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Say for instance i have a coefficient matrix of 3 x 3. How do i enter it into the scientific calculator to work out it's inverse when you need that say for working out a system of equations?

Just need a heads up on what buttons to hit as the SC has so many options i'm lost.

I'm currently studying an Algebra book and am looking at the systems of linear equations section. Now when it comes to equations, where you have more than three or four, it becomes too involved working things out by elimination or substitution. So that's why i'm wanting to use the calculator. Plus the calculator doesn't make errors!

I was practicing finding the inverse of a matrix using the coefficient matrix and then putting an identity matrix side by side with it and doing row operations. That's fine with small numbers but gets very complicated when you start dealing with large numbers. Plus it takes too long!

Anyway hope someone can give me some pointers to make life easier!

Gary

Equation Aug 17, 2014

#1**+8 **

I don't think the calculator on this site has the capability to do this, and I'm not sure what type of calculator you're using, so I can't tell you how to do this......(maybe look at your manual???)

However, here's a pretty good website, WolframA;pha, that will calculate an inverse of an invertible square matrix........

http://www.wolframalpha.com/input/?i=inverse+matrix&a=*C.inverse+matrix-_*Calculator.dflt-&f2={{5%2C4%2C9}%2C{3%2C2%2C1}%2C{2%2C1%2C3}}&f=MatrixInverse.invmatrix_{{5%2C4%2C9}%2C{3%2C2%2C1}%2C{2%2C1%2C3}}

I don't know if that helps you, or not.

CPhill Aug 17, 2014

#2**+13 **

You *can* do it on the calculator here. First select vectors/matrices from the right side of the input bar. Then choose 3x3 matrix for A and B. Enter your coefficients in A. keep zeros in B. Select ^-1 for A and + for combining the matrices, then press =. A little cumbersome, but it works!

Alan Aug 17, 2014

#3**+3 **

Thanks CPhill for taking the time out to try to help. Thanks for the link as well, i'll check it out.

Alan: You came through for me again mate! I didn't know if it was possible to do it through the SC but i had a hunch it was but just didn't have a clue what to do or how to go about it!

You've helped me so much with that answer.

The book we talked about before 'Electronics: A Systems Approach' page 62. The coefficient matrix on the left there is the one i'm talking about calculating. From the Algebra book i'm studying it says get the inverse matrix and then multiply it with the constants matrix to get the solution to the systems of equations problem but using a matrix instead of doing it by hand.

Thanks Alan you're a great help.

Gary

Equation Aug 17, 2014

#5**+3 **

".*..get the inverse matrix and then multiply it with the constants matrix to get the solution to the systems of equations problem but using a matrix instead of doing it by hand..*.."

You could do this all in one step here, by changing B to vector and entering the appropriate values, then changing the + sign to a * sign.

No! That doesn't work! For some reason the calculator here tries to calculate A^{-B} if you do that!! You'll have to do it in two separate stages (i.e. find A^{-1} first then reenter that as a new A and multiply by B).

Alan Aug 18, 2014

#6**+3 **

Cheers Alan. I just noticed you can't do a 3 x 1 for the constants matrix. Unless i'm missing something (which is very possible!) 2 is the lowest i see in the options.

Not too much of a problem multiplying the inverse with the constants and adding everything up to get the answers just by hand. But it would be nice to just have that constant matrix with the 3 rows 1 column option and multiply it straight with the SC.

What do you think is the best route for calculating these systems of equations say if you have more than 3 equations? There are so many options, sometimes as a relative novice you're in a bit of a quandary as to what to use.

I was reading about Cramer's rule but the Algebra book i'm looking at says that get's more involved and is a bit more work. I think it looks quite a good way to calculate though.

For me i think after you know the steps to take to set up the equations in matrix form, using a calculator like the SC would be best as it saves you a major headache!!!

