Question about Casio FX-9750GPlus Calculator

Suppose square matrix A and compatible matrix B, and that Mat A is non-singular.

- Press [MENU] key and go to RUN screen.
- Press [F2:MAT] to access the matrix tabs
- Press F1:Mat]. The Mat identifier is entered on the command line.
- Press [ALPHA] key then A
- Command line now shows Mat A.
- Press [SHIFT][ )] to activate [X^-1]
- Press [*] multiplication key then [ALPHA] and B.
- Command line now shows Mat A^-1 * Mat B.
- Press [EXE] to display the result.

Posted on Apr 07, 2011

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Posted on Jan 02, 2017

What do you mean by "solve matrix"? Do you want its determinant? Its inverse? The eigenvalues and eigenvector? Solve a system of linear equations?

Jun 11, 2013 | Casio FC-200V Scientific Calculator

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.

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.

Mar 17, 2012 | Casio FX-9860G Graphic Calculator

Select the EQN computational mode 5:EQN

For system of linear equations in 3 unknowns select 2: in the screen below.

Arrange the variables in your equations in the same order (x first, y second, and z third).

The coefficients to enter in the editor are the factors of the variables: In your case a_1=10, b_1=-3, c_1=10, d_1=5, a_2=8, b_2=-2, c_2=9, d_2=3; a_3=8, b_3=1, c_3=-10, d_3=7.

To enter d_1, d_2, and d_3 you will have to scroll to the right to reach the cells where they should go.

Once finished entering the coefficients, press EXE to get the solutions. You may have to use the arrow Down to display the y- and z-solutions.

I verified that the matrix is non-singular and the system has a solution. In fraction form

x=27/53, y=-1 and 22/53; finally z=-23/53

For system of linear equations in 3 unknowns select 2: in the screen below.

Arrange the variables in your equations in the same order (x first, y second, and z third).

The coefficients to enter in the editor are the factors of the variables: In your case a_1=10, b_1=-3, c_1=10, d_1=5, a_2=8, b_2=-2, c_2=9, d_2=3; a_3=8, b_3=1, c_3=-10, d_3=7.

To enter d_1, d_2, and d_3 you will have to scroll to the right to reach the cells where they should go.

Once finished entering the coefficients, press EXE to get the solutions. You may have to use the arrow Down to display the y- and z-solutions.

I verified that the matrix is non-singular and the system has a solution. In fraction form

x=27/53, y=-1 and 22/53; finally z=-23/53

Jun 29, 2011 | Casio FX-115ES Scientific Calculator

This calculator cannot handle matrices that have anyone dimension larger than 3.

How to solve your problem?

use a computer program devoted to solving linear equations

OR buy another calculator that can handle larger systems of equations (a graphing calculator, maybe).

How to solve your problem?

- Set up the problem by hand (pencil paper): use Cramer's rules. Make use of the calculator to perform the various calculation.

- Sacrifice one equation which you will use to eliminate one variable of the 4 by expressing it in terms of the other three. Do the algebra to obtain a system of 3 equations in three unknowns.
- Use the calculator to find the solutions ( the three variables that are left).
- Use the 4th equation (the one you scacrificed) to obtain the value of the 4th unknown.

use a computer program devoted to solving linear equations

OR buy another calculator that can handle larger systems of equations (a graphing calculator, maybe).

Feb 16, 2011 | Casio FX-115ES Scientific Calculator

Let your system of linear equations be [A][X]=[B] where [A] is a 3x3 matrix, [X] is column vector ( a 3x1 matrix) , and [B] is also a column vector or (3x1) matrix.

Assuming you are able to invert the [A] matrix to find [A^-1].

Multiplying on the keft by [A^-1] the system of equations you get

[A^-1].[A][X]=[A^-1][B].

But since the product [A^-1][A] is the (3x3) Identity matrix, the left side of the equation is jut [X], while the right side is [A^-1][B]. As you can see it is not [B][A^-1] .

To solve your problem, you should define your [A] matrix, define the [B] column vector (3x1) matrix, then perform the operation [A^-1][B] using the [X to -1] power key to calculate the inverse [A^-1].

Assuming you are able to invert the [A] matrix to find [A^-1].

Multiplying on the keft by [A^-1] the system of equations you get

[A^-1].[A][X]=[A^-1][B].

But since the product [A^-1][A] is the (3x3) Identity matrix, the left side of the equation is jut [X], while the right side is [A^-1][B]. As you can see it is not [B][A^-1] .

To solve your problem, you should define your [A] matrix, define the [B] column vector (3x1) matrix, then perform the operation [A^-1][B] using the [X to -1] power key to calculate the inverse [A^-1].

Jan 10, 2011 | Casio FX-115ES Scientific Calculator

What do you mean by "solve a matrix"? Find the inverse? Find the determinant? Find the solution to a system of linear equations? Within limits the fx-115ES can do any of these and more.

The procedures are described in the "Matrix Calculations" section of the manual. If you still have problems, please reply to this post specifying what you want to do.

The procedures are described in the "Matrix Calculations" section of the manual. If you still have problems, please reply to this post specifying what you want to do.

Nov 18, 2010 | Casio FX-115ES Scientific Calculator

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.

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.

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.

- define X = a+i*b
- Define Y=c+i*d
- Rewrite the first equation substituting a+i*b for X and c+i*d for y.
- Gather all real terms together, and all imaginary terms together on the left.
- You will have (8a-15b-8c) +i(15a+8b-8d) = 10 +0i
- 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

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.

Feb 10, 2010 | Texas Instruments TI-84 Plus Calculator

Hello,

Let us assume you have two simultaneous linear equations :

**a_1*x+ b_1*y+c_1=0**

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Let us assume you have two simultaneous linear equations :

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Aug 12, 2009 | Sharp EL-531VB Calculator

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

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

May 03, 2009 | Texas Instruments TI-84 Plus Calculator

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