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sai dharini Posted on Feb 04, 2014
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Complex matrix equation involving complex numbers how to solve

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k24674

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  • Casio Master 8,093 Answers
  • Posted on Feb 04, 2014
k24674
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Is it complex or matrix? You can have only one mode at a time.What kind of equations? The solve function uses real-valued equations.
Complex numbers: You can add, subtract, multiply and divide. You can convert a number from rectangular to polar forms and vice-versa; you can calculate the absolute value of a complex number, or a conjugate.
All the operations are done in the Mode CMPLX.

5 Related Answers

Anonymous

  • 106 Answers
  • Posted on Dec 20, 2009

SOURCE: how to solve equation involving angles

use complex mode

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k24674

  • 8093 Answers
  • Posted on Jul 25, 2010

SOURCE: In complex number mode, how to get out?

"Is there anything that I can do about this to stop my answers being complex numbers when solving cubic equations?"
You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be
4591a1b3e15019ae25ad5813d5744ca9.png
If coefficients are complex you should expect some complex roots. Right?
If the coefficients are REAL then depending on the discriminant
80efae5ccea2ac0702e1a9c3b25afd2b.png
you can have three cases
DELTA positive : three distinct real roots
DELTA=0 , the equation has a multiple root and all roots are REAL
DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

d64abc4.jpg










k24674

  • 8093 Answers
  • Posted on Aug 29, 2011

SOURCE: Hello How to solve complex

The simultaneous equation solver requires the coefficients to be real. Similarly matrices must have real coefficients.
Your only alternative is to express each of A and B as a real part and imaginary part.
A= x1+iy1
B=x2+iy2.
Substitute x1+iy1 for A in the two equations. Substitute x2+iy2 for B in the two equations. Do the algebra. Gather real parts and gather imaginary parts. Split each original equation into two equations: One equation comes from setting Real Part of left side = real part of right side (1); the other equations comes from setting the imaginary part of left side= imaginary part of right side (here 0).

Do the same procedure for the 2nd original equation.
At the end of the process you will have 4 coupled linear equations in the 4 unknowns (x1,y1,x2,y2).

Then you might want to use the calculator to solve the derived system. Once you have x1,y1,x2,y2 you reconstruct A=x1+iy1, etc.

Anonymous

  • 122 Answers
  • Posted on Sep 05, 2011

SOURCE: solving linear equations in 3

Press key Mode, first select option 5:EQN (by pressing key 5), then option 2: a_nX+b_nY+c_nZ=d_n (by pressin 2).
Now input coefficients from equations in table (each column is for one variable), just type in number and press =, then use arrows to move to another cell.

k24674

  • 8093 Answers
  • Posted on Mar 18, 2012

SOURCE: Edit Casio FX - 9860G SD have complex equations??

Thhe Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to a+ib, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.
The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).
Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.

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Related Questions:

0helpful
1answer

Hi I have Casio fx -991es plus calculator its give me wrong answer in complex number equations

It does not solve complex number equations. You can do simple calculations on complex numbers: addition, multiplication, etc. But nothing fancy.
0helpful
1answer

I want to solve an complex equation on casio fx-570 ES Plus

Your scientific calculator is unable to solve complex equation with complex coefficients. You should try to solve by hand directly using the quadratic formula or by factoring the polynomial in z
failing that, another way would be to set z=x+iy, substitute this for z, carry out the algebra and try to separate real and imaginary parts. But your two equations will constitute a system of two quadratic equations. I am not aware of any general method to solve coupled nonlinear equations.
Good luck.
1helpful
1answer

Edit Casio FX - 9860G SD have complex equations??

Thhe Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to a+ib, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.
The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).
Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)

Alternatively, after you create the system of 4 linear equations you can use the matrix utility to find Real(x), Im(x), Real(y) and Im(y) and recompose the x and y.
1helpful
1answer

Casio FX - 9860G SD have complex equations?? edit

The Casio FX-9860G SD can solve a polynomial equation of degree 2 or 3 with REAL coefficients. If the complex MODE is set to REAL it will find the real roots. If the complex mode is set to a+ib, it will find the real and complex roots.

Apparently it will take coefficients that are real, and will give a Ma Error if any coefficient is complex.

