Linear Systems in Matlab

Solving Linear Systems of Equations in MATLAB

$[A](x)=(b)$

See the discussion of linear algebra for help on setting up a linear system of equations. 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;

For example, if we wanted to solve the system of equations

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

we would first rewrite these in a matrix-vector form as

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

This is implemented in MATLAB as

  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.

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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