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.

  • 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

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 $\vec{b}$.


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