For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

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## Table of Contents

## Transcription

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### Using Matrices to Solve Systems of Linear Equations

- We can represent an entire linear system with an an
*augmented matrix*:x + y + z = 3 2x −2y +3z = −4 −x −z = 0 ⇒ ⎡

⎢

⎢

⎢

⎣1 1 1 3 2 −2 3 −4 −1 0 −1 0 ⎤

⎥

⎥

⎥

⎦

- Each row represents an equation of the system,
- Each left-side column gives a variable's coefficients,
- If a variable does not appear, it has coefficient 0,
- The vertical line represents `=',
- The right-side column gives the constant terms.

- There are three
*row operations*that we can perform on an augmented matrix:- Interchange the locations of two rows.
- Multiply (or divide) a row by a nonzero number.
- Add (or subtract) a multiple of one row to another.

__lot__of arithmetic and steps. With that much calculation going on, it's easy to make mistakes. Counter this by__noting__each step you take and what you did. This will make it easier to avoid making mistakes and will help you find any that manage to creep through. - We can use row operations to put a matrix in
*reduced row-echelon form*. While it has a formal definition, it will be enough for us to think of it as a__diagonal of 1's__starting at the top-left with__0's above and below__(entries on the far right can be any number). If we can use row operations to put an augmented matrix in reduced row-echelon form, we will have solved its associated linear system. -
*Gauss-Jordan elimination*is a method we can follow to produce reduced row-echelon form matrices through row operations and thus solve linear systems.- Write the linear system as an augmented matrix.
- Use row operations to attain a 1 in the top left and zeros below. Then move on to creating the next 1 diagonally down, with zeros below. Repeat until end.
- Now, work from the bottom right of the diagonal, canceling out everything above the 1's. Continue up the diagonal until you have only 0's above as well.
- The matrix should now be in reduced row-echelon form, giving you the solutions to the system. [Note: If there is no solution or infinitely many, you will not be able to achieve reduced row-echelon form.]

- We can also find the solutions to a linear system by using determinants and
*Cramer's Rule*. Let A be the*coefficient matrix*for our linear system. [The coefficient matrix is the left-hand side of an augmented matrix. It is all the coefficients from the linear system arranged in a matrix.] Let A_{i}be the same as A__except__the i^{th}column is replaced with the column of constants from the linear system (the right side of the equations). Then, if det(A) ≠ 0, the i^{th}variable x_{i}is

If det(A) = 0, then the system has either no solutions or infinitely many solutions.x _{i}=

det

A _{i}

det

A

. - Using matrix multiplication, we can write a linear system as an equation with matrices:
AX = B, where - A is the
__coefficient__matrix of the linear system, - X is a single-column matrix of the
__variables__, - B is a single-column matrix of the
__constants__.

- A is the
- If A is invertible, then there exists some A
^{−1}we can multiply by that will cancel out A above, allowing us to get X alone. By computing A^{−1}and A^{−1}B, we will have solved the system. (Notice that we have to multiply A^{−1}from the left on both sides of the equation because matrix multiplication is affected by direction.) If A is not invertible (det(A) = 0), then the system either has no solutions or infinitely many solutions. - While all of these methods work great for solving linear systems, they all have the downside of being tedious: they take lots and lots of arithmetic. Good news! Almost all graphing calculators have the ability to do matrix (and vector) operations. You can enter matrices, then multiply them, take determinants, find inverses, or put them in reduced-row echelon form. Even if you don't have a graphing calculator, there are many websites where you can do these things for free.

__Note:__The ideas in this lesson can be rather difficult to follow with just words. The video will help explain this a lot, as it has a lot of visual diagrams to show what's going on step-by-step.

