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>
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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}

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.