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Yesterday I was doing an structures exam with my brand new HP prime.
At the end of one of the problems i had to solve a linear system of equations (KX=F) where K is the stiffness matrix. the problem came when i tried to solve it using the linear solver app, the system was something like:
5e11X+6e11Y=50000
6e11X+12e11Y=6000
the calculator just gave me X=0 Y=0.
I tried to solve it manually by calculating the inverse matrix of [[5e11,6e11],[6e11,12e11]] and the result was a matrix of zeros [[0,0],[0,0]].

after trying a couple of times I just took my old 50g from my backpack and solved the system without problems.

later in my house i tried again and always the result was a matrix of zeros when trying to invert a matrix with big coefficients, and also the same in my computer´s HP prime virtual calculator. reading the prime´s manual i found that the calculator does not manage numbers smaller than 1e-49 but its far away from the expected results of the equation system (values around 1e-4). also i found that there is no problem using the CAS mode or using approx([5e11,6e11],[6e11,12e11]]^-1) but i don't know why.

thanks god i had my 50g but i don't understand why i can't make the inverse matrix in the prime if i can do it even with a casio basic calculator.

pedro
civil engineering student
(12-16-2016 02:57 PM)pedroHPsp Wrote: [ -> ]Yesterday I was doing an structures exam with my brand new HP prime.
At the end of one of the problems i had to solve a linear system of equations (KX=F) where K is the stiffness matrix. the problem came when i tried to solve it using the linear solver app, the system was something like:
5e11X+6e11Y=50000
6e11X+12e11Y=6000
the calculator just gave me X=0 Y=0.
I tried to solve it manually by calculating the inverse matrix of [[5e11,6e11],[6e11,12e11]] and the result was a matrix of zeros [[0,0],[0,0]].

after trying a couple of times I just took my old 50g from my backpack and solved the system without problems.

later in my house i tried again and always the result was a matrix of zeros when trying to invert a matrix with big coefficients, and also the same in my computer´s HP prime virtual calculator. reading the prime´s manual i found that the calculator does not manage numbers smaller than 1e-49 but its far away from the expected results of the equation system (values around 1e-4). also i found that there is no problem using the CAS mode or using approx([5e11,6e11],[6e11,12e11]]^-1) but i don't know why.

thanks god i had my 50g but i don't understand why i can't make the inverse matrix in the prime if i can do it even with a casio basic calculator.

pedro
civil engineering student

The first thing to check is if the firmware is up to date.
hello Han!

I updated the firmware about one week ago. but just in case i re-installed HP Prime virtual calculator in my computer set it to factory settings and updated the firmware (to see if the problem was my calculator or the settings) , it seems that if i use approx() it works but still don't know why this happens
(12-16-2016 03:59 PM)pedroHPsp Wrote: [ -> ]hello Han!

I updated the firmware about one week ago. but just in case i re-installed HP Prime virtual calculator in my computer set it to factory settings and updated the firmware (to see if the problem was my calculator or the settings) , it seems that if i use approx() it works but still don't know why this happens

The issue has to do with "tiny elements," which are normally "converted" to 0 in the non-CAS environment. If you use the CAS view, your inverse will work fine. I hope that in the future the developers might consider a "tiny element" setting (convert to 0 or leave as-is) so that this issue can be resolved in the Home view as well.
Using Cas-view with standard settings I get the desired result, in Home some rounding seems to be done.
Arno
I have the same issues in with matrices in electrical circuits where the resistors/ impedances tends to be in orders of mega ohms.... and inductance in milli and capacitance in orders on nano.
Another way to avoid zeros as the answer is to scale the A matrix. (Ax=b). or normalize the A matrix hence putting the A matrix in unity. (k*Ax=b) where k is a scalar.
in your case k=1e11 and new A=[[5 6],[6 12]]. and x=(1/k)*(b/A). As close you get in with your linear equations use any of the over-abundantly ways of solving equations in Prime.