Addendum.
The calculator CANNOT solve equations with complex coefficient. YOU can however convert the system of linear equations with ccomplex coefficients ( of the type you show) as a system of 4 linear equations in 4 unknowns; Split x into a real and an imaginary part, split y into a real and an imaginary part. Substitute Real(x)+iIm(x) for variable x in the equations; substitute Real(y)+iIm(y) for y in the two equations; do the algebra. In each of the original equations split the Real and imaginary parts. You should be able to derive 4 linear equations in unknowns Real(x), Im(x), Real(y), and Im(y).
Use the linear equation solver to obtain the solutions. Recompose x=Real(x)+iIm(x), and y=Real(y)+iIm(y)
3helpful
1answer

Solve a matrix of complex numbers using ti89 or any other ti calculator

Well, the equation is Ax=b and we want so solve for x
There are several ways of entering the data, but this is one way
Code: A=[15+35i 29+1i; 46+13i 62+3i]
b=[0;0]
x=A\b which has the trivial solution x=[0;0] (the ";" means "next row", a space next column.) To input a matrix use APPs then Data/Matrix Editor then New. Then change Data to Matrix. Put in the name such as "AA" in the variable field. Then enter row and column dimensions. Then edit as a spread sheet. Use F6(util) to delete a column or row. To manipulate matrices, exit the matrix application, then go to MATH then Matrix. Notice that J(RowOpps) is available. Good luck and thank you for using FixYa! We would sure appreciate a 4 "thumbs-up" rating for this answer. Happy holidays!
0helpful
1answer

In complex number mode, how to get out?

"Is there anything that I can do about this to stop my answers being complex numbers when solving cubic equations?"
You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be
4591a1b3e15019ae25ad5813d5744ca9.png
If coefficients are complex you should expect some complex roots. Right?
If the coefficients are REAL then depending on the discriminant
80efae5ccea2ac0702e1a9c3b25afd2b.png
you can have three cases
DELTA positive : three distinct real roots
DELTA=0 , the equation has a multiple root and all roots are REAL
DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

d64abc4.jpg










4helpful
1answer

I am trying to solve imaginary number equations with matrix using the TI-84 For example: (8+15i)x-8y = 10 -8x-(8-10i)y = 0 8+15i -8 10 -8 8-10i 0 should i convert to another form?

I am afraid you cannot use the TI8xPlus family of calculators to solve linear systems in matrix form. In this calculator, matrices must have real coefficients.
You can however separate (expand) the problem into a linear system of 4 equations in 4 unknowns and try to solve it with the calculator.
  1. define X = a+i*b
  2. Define Y=c+i*d
  3. Rewrite the first equation substituting a+i*b for X and c+i*d for y.
  4. Gather all real terms together, and all imaginary terms together on the left.
  5. You will have (8a-15b-8c) +i(15a+8b-8d) = 10 +0i
  6. This equation can be split into two equations by saying that
  • the real part pf the left member is equal to the real part of the right member, this gives 8a-15b-8c=10, and
  • the imaginary part of the left member is equal to the imaginary part of the right member, this gives 15a+8b-8d=0
Similarly, you rewrite the second complex equation substituting a+ib for X and c+id for Y, then expand the binomial products, gather the real parts on the left together, and the imaginary parts on the left together. After that you equate the real part on the left to the real part on the right, and do the same for the imaginary parts.

If I did not make mistakes during the expansions and the gathering of terms you should get the following equation
-(8a+8c+10d) +i*(-8b+10c-8d) =0+i*0 from which you extract an equation for the real parts, -(8a+8c+10d)=0 and another for the imaginagy parts i*(-8b+10c-8d) =i*0

If I did not make mistakes (you should be able to find them, if any) your system of two linear equations with complex coefficents has been converted to a system of 4 linear equations with real coefficients.

8a-15b-8c=10
15a+8b-8d=0
8a+8c+10d=0
-8b+10c-8d =0

Now, you can in theory solve this system with help of the calculator, to find a, b, c, and d. When these are found, you can reconstruct the X and Y solutions.

Now get to work: Ascertain that my extracted equations are correct, then solve for a, b,c, and d, and reconstruct X and Y.
I am no seer, but my hunch is that this system is degenarate. I will not explain what that means.
1helpful
2answers

Solve comlex equation

Hello,
I do not understand what you mean by solve complex equations, but if you want to manipulate complex numbers here are the tools you need to enter your complex numbers. Operations and function evaluations do not need special symbols.

b4c488c.jpg Hope it helps.
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