### Using Matrices to Solve Systems of Linear Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction
- Augmented Matrix
- Row Operations
- Interchange the Locations of Two Rows
- Multiply (or Divide) a Row by a Nonzero Number
- Add (or Subtract) a Multiple of One Row to Another
- Row Operations - Keep Notes!
- Gauss-Jordan Elimination - Idea
- Gauss-Jordan Elimination - Method
- Cramer's Rule - 2 x 2 Matrices
- Cramer's Rule - n x n Matrices
- Solving with Inverse Matrices
- The Mighty (Graphing) Calculator
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Introduction 0:12
- Augmented Matrix 1:44
- We Can Represent the Entire Linear System With an Augmented Matrix
- Row Operations 3:22
- Interchange the Locations of Two Rows
- Multiply (or Divide) a Row by a Nonzero Number
- Add (or Subtract) a Multiple of One Row to Another
- Row Operations - Keep Notes! 5:50
- Suggested Symbols
- Gauss-Jordan Elimination - Idea 8:04
- Gauss-Jordan Elimination - Idea, cont.
- Reduced Row-Echelon Form
- Gauss-Jordan Elimination - Method 11:36
- Begin by Writing the System As An Augmented Matrix
- Gauss-Jordan Elimination - Method, cont.
- Cramer's Rule - 2 x 2 Matrices 17:08
- Cramer's Rule - n x n Matrices 19:24
- Solving with Inverse Matrices 21:10
- Solving Inverse Matrices, cont.
- The Mighty (Graphing) Calculator 26:38
- Example 1 29:56
- Example 2 33:56
- Example 3 37:00
- Example 3, cont.
- Example 4 51:28

### Math Analysis Online

### Transcription: Using Matrices to Solve Systems of Linear Equations

*Hi--welcome back to Educator.com.*0000

*Today, we are finally going to see why we have been studying matrices--just how powerful they are.*0002

*We are going to use matrices to solve systems of linear equations.*0006

*Consider the following system of linear equations: x + y + z = 3; 2x - 2y + 3z = -4; and -x - z = 0.*0010

*Notice that, as long as we keep the variables in the same order for each equation (we can't swap it*0019

*to y + x + z; we keep it in xyz order every time), we could write these coefficients to all of the variables as a coefficient matrix.*0023

*What we have in front of this x is just a 1; in front of this y is just a 1; and in front of this z is just a 1.*0031

*So, we can write a first row of 1, 1, 1, all of these coefficients that are on that row right there of the equation.*0037

*For the next one, we have 2, -2, 3; so we put them all down here; so we have done that equation as the coefficients in that row, showing up there.*0045

*And since we can always trust that we are going to have the x, the y, and the z here,*0055

*because we are always staying in this order of x, y, z (that is why we have to keep the variables in the same order each time),*0060

*we can create this coefficient matrix; and finally, we have -1 here and -1 here.*0068

*And why do we have a 0? Well, if y doesn't show up, it must be because we have a 0y; so that is why we get a 0.*0074

*So, we have done all 3 equations, the coefficients to the variables in all three equations.*0081

*This idea of converting the information in a linear system into a matrix*0085

*will allow us to explore ways that we can have linear systems interact with matrices and vice versa.*0089

*How can a matrix allow us to solve a linear system of equations?*0094

*Our first idea is the augmented matrix; we can take this idea of the coefficient matrix and expand on it.*0098

*Instead of just representing coefficients for the equations, we can represent an entire linear system,*0104

*the solutions included, with an augmented matrix.*0109

*So, previously we didn't have the constants for the equations, what was on the right side of the equals sign.*0112

*So now, we have that show up on the right side over here.*0118

*So now, we have the coefficients, and we have what each of those equations is equal to.*0121

*So, each row represents an equation of the system: x + y + z = 3 is 1, 1, 1, 3, because that first 1*0126

*represents the coefficient on x; the first one on the y; the first one on the z; and it all comes together to equal 3,*0136

*because we know that just addition is what is going on between all of those coefficients, because it is a linear system.*0142

*Each left-side column gives a variable's coefficients: all of the coefficients to x show up here.*0147

*We have 1, 2, and -1 on the x, and we have 1, 2, and -1 in that column, as well, on the left side.*0153

*Where the variable does not appear, it has a coefficient of 0.*0160

*The 0 is to show that we have nothing for y here, because if you have nothing, we can think of it as just 0 times y.*0163

*The vertical line represents equalities; since this vertical line here is representing each of these, there is an equals sign here.*0171

