MatrixSolutions: Difference between revisions
No edit summary |
m (6 revisions) |
||
(3 intermediate revisions by one other user not shown) | |||
Line 3: | Line 3: | ||
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> | ||
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> | c & d\end{array}\right]</math>, | ||
<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} | ||
Line 35: | Line 37: | ||
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. |
Latest revision as of 03:43, 16 January 2014
The general idea of a matrix inverse is that if A is a matrix, its inverse, 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:
find
I think that the easiest solution is to solve as a set of simultaneous equations. If ,
Thus, the rules of matrix multiplication give us this set of four equations:
which is very easy to solve by substitution:
- 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.
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.