Equation Aug 18, 2014

#7**+3 **

You should be able to do a 3x1 vector - it's the default when you switch into the vector/matrix mode!

I think there's a bug in the calculator that prevents the A^{-1}*B option working directly. I've sent a message to Admin about it. ADMIN HAS NOW FIXED THIS!

I wouldn't use Cramer's rule for anything more than a 3x3. I wouldn't expect anyone to do a 4x4 or bigger by hand! Get hold of some software that will do this for you. There is a free download program called SMath Studio Desktop that should do this.

Alan Aug 18, 2014

#8**+3 **

Thanks for the tips Alan. For the 3 x 1 vector i've never worked with those yet or studied anything about them so i didn't know i could use that option for doing the final multiplication of the inverse and constant matrices. I thought i always had to have matrix selected for both sides!!

I'll look into the SMath tip and look it up via google as it sounds like what i'm looking for.

Thanks again for all your help.

Equation Aug 18, 2014

#9**+3 **

Tried it out but not getting the answer from the book. Answer to the problem for the three unknowns is 326, 34, 53. All ma measurements. 0.326, 0.034, 0.053 Amps converted from the ma's. That's the results on page 61 of 'Electronics A Systems Approach'

Probably error on my part, doing something wrong! But the figures i'm getting are nowhere near the answers.

Equation Aug 18, 2014

#10**+13 **

See if the following help:

To solve these equations using this website's calculator enter the following in the vector/matrix section (Note, you only enter the values in the table part, not in the top line. You also select the ^-1 next to the A):

Press the equals sign to get the following:

Hmmm! Seems to be a sign error for I_{2} here.

Alan Aug 18, 2014

#11**+3 **

Thanks Alan for taking the time to write all that out. It pretty much explains it all for me!!

I think there is an error in one of the equations there in the book. I think he's made one negative when it shouldn't be or switched something around by mistake. It happened before with the other problem you helped me solve where he had reversed two figures and put them in the wrong sequence.

Easy done when dealing with all that data.

Thanks for all the help i appreciate it.

Gary

Equation Aug 19, 2014

#12**+13 **

Best Answer

*"I think there is an error in one of the equations there in the book. "*

No, I think the error in the sign is a bug in the website calculator. Mathcad gets the sign the same as the text book.

Alan Aug 19, 2014

#13**+3 **

Hi Alan yeah i see what you mean about a bug in the calculator. I was practicing finding matrix inverses today and put in a matrix from the book i'm reading. Put in all 0's on the B matrix and i got the same answer as the book but two of the numbers were signed as being - instead of +. I don't know why that is apart from there's something wrong there.

Anyway when i corrected the signs and took the negatives off and multiplied with the constants matrix i got the correct answer.

It's all fine but the problem would be if you didn't recognize that there was an issue there. You'd be getting the answers all wrong probably and racking your brain to work out why!!

Gary

Equation Aug 19, 2014

#14**+8 **

Thanks Alan!

$$\left({\begin{bmatrix}

{-{\mathtt{160}}} & {{\mathtt{20}}} & {{\mathtt{30}}}\\

{{\mathtt{20}}} & {-{\mathtt{210}}} & {{\mathtt{10}}}\\

{{\mathtt{30}}} & {{\mathtt{10}}} & {-{\mathtt{190}}}\\

\end{bmatrix}

}^{-{\mathtt{1}}}\right){\mathtt{\,\times\,}}\begin{pmatrix}

-{\mathtt{50}}

\\

{\mathtt{0}}

\\

{\mathtt{0}}

\end{pmatrix}

= \begin{pmatrix}

{\mathtt{0.326\: \!711\: \!541\: \!618\: \!781\: \!8}}

\\

{\mathtt{0.033\: \!656\: \!214\: \!086\: \!356\: \!9}}

\\

{\mathtt{0.053\: \!357\: \!412\: \!575\: \!931\: \!7}}

\end{pmatrix}$$

admin Aug 19, 2014