MatrixSolutions: Difference between revisions

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Line 7: Line 7:
<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}
Line 15: Line 15:
<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}

Revision as of 01:11, 13 January 2009

The general idea of a matrix inverse is that if A is a matrix, its inverse, 𝐗1 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

𝐗=[1412]1

I think that the easiest solution is to solve as a set of simultaneous equations. If 𝐗=[abcd],

[1412][abcd]=[1001]

Thus, the rules of matrix multiplication give us this set of four equations:

a+4c=1
b+4d=0
a+2c=0
b+2d=1

which is very easy to solve by substitution:

b=4d
a=2c
2c+4c=1
c=1/6
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.