*And finally, the right-side column, this column here, gives us...*0179

*This column of our matrix is the constant terms from our equations.*0183

*So, we have a way of converting that entire set of equations into a single matrix.*0187

*So now, we can look at how we can play around with this matrix to have it give up what those variables are equal to.*0192

*Since we can represent a linear system as an augmented matrix, we can do operations on the matrix the same way we interact with a linear system.*0199

*Anything that would make sense to do to a linear system, to an equation in a linear system,*0206

*should make sense--there should be a way to do it over on the matrix version,*0209

*because since we can do it to a linear system, and our augmented matrix is just showing us a linear system--*0213

*it is a way to portray a linear system--then if we can figure out a way to interact with our matrix*0220

*that is the same as interacting with a linear system, we know that it is just fine.*0224

*So, that gives us the idea of 3 row operations.*0228

*The first one is to interchange the locations of two rows.*0231

*If I have row 1 and row 2, I can swap their places.*0235

*Our next idea is to multiply or divide a row by a non-zero number.*0239

*I can multiply an entire row by 2 or by 5 or by -10--whatever I want to--*0243

*or divide by 2, because that is just the same as multiplying by 1/2.*0249

*And then finally, add or subtract a multiple of one row to another.*0253

*If I have row 2 here and row 5 here, I can have row 5 subtract twice on itself; so it is row 2 - 2(row 5).*0257

*I can have a multiple of one row subtract from another one.*0267

*So, why does this make sense--how is this like a linear system?*0270

*Well, all of these operations are completely reasonable, if we had a linear system.*0274

*If we have an equation here and an equation here, it is totally meaningless for us to swap the order that the equations come in.*0278

*The location of the equation isn't an important thing when we are working with a linear system.*0284

*We just have to look at all of them, so it doesn't matter which came first and which came last.*0289

*So, we can move them around, and it doesn't matter.*0292

*That means that we can move our rows around, and it doesn't matter, because it is just representing a linear system.*0294

*Similarly, multiplying both sides of an equation is just algebra.*0298

*If I have an equation, I can multiply 2 on the left side and 2 on the right side, and that is just fine.*0302

*You can think of that as multiplying each of the numbers inside of the equation.*0308

*If we multiply each of the numbers inside of the equation, that is the same thing as just multiplying an entire row of our augmented matrix.*0313

*Finally, adding a multiple of an equation is elimination.*0318

*Remember: when we were talking about linear systems at first, elimination was when we can add a multiple of one equation to another equation.*0322

*Well, that is the same thing as adding a multiple of one row to another row over in the augmented matrix version.*0327

*So, everything here has a perfect parallel between the two ideas.*0333

*Each one of our row operations makes total sense, if we were just working with a linear system.*0336

*And since our augmented matrix is just representing a linear system, they make sense over here with the augmented matrices, as well.*0341

*While row operations are a simple idea (each one of these is pretty simple--swap; multiply; or add a multiple--not that crazy)--*0348

*working with them will involve a lot of arithmetic and steps.*0356

*You are going to have a lot of calculations going on, and while none of them will probably be very hard calculations,*0360

*you are going to be doing so many that it is easy to make mistakes.*0364

*So, when you are working on row operations, when you are working on doing this stuff,*0368

*I want you to be careful with what you are doing, and counter the fact that you are likely to make mistakes*0372

*by noting each step: note what you just did for each step.*0376

*In addition to this being a good way to keep you from making mistakes, some teachers will simply require it,*0380

*and won't give you credit if you don't do it; and that is pretty reasonable,*0384

*because you will definitely end up making mistakes sooner or later if you don't do this sort of thing.*0386

*So, take a note of what you did on each step; show what you did between the first one, and then the second one,*0391

*and then the third one, by writing on the side what you just did.*0397

*This will make it easier to avoid making mistakes, and it will help you find any that manage to creep through.*0401

*If you get to the end, and you see that this doesn't make sense--something must have gone wrong--*0405

*you can go up and carefully analyze each of your steps, and figure out where you made a mistake.*0408

