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Matrix & Simultaneous Equations Help


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#1 Ladekhan

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Posted 13 February 2011 - 11:25 AM

Firstly i want to solve find the determinant on this 3x3 matrix:

[ 1 2 -1 ]

[ 2 5 -(a+2) ]

[ -1 a-5 1 ]

so i typed:

[ 1 2 -1 ]

det([ 2 5 -(a+2) ])


[ -1 a-5 1 ]


but i got an error message: "Incorrect Argument"


-------------------------------------------------

Secondly, I wish to know how to solve these three simultaneous equations for "a" on the classpad

x + 2y -z = 2

2x + 5y -(a+2)z = 3

-x + (a+5)y + z = 1

I tried using the solve function on the 2D menu, but i got the error message: "Invalid Dimension"


Thanks in advance for any help.

#2 MicroPro

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Posted 13 February 2011 - 11:57 AM

Hello.
Question 1 works ok on my classpad, maybe a is already defined in your classpad: Type Clear_a_z and hit EXE to delete it if it already exists.
Question 2: The number of variables on the right of the vertical line must be the same as the number of equations, which means on the right you should write x,y,z. Then you will have the answer. And, you can't solve 3 equations with 4 variables so there is no explicit answer for a here. x, y & z are all functions of a in your problem.

#3 Ladekhan

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Posted 13 February 2011 - 12:20 PM

Thanks, your solution worked for question 1,

but for question 2 i know u can't solve for 4 variables with 3 equations, hence i was looking to express "a" in terms of "x", "y", and "z"

I tried adding the x, y, and z but it still says "Invalid dimension".

Edited by Ladekhan, 13 February 2011 - 12:22 PM.


#4 pan.gejt

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Posted 13 February 2011 - 10:53 PM

Have you tried Gauss elimination method (the rref command)? I tried it (in Mathcad not Classpad) and I have obtained 3 equations:

1. x = (2*a^2+9a+10)/(a^2+7a)

2. y = 3/(a+7)

3. z = (a+10)/(a^2+7a)

According to this equations this system has 2 solutions for x and z and 1 solution for y (in terms of a).
But be careful, if you expressed "x, y, z in terms of a" you cannot choose the x , y and z randomly- you must still keep condition x + 2y -z = 2
Maybe, I am completely wrong.

#5 Ladekhan

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Posted 16 February 2011 - 12:17 PM

Yes I did end up using the Gauss method, however I was hoping that I would not have to resort to that method, seeing as though on my CAS mathematics examination they would likely give me equations which would require time to convert into matrices, which puts me at a disadvantage against those using the TI-Nspire since the TI-Nspire's solve function is able to solve simultaneous equations in terms of variables such as "a".

Is there a way to do it without using matrices?

#6 pan.gejt

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Posted 16 February 2011 - 08:10 PM

You are right, the TI-Nspire can solve this equations very fast. I solved it and I obtained 3 result- one for "a" and two for "x"
Try the following method on Classpad:
Solve the 2nd and the 3rd equations in terms of "a" and "x"- "solve ({2x + 5y -(a+2)z = 3,-x + (a+5)y + z = 1},{a,x}".
You will obtain two solutions- one for "a" and one for "x". But I do not know how to get the 2nd solution for "x".

#7 MicroPro

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Posted 18 February 2011 - 12:51 PM

Is there a way to do it without using matrices?

Couldn't you solve Question 1 with ClassPad's multiple equation symbol? That way you can solve the equations of your question 2, too: You need to write x,y,z to the right of the vertical line. And ClassPad solves them fast. Could TI solve for a? Solving for a requires additional steps to be done.

I'd tried solving equations for x, y, z but unfortunately solving them in a system causes Insufficient Memory error on CP.
And using parametric 3d graphing you get a somewhat discontinuous spiral so i'm not sure whether a can be expressed as a function of x, y and z.

BTW ClassPad CAN solve matrices. Look at this old topic and this new one as examples to learn some tricks for doing this.

#8 tfpp

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Posted 08 March 2020 - 03:03 AM

I was trying to solve 4 unknowns with 3 equations and found a trick to do it with the ClassPad II. You just need to treat a as an unknown and write the 4th equation as a=a and solve 4 equations. So, you can enter

{x+2y-z=2

 2x+5y- (a+2)z=3

 -x+(a+5)y+z=1

 a=a                     | x,y,z,a

to get the answer

x=10+9a+2a^2/(7a+a^2), y=3/(7+a), z=(10+a)/(7a+a^2), a=a

Hope that helps other who encounters this problem. It's 2020 so, I guess the OP does not need to know this anymore :)






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