MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
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MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Solving a system of equations in three variables could be more easily done uses matrices as opposed to using substitution or elimination. First, make sure that the variables in all three equations are in the same order and that it is equal to the constant. Next, write out two matrices, A and B. Matrix A will be a 3 by 3 matrix comprised of the coefficients of the equations and matrix B will be a 3 by 1 matrix comprised of the constants. Next, input the values in your graphing calculator. Under Matrix, edit matrix A by inputting the dimensions, then inputting the values. Do the same for matrix B. You want to calculate the product inverse of matrix A, so type matrix A and the inverse sign and multiply that by matrix B. The solution will be written in the same order of your equation -- x, y, and z.
When you have systems of equations that are two by two meaning two variables and two equations it's usually quick enough that you can solve it using substitution or elimination or graphing. But when you get into systems like this that are three by three meaning three equations and three variables, a lot of times using matrices can be a lot quicker than doing it by hand.
The first thing I want to do is think about what my matrices are going to look at on my graphing calculator. I'm going to want to set up my A matrix, my coefficient matrix, using the coefficient or x y and z.
The first equation is 1, 1, 1 then I have 1, -1, 1 and this last one be careful I don't have a y term. So it's going to be 3, 0, -1. That's my coefficient matrix that's called A. Then that's going to be also multiplied, oh sorry, I'm going to inverse A and multiply it by my constant matrix B. B is 41, 15, 4.
Okay let's get those guys on to the graphing calculator and figure out the solutions. When you turn on your graphing calculator you start at your home screen. I'm going to go into where the matrices are so I can put it in my A and B matrices. I'm going to edit matrix A, it starts by asking me the dimensions. Remember I have three equations and three variables so my dimensions are 3 by 3. Then I'm going to enter all of my coefficients for x, y and z there's my first equation done 1, 1 and 1. My second equation was 1, -1 and 1 and then my third equation was 3, 0, -1.
Okay that's my matrix A it represents the coefficients from my three equations and the three variables. Now I need to go in and edit matrix B. Matrix B it's going to ask me just like before to enter the dimensions. My dimensions now are 3 by 1, so when I'm putting in my values my first equation was equal to the constant 41, my second equation was equal to the constant 15 and my third equation was equal to the constant 4.
Great, now it's just a matter of pushing the right buttons. Matrix A and matrix B are in there, I want to calculate the product of matrix A inversed, so I'm going to use this button here for inversed there it is right there see the inverse, that's going to be multiplied by matrix B. So here comes matrix B. Make sure I have it set up the way I want it in the home screen and when I hit enter it gives me my answers x is equal to 8, y is equal to 13, and z is equal to 20. And I know it's that order because x, y, z is the order in which I entered my coefficients.
So you guys matrices can be really effective tools for solving systems especially when they are larger than 2 by 2. This guy was 3 by 3 I could have done it by hand using substitution but it would have taken me a long time. This took me 2 minutes not bad. This is just another tool for you guys to have in your algebra tool bell.
Unit
Solving Systems of Equations