*Or maybe things actually do come out to be weird for some reason.*0413

*So, here are some suggested symbols for each of the three row operations.*0416

*You don't have to use them; use whatever makes sense to you (and if your teacher cares about it, whatever makes sense to your teacher).*0420

*But these work well for me, and I think they make sense, pretty clearly.*0424

*If we are interchanging row i and row j (we are swapping their locations), I like just a little arrow, left-right, between them.*0428

*We show some row by using a capital R to talk about a row, and then the number of it.*0435

*For example, if we want to talk about the second row, we would just talk about it as R _{2}.*0440

*If we wanted to talk about the ninth row, it would be R _{9}; and so on, and so on.*0444

*So, if we are swapping row i and row j, we have R _{i}, little arrow going back and forth, R_{j}.*0450

*If we want to multiply row i by the number k (we are multiplying by k), we just have k times row i--as simple as that.*0455

*And finally, if we are adding a multiple of row i to row j, then we have kR _{i} (k times row i,*0463

*the thing that we are adding the multiple of) to what it is being added to.*0472

*We will see these pretty soon, when we are actually starting to see how this stuff gets done.*0477

*Gauss-Jordan elimination: here is an idea: since all of our row operations make sense for solving a linear system--*0482

*they all make sense for a way to do it--we can apply them to find the value of each variable.*0489

*If we manage to get our augmented matrix in a form like this form right here, we would immediately know what each variable is; why?*0494

*Well, notice: this first row here has 1, 0, 0; well, 1 here would correspond to the x here.*0505

*And then, this one here would be non-existent, because there are two 0's there; so there would be no y; there would be no z.*0511

*And we know that it is equal to -17.*0516

*The exact same thing is going on here; we know that y, because that is the y column, must be equal to 8 (nothing else shows up),*0520

*and that z (since that is the z column) must be equal to 47.*0528

*We have managed to solve the system by just moving stuff around in this augmented matrix.*0531

*We know that that has to be true, because that augmented matrix must be equivalent to the linear system,*0536

*because we turned our linear system into an augmented matrix, and then we had all of these row operations*0541

*that are just the same as working with a linear system.*0546

*So, what we have here is still the same as our original linear system.*0548

*So, we can convert back to the linear system and see what our answers are.*0551

*We call this format for a matrix reduced row echelon form; it is a mouthful, and it is kind of confusing at first.*0556

*Echelon has something to do with a triangle shape; it is "row echelon" because the rows are kind of arranged in a triangle,*0563

*and "reduced" because they all start with 1's, and there are 0's...*0569

*Honestly, don't really worry about it; just know reduced row echelon form, that really long name, and you will be fine.*0572

*While it has a formal definition (there is a way to formally define it), it is going to be enough*0577

*for us to just think of it in this casual way, where it is a diagonal of 1's.*0581

*That main diagonal will have all 1's on it; it will start at the top left, and it will continue down diagonally.*0585

*And it will have 0's above and below.*0591

*So, we have 1's along the main diagonal, and above and below our 1's, there will be 0's.*0593

*So, we have a 0 here above the 1's and a 0 here below the 1's.*0603

*We have 0's here above the 1's and a 0 here below the 1's.*0607

*Finally, the numbers on the far right can be any number.*0611

*Over here, -5, 8...it doesn't matter; over here, we manage to have two columns,*0616

*because the 1 diagonal's part of the reduced row echelon form is just an identity matrix.*0620

*So, it can't get any farther than whatever the square portion of it would be.*0628

*So, we can end up having multiple columns, as well.*0632

*But in our case, when we are working on this elimination to solve linear systems, we will always only have a single column on the far right.*0634

*Anyway, the point is that we have this diagonal of 1's with 0's above and below; and the stuff on the far right can be any number.*0641

*Now, from what we have just discussed, if we can use row operations to put an augmented matrix in reduced row echelon form,*0648

*like this one right here, we will solve its associated linear system,*0653

*because we will know that, since there is only one of a variable in that row, it must be equal to the constant*0657

