(4.1) | |||

(4.2) | |||

(4.3) |

In matrix notation, we can rewrite the problem as as follows.

The solution is made by taking the inverse of to give

Linear systems like this are easy to solve in MATLAB using the left division operator,

`\`

. Do this example.
% Solving a linear system A = [3 2 -1 -1 3 2 1 -1 -1] y = [10 5 -1]' % Note the use of the traspose operator! x = A\y % x should now contain the solution % if that's true A*x should give back 'y' A*xMATLAB gives which we can verify is the correct answer by trying and seeing that it yields the original . Note that in this example we used the matrix multiplication operator

`*`

(not `.*`

). This
time it worked because the number of columns in is equal to the number
of rows in .
An alternative way to solve the problem is to use the inverse of
directly, as shown below.
% Solving a linear system using the inverse of A x = inv(A)*y % x should now contain the solution % If that's true A*x should give back 'y' A*xThis approach may seem more straightforward but it requires more CPU time since the inverse of must be computed, whereas in the first approach MATLAB doesn't have to find the inverse directly. While CPU time is not an issue for our small examples, it can become the most important issue for large problems. Several special functions also exist for finding the dot product, cross product and norm of vectors. Doing the example below should be sufficiently instructive.

% Dot, cross, and all that v = [1 0 -1] w = [2 1 1] dot(v,w) cross(v,w) cross([1 0 0], [0 1 0]) % i.e., i x j = k dot(v,v) norm(v) % length of v, i.e, sqrt(v.v)