Difference between revisions of "Linear Systems in Matlab"

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m (Solving Linear Systems of Equations in MATLAB)
(Example)
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=== Example ===
 
=== Example ===
If we wanted to solve the system of equations
+
 
 +
Consider the following set of equations:
 
<center><math>
 
<center><math>
 
\begin{cases}
 
\begin{cases}
 
   6y = -5x \\
 
   6y = -5x \\
 
   y = -3x -5
 
   y = -3x -5
\end{cases}
+
\end{cases}.
 
</math></center>
 
</math></center>
we would first rewrite these in a matrix-vector form as
+
These can be easily solved by hand to obtain
 +
<math>(x,y) = \left( \tfrac{30}{13}, \tfrac{25}{13} \right) </math>.
 +
 
 +
To solve the system of equations using MATLAB, first rewrite these in a matrix-vector form as
 
<center><math>
 
<center><math>
 
  \begin{align}
 
  \begin{align}
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  \left( \begin{array}{c} 0 \\ -5 \end{array} \right) .
 
  \left( \begin{array}{c} 0 \\ -5 \end{array} \right) .
 
</math></center>
 
</math></center>
We could then easily solve this system in MATLAB by using the following commands:
+
Once in matrix-vector form, the solution is obtained in MATLAB by using the following commands:
 
<source lang="matlab">
 
<source lang="matlab">
 
   A = [ 5 6; 3 1 ];  % define the matrix
 
   A = [ 5 6; 3 1 ];  % define the matrix
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   solution = A\b;    % solve the system of equations.
 
   solution = A\b;    % solve the system of equations.
 
</source>
 
</source>
In this example, <tt>solution</tt> is a column vector whose elements are <tt>x</tt> and <tt>y</tt>.
+
In this example, <tt>solution</tt> is a column vector whose elements are <tt>x</tt> and <tt>y</tt>:
 +
<center><math>
 +
\mathrm{solution} = \left[ \begin{array}{c} -2.3077 \\1.9231 \end{array} \right],
 +
</math></center>
 +
which is consistent with the answer we obtained by hand above.  The figure shows plots of the two equations, along with a circle indicating the solution (where the lines intersect).  [[image:linsys_example.png|right|400px]]
 +
 
 +
 
 +
==== Matrix Inverse ====
  
 
Note that we can also form the inverse of a matrix,
 
Note that we can also form the inverse of a matrix,
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   Ainv = inv(A);  % entirely equivalent to A^-1.
 
   Ainv = inv(A);  % entirely equivalent to A^-1.
 
</source>
 
</source>
 
  
 
=== Sparse Systems ===
 
=== Sparse Systems ===

Revision as of 08:46, 25 August 2008


Solving Linear Systems of Equations in MATLAB

This section discusses how to solve a set of linear equations [A](x)=(b) in MATLAB. See the discussion of linear algebra for help on writing a linear system of equations in matrix-vector format. There is also help on creating matrices and vectors in MATLAB.

The simplest way of solving a system of equations in MATLAB is by using the \ operator. Given a matrix A and a vector b, we may solve the system using the following MATLAB commands

   x = A\b;

Example

Consider the following set of equations:


\begin{cases}
  6y = -5x \\
  y = -3x -5
\end{cases}.

These can be easily solved by hand to obtain (x,y) = \left( \tfrac{30}{13}, \tfrac{25}{13} \right) .

To solve the system of equations using MATLAB, first rewrite these in a matrix-vector form as


 \begin{align}
   5x + 6y = 0 \\
   3x + 1y &= -5 \\
 \end{align}
 \quad \Leftrightarrow \quad
 \left[ \begin{array}{cc} 5 & 6 \\ 3 & 1 \end{array} \right]
 \left( \begin{array}{c} x \\ y \end{array} \right) =
 \left( \begin{array}{c} 0 \\ -5 \end{array} \right) .

Once in matrix-vector form, the solution is obtained in MATLAB by using the following commands:

  A = [ 5 6; 3 1 ];   % define the matrix
  b = [ 0; -5 ];      % define the vector
  solution = A\b;     % solve the system of equations.

In this example, solution is a column vector whose elements are x and y:


 \mathrm{solution} = \left[ \begin{array}{c} -2.3077 \\1.9231 \end{array} \right],
which is consistent with the answer we obtained by hand above. The figure shows plots of the two equations, along with a circle indicating the solution (where the lines intersect).
Linsys example.png


Matrix Inverse

Note that we can also form the inverse of a matrix,


  [A] (x)=(b) \quad \Leftrightarrow \quad (x)=[A]^{-1}(b).

This can be done in MATLAB as illustrated by the following:

  A = [ 5 6; 3 1 ];
  b = [ 0; -5 ];
  Ainv = A^-1;       % calculate the inverse of A
  solution = Ainv*b; % calculate the solution

We could also calculate A-1 by

  Ainv = inv(A);   % entirely equivalent to A^-1.

Sparse Systems

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Linear Systems using the Symbolic Toolbox

Occasionally we may want to find the symbolic (general) solution to a system of equations rather than a specific numerical solution. The symbolic toolbox provides a way to do this.

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