*on the other side of the augmented matrix--one the right side of the augmented matrix, past that vertical line that shows equality.*0662

*We will have solved its associated linear system.*0669

*Gauss-Jordan elimination is just named after the people who created it.*0672

*It is a method we can follow to produce reduced row echelon form matrices through row operations to solve linear systems.*0676

*It is just a simple method of being able to get 0's to show up and 1's to show up on the diagonal, and then we are done.*0683

*So, it is just a method that we can follow through that will always end up resulting in a reduced row echelon form.*0689

*All right, so let's see how it is done.*0695

*The very first thing you do for Gauss-Jordan elimination is: you take your linear system, and you write it in augmented matrix form.*0697

*So, we have our augmented matrix form over here.*0705

*We look at the coefficients; we convert them over; the line to show the equals signs is here, and then our constants are on the far right side.*0707

*All right, the next thing: you use row operations to attain a 1 in the top left*0715

*(we start in the top left here), and then 0's below.*0722

*We get a 1 here; and then we wanted 0's below it, because it is a 1, and 0's are above and below.*0726

*So, we start working with 1's, creating 1's...well, sorry, from your point of view...creating 1's and creating 0's underneath them.*0731

*So, we first get a 1 up here; and then we create the 0's underneath it.*0740

*Once we have done that, we move on to creating the next one diagonally down and doing 0's below that.*0746

*And we just keep repeating until we have made it all the way down the diagonal.*0751

*So, you just keep going until you are all the way down the diagonal, creating 1's and creating 0's below.*0754

*First, we started with a 1 up here; we already have the first thing done.*0760

*So, our next step is...how do we get this stuff to turn into 0's?*0764

*We do row operations: we want to get rid of this 2, so since row 1 has a 1 there, we subtract by 2(row 1).*0767

*2 row 1 gets us -2 here, -2 here; this becomes a 0; -2 times 1 on -2 gets us -4; 1 times -2, added to 3, gets us 1; and -2 times 3, added to -4, gets us -10.*0776

*Next, we are adding row 1, because we need to get rid of this -1 here.*0795

*So, we add row 1, because it is positive 1; add +1 to -1; we get 0; add +1 to 0; we get 1; add +1 to -1; we get 0; add +3,*0800

*because it is just a multiple--how many times we are adding whatever is on that first row, because it was just 1 of row 1-- 3 on 0 gets us 3.*0813

*All right, at this point, we have 0's below; great.*0821

*So now, we are ready to move on to the next step in the diagonal.*0824

*All right, our next step in the diagonal is going to be this -4 here.*0828

*We want to get that to turn into a 1.*0832

*We could do this by manipulating it with canceling things out, or by dividing both sides of that entire row by -4,*0834

*or multiplying the entire row by -1/4; but we might notice that we already have a 1 here.*0844

*Well, let's just use that: if we swap these two rows, we will manage to have a 1 in our next location; great--we do that instead.*0850

*So, we make row 2 swap with row 3, because up here, they used to be row 2 and row 3.*0858

*And now, they take their new spots; they swap locations; and we have our new matrix right here.*0864

*The next step: we see that we have the 1 here, so our next step is...we need to turn everything below the 1's to 0's on this first portion.*0871

*So, how do we do that? Well, we can add +4 times row 2, because we have a 1 here.*0882

*So, we add 4 times that, and that will cancel out the -4.*0887

*4 times 1, added to this, gets us 0; 0 times 4, added to 0--we still have 0.*0891

*0 times 4, added to this--we just have 1 still; 3 times 4, added to -10, gets us 2.*0897

*So, at this point, notice that we have nothing but 1's on our main diagonal, and we have 0's all below it.*0903

*However, we don't have 0's above it yet; there are things other than that; so that is the next step.*0910

*Once you have 1's all along the main diagonal and 0's all along below that main diagonal,*0917

*the next step is to cancel out the stuff above it.*0927

*Our first thing is: we work from the bottom right of the diagonal, and we cancel out above the 1's.*0930

*You work your way up: so our first step is to cancel out everything here and here.*0940

*But we notice that, because this row is 0, 1, 0, we can actually do it in one step,*0944

