0
$\begingroup$

For example If I was doing row reduction to an augmented matrix to find a solution or something, I could just plug the unknowns into one of the smallest equation of the linear system to see if its correct (99.9% this would work, well has worked for me thus far). For inverse how would I check? (This is for test purposes so like it doesn't take up too much time).

  • 0
    just multiply your original matrix with your inverted matrix, the standard basis matrix should come out – 2017-02-18
  • 0
    Multiply your test-inverse with your original matrix. This should give the identity matrix. Check it for more and more of the rows for increasing level of certainty. – 2017-02-18
  • 0
    That takes too much time if my numbers are like 2/14 throughout the entire inverse. Wait could I multiply my original matrix with the inverse and see if the first one or two numbers are the same ? (If they are the same most likely the rest are right too, right?) – 2017-02-18
  • 0
    Multiply your inverse by the original matrix and see if the result is an identity matrix. – 2017-02-18

1 Answers 1

0

You started with a matrix $A$ (whose determinant is non-zero). You've calculated a matrix $B$ which you claim is equal to $A^{-1}.$ To check, just calculate $AB$ and make sure it's equal to the identity matrix.

True, this takes about $n^{3}$ operations to do by hand, for an $n\times n$ matrix, but it's basically fool-proof, and if you're calculating the inverse by hand then $n$ can only be as large as $3$ or $4.$

  • 0
    I only asked for test purposed if it takes $n^3$ operations by hands its really pointless, I could just google up a inverse checker with ease. For example for my RREF example. If there are 3 equations in a linear system and in a test I just check my solution with one of them, most likely my entire solution is right though there could be a rare case were only that one equation is right but I don't have that much time to waste and check them all. – 2017-02-18
  • 1
    @user349557: You appear to be complaining about how long something is taking when you are doing it *by hand.* If you want to do computations super-quickly then of course you should get a computer to do it. Presumably you care about this because you want marks in an exam? In that case it's entirely up to you whether or not those marks are worth the few minutes you could spend checking that your answer is correct. Note: if you do multiply them, do it properly. Don't just do a single row and assume that will do. I know from experience. – 2017-02-18
  • 0
    We might not be on the same page. I wrote a quiz yesterday and It felt tedious going through all my steps for the inverse inorder to check it. The inverse was a 4x4 matrix all fractions with double to single digits numerator / denominator. How does one multiply that with the original matrix in a few minutes? This may not be fool proof but I think it'd be better if I multiply both matrixes $A^{-1}$ and $A$ and just see if its 1 0 0 0. I tried this too and it did work for several questions. Its not fool proof but much faster since inverses get really ugly. – 2017-02-19
  • 0
    @user349557: Yes, I'm sure it did feel tedious; frankly, it *is* tedious! That's why people invented computers: To do tedious calculations, preferably quickly (and as it turns out, much faster and more accurately than humans). Multiplying $4\times4$ matrices *does* only take a few minutes - if it's taking you more than, say, 10 minutes, then you just need practice. If the entries are "complicated": Presumably they're all rational numbers, just factor out the denominators to get integer matrices. If the numbers are big: Then you probably should just have a calculator on hand. – 2017-02-19
  • 0
    Oh if I had a calculator allowed this topic wouldn't even be made. Yes my mental multiplication sucks after like multiples of 12+ where I start using long multiplication which is obviously very tedious and time consuming in my case. You could probably do even double digit in a few minutes but I cant say for me. I'll stick with this shortcut. – 2017-02-19