=0</math><br>
	<math>-b+2d=1</math><br>		:<math>-a+2c=0</math><br>
			:<math>-b+2d=1</math><br>

	which is very easy to solve by substitution:		which is very easy to solve by substitution:
	<math>b = -4d</math><br>
	<math>a = 2c</math><br>		:<math>b = -4d</math><br>
	<math>2c+4c=1</math><br>		:<math>a = 2c</math><br>
	<math>c = 1/6</math><br>		:<math>2c+4c=1</math><br>
	''etc,...''		:<math>c = 1/6</math><br>
			::''etc,...''

			Of course, you can use Cramer's rule if you remember it, or some other solving algorithm. I think you can get a matrix solution app for your iPhone, as well.

Mscohen: New page: The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, $\mathbf^{-1}$ is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I...

2009-01-13T00:57:21Z

New page: The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I...

New page

The general idea of a matrix inverse is that if '''A''' is a matrix, its inverse, <math>\mathbf^{-1}</math> is the matrix which, when multiplied by '''A''' yields the identity matrix, '''I'''. '''I''', in turn, is the matrix which, when multiplied by any matrix '''B''' just gives bavk '''B'''.

For the problem posed in the quiz:

find <math>\left[\begin{array}{cc}
1 & 4\\
1 & 2\end{array}\right]^{-1}</math>

I think that the easiest solution is to solve as a set of simultaneous equations. If <math>\mathbf{X} = \left[\begin{array}{cc}
a & b\\
c & d\end{array}\right]</math> is the inverse,

<math>\left[\begin{array}{cc}
1 & 4\\
1 & 2\end{array}\right] * \mathbf{X} = \left[\begin{array}{cc}
1 & 0\\
0 & 1\end{array}\right]</math>

Thus, the rules of matrix multiplication give us:
<math>a+4c=1</math> 
<math>b+4d=0-</math> 
<math>-a+2c=0</math> 
<math>-b+2d=1</math> 

which is very easy to solve by substitution:
<math>b = -4d</math> 
<math>a = 2c</math> 
<math>2c+4c=1</math> 
<math>c = 1/6</math> 
''etc,...''