MatrixSolutions: Difference between revisions
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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> | ||
Line 9: | Line 11: | ||
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} |
Revision as of 01:04, 13 January 2009
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.