Question about Texas Instruments TI-84 Plus Calculator

# 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?

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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.

Posted on Feb 11, 2010

• k24674 Feb 11, 2010

Sorry, I am indeed no seer and apparently the matrix is non singular. Again, if I did not make mistakes its determinat is -78884, so it can be inverted. Good Luck.

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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.

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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The TI-84 works with real fractions. It will not convert complex numbers to fractions.

You can convert the real and imaginary components to fractions separately. For your example, 1/(1+i) is .5-.5i. Converting .5 and -.5 to fractions, you get 1/2 + (1/2)i.

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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
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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.

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### 2c + 6 + 7c = 8

To solve this with a ti-84 press [math] and select solve from the menu. This will open the equation solver. Type in your equation
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Mar 30, 2011 | Texas Instruments TI-84 Plus Calculator

### 5/4x-18=-3 Ti 84 plus One: just press enter/solve Two: press 2nd Test (Math) then press enter Every time I put in and equation using the above method for the equal sign. it either takes me back to...

If you wan to solve an equation using the SOLVE feature you enter it as follows (your example)
You should rewrite it as 5/4*X-18+3 (= 0). Implicitly, the calculator assumes that the right side is ZERO and you do not enter any right side nor an equal sign.
Type in
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I assume you are speaking of solving a system of equations with a number of unknowns. If not, please correct me. Here's an example in practice:

If you have a system of 3 equations with 3 unknowns, you would set up your matrix so that the coefficients of each variable for a particular equation are on one row. So, given equations x + y + z = 0, 2x + 3y - 4z = 1, x + -z = -1 you would type the following into your calculator: [[1,1,1,0][2,3,-4,1][1,0,-1,-1]] and press enter to make sure you typed it correctly. notice that in the third row there is a zero, since we have zero time y for the third equation. Then row-reduce the matrix (2nd > 5 > 4 > 4 or in the CATALOG as rref). You should get out the matrix [[1,0,0,-1][0,1,0,1][0,0,1,0]]. This says that x=-1 y = 1 z=0 since my first column contained the coefficients for the x variable, the second for the y variable, and the third for the z variable. The last column contains the solution, the part on the other side of the equals sign.

Hope this helps! For more reading (from someone else; I just made this one up), check out the Wikipedia articles on Gaussian elimination and Systems of linear equations

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If you want the answer in polar form, set MODE to re^xi and and DEGREE. Tap ENTER and QUIT. Tap the square root sign, tap (-)4, close the parenthesis, and tap ENTER. The answer shown is 2e^90i. (2 at ninety degrees). If you selected RADIAN mode, the answer would be 2e^1.57i (2 at 1.57 radians; or 2 at pi/2 radians)

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