Azzera filtri
Azzera filtri

For loop for system of equations

1 visualizzazione (ultimi 30 giorni)
Vincent Gambuzza
Vincent Gambuzza il 10 Feb 2020
Risposto: Jyothis Gireesh il 13 Feb 2020
I need to solve this system of equations with a for loop so that the dimensions match. I would like to loop through the “b” matrix
  1 Commento
darova
darova il 10 Feb 2020
Can you please show your attempts? What have you tried?

Accedi per commentare.

Risposte (1)

Jyothis Gireesh
Jyothis Gireesh il 13 Feb 2020
The methodology outlined in the image attached may not be the most suitable one for this case. The recommended way to solve the system of equations given above may be to use the “mldivide” operator in MATLAB. So, the matrix may be calculated as
x = A\b;
But using the above methods yields the following warning message
‘Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 2.202823e-18’.
One possible work-around in this case is to use the pseudo-inverse of A. The following code may be used to compute “x”
x = pinv(A)*b;
which gives
x = -1.0556 -2.5556 -4.0556
0.1111 0.1111 0.1111
1.2778 2.7778 4.2778
Please refer to the following documentation link on “pinv” to get further clarification on calculating pseudo-inverse.

Categorie

Scopri di più su Loops and Conditional Statements in Help Center e File Exchange

Tag

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by