*where we can subtract one of row 3 and one of row 2 (it is getting kind of hard to see, with all of those colors there).*0948

*So, subtracting one of row 3: we have 1 here, minus 1; so now we subtract 2 on this, so we have 3 - 2 so far,*0954

*for what is going to show up here; let's also subtract by row 2; row 2...minus here...*0964

*1 - 1 comes out to be 0; since it is zeroes everywhere else on that row, we don't have to worry about them interfering,*0971

*except for over here; we have it subtracting by another 3; so 3 - 2 - 3 comes out to be equal to -2, and we get -2 here.*0976

*At this point, we have reduced row echelon form.*0986

*We have 1's on the main diagonal and 0's above and below, so we can convert this.*0990

*Our x here becomes -2; the representative -1 of y here becomes 3, so we have y = 3.*0995

*And the representative 1z becomes z = 2; so now we have solved the thing.*1002

*And I want you to know: if there is no solution, or infinitely many--if our linear system can't be solved,*1007

*or it has infinitely many solutions--this method will end up not working.*1013

*You will not be able to achieve reduced row echelon form if there is no solution or infinitely many solutions.*1017

*So, that is something to keep in mind.*1023

*All right, a new way to do this: we are going to look at a total of 3 different ways to solve linear systems, each using matrices.*1026

*Another way to do this is through Cramer's Rule.*1033

*We can find the solutions to a linear system by this rule.*1035

*Given a two-variable system (we will start with 2 x 2, until we get a good understanding of what is going on),*1037

*where we have the a's (each of our a's here is just a constant), a _{11}x, a_{12}y, a_{21}x, a_{22}y,*1042

*is equal to constants on the right side (b _{1} and b_{2} are also constants),*1050

*we can create a normal coefficient matrix (A is the normal coefficient matrix).*1055

*All of our coefficients, a _{11}, a_{12}, a_{21}, a_{22}...*1063

*show up just like they would normally in a coefficient matrix.*1068

*Now, A _{x}...what it is going to do is take this column here on a, and it is going to replace it with the constants to the equation.*1071

*It is going to replace a _{1}, a_{21} with this; and so, we have b_{1}, b_{2} for that column.*1083

*And then, the rest of A is like normal.*1089

*Similarly, for y, we are going to swap out the y constants from A.*1092

*And so, we are going to have A like normal, except we swap out the constant column for where the y variables were occupying.*1098

*The y column gets swapped to the constant column.*1105

*So, the constant column goes in there.*1108

*Notice that A _{x} is just like A, except that it has this constant column replacing the x column.*1111

*Similarly, A _{y} has the constant column replacing the y column from our normal coefficient matrix.*1119

*All right, that is the idea; now if the system has a single solution, if it comes out to be just one solution--*1125

*it isn't infinitely many; it isn't no solutions at all; if the system has a single solution,*1131

*then x will be equal to the determinant of A _{x}, over the determinant of A,*1137

*the determinant of its special matrix, divided by the determinant of the general coefficient matrix.*1143

*Similarly, y is going to be equal to the determinant of its special matrix, divided by the determinant of the general coefficient matrix.*1150

*OK, this method can be generalized to any linear system with n variables.*1160

*Let A be the coefficient matrix for this n-variable linear system.*1165

*Let A _{i} be the same as A, except the i^{th} column; the column that represents*1169

*the variable we are currently working with--the variable that we want to solve for--that column*1177

*will be replaced with the column of constants.*1182

*So, we replace the column for the variable we are interested in solving for with the column of constants.*1185

*And that makes A for whatever variable we were looking for: A _{i} in this case,*1191

*if we are looking for the i ^{th} variable, up until we are looking for the i^{th} one.*1195

*We replace it with the column of constants from the linear system, just the right side of the equations,*1202

*the b _{1}, b_{2} on our previous 2 x 2 example.*1206

*OK, with that idea in mind, if the determinant of A, the determinant of our normal coefficient matrix, right up here,*1210

*is not equal to 0, then the i ^{th} variable, x_{i}, this variable we are trying to solve for,*1216

