0
$\begingroup$

To solve the system of equations:

$2x-3y-4z = 2$

$-z = 5$

$x-2y+z = 3$

you could use matrices, the two primary methods I have come across involve using the inverse (method 1) or just using Gaussian elimination straight away on an augmented matrix (method 2). I have attached the two methods below:

method 1

method 2

My question is whether there is any advantage of using method 1, as you have to carry out the same process as method 2 of Gaussian elimination, but then also have to multiply matrices afterwards, meaning method 2 is surely always quicker? Thanks.

$\endgroup$
  • 1
    $\begingroup$ Personal opinion: Gaussian elimination is better. Fewer intermediate steps so less likely to go wrong. $\endgroup$ – Sonal_sqrt Aug 30 '19 at 10:49
  • $\begingroup$ since $z$ is given, you can easily substitute and solve 2 equations $\endgroup$ – farruhota Aug 30 '19 at 11:05
  • $\begingroup$ @farruhota yeah, I was just using the example to specifically learn the matrix method, thanks for pointing it out though! $\endgroup$ – Jamminermit Aug 30 '19 at 11:39
  • $\begingroup$ There will be no inverse if the system has an infinity or no solutions. $\endgroup$ – Bernard Massé Aug 30 '19 at 11:45
  • $\begingroup$ There are other ways of computing the inverse besides using Gaussian elimination, which is what you’re doing. $\endgroup$ – amd Aug 30 '19 at 20:23
1
$\begingroup$

If all you want is to solve $A \vec{x} = \vec{b}$, where $A$ is an invertible matrix, by either

1) Determining $A^{-1}$ using Gaussian elimination on $A$ augmented with the identity matrix, then computing $A^{-1} \vec{b}$, or

2) Using Gaussian elimination on $A$ augmented with $\vec{b}$,

then yes. Method two is the more straightforward method. However, I can still think of several reasons for teaching and learning method one.

Probably the main benefit of method one is that you have $A^{-1}$ to work with at the end of your problem. Suppose your teacher gives you several problems with the same matrix but different $\vec{b}$ vectors. Using method one you only have to use Gaussian elimination once, to find $A^{-1}$. After that all you have to do is compute $A^{-1} \vec{b}$ for each different $\vec{b}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.