*is equal to the determinant of its special matrix that has that column replacing it,*1223

*divided by the determinant of the normal coefficient matrix; that is what we have right here.*1227

*It is just like in the 2 x 2 form, except we can do it on a larger scale, as well.*1232

*You swap out this one column; you take the determinant of that special matrix; you divide it by the determinant of the normal coefficient matrix.*1236

*And that gives you the variable for whatever column you had swapped.*1242

*We will get the chance to see this done on a more confusing scale (which is the sort of thing*1247

*that we want to be able to understand--this on a larger scale) in Example 3.*1251

*And we will just see this get applied normally in Example 2 for a 2 x 2 matrix.*1255

*Also notice: if the determinant of A is equal to 0, then the system will have either no solutions or infinitely many solutions.*1259

*All right, the final method to do this: we can solve with inverse matrices.*1266

*This one is my personal favorite for understanding how this stuff works; I think it is the easiest to understand.*1271

*But that is maybe just me.*1276

*Using matrix multiplication, we can write a linear system as an equation with matrices.*1278

*How can we do this as an equation?*1282

*It made sense with an augmented matrix, because we talked about the special thing.*1284

*But how is 1, 1, 1, 2, -2, 3, -1, 0, -1--notice that that is just our normal coefficient matrix A showing up here--*1287

*if we multiply it by the column matrix x, y, z equals our coefficient column here, that ends up being just the same--*1298

*it is completely equivalent; these two ideas here are completely equivalent--multiplying the matrices*1313

*versus the linear system, and the linear system versus the matrices being multiplied together; they are completely the same.*1319

*Let's see why; let's just do some basic matrix multiplication on this.*1323

*What is going to come out of this? We have a 3 x 3 (3 rows by 3 columns), 3 rows by 1 column;*1327

*so yes, they match up, so they can multiply; that is going to produce a 3 x 1, 3 row by 1 column, matrix in the end.*1335

*So, let's see what is going to get made out of this.*1342

*We are going to have a 3 x 3; our first row times the only column is 1, 1, 1, 1; so I'll make this a little bit larger,*1346

*so we can see the full size of what is going to go in...1 times x + 1 times y + 1 times z is x + y + z.*1356

*Next, 2, -2, 3 on x, y, z gets us 2x - 2y + 3z.*1369

*Finally, -1, 0, -1 on x, y, z gets us -x + 0y (so let's just leave it blank) - z.*1378

*Now, if we know that that is equal to our coefficient matrix, because we said it from the beginning,*1389

*then all we are saying...3, -4, 0...well, for two matrices to be the same thing, for them to be equal to each other,*1395

*every entry in the two matrices has to be equal to its entry in the same location.*1403

*So, the top one, x + y + z, equals 3; that is just the exact same thing as this.*1409

*2x - 2y + 3z has to be equal to -4; well, that is the same thing as saying 2x - 2y + 3z = -4.*1416

*And the same thing: -x - z is saying it is equal to 0 through the matrices; and that is the same thing it was saying by the linear system.*1424

*So, the linear system, taken as a whole, is just the same thing as taking the coefficient matrix,*1430

*multiplying it by this coefficient column matrix...and that is going to come out to be equal to our constant matrix,*1434

*which was the constants for the equations; that is the idea that is going to really be the driving force behind using inverse matrices.*1444

*All right, we can symbolically write this whole thing as Ax = B; A times x equals B,*1451

*where A is the coefficient matrix right here; then X is a single-column matrix of the variables;*1463

*that is our x, y, z; whatever variables we end up using are going to go like that; and then finally,*1476

*B is a single-column matrix of the constants (3, -4, 0 gets the same thing right here); OK.*1484

*Notice: if we could somehow get X alone, if we could get our variable matrix, our variable column, alone on one side,*1492

*whatever it was on the other side, if it equals numbers on the other side in a matrix,*1499

*then we would have solved for it, because we would say that x is equal to whatever the corresponding location is on the other side;*1503

*y is equal to whatever its corresponding location is on the other side; z is equal to whatever its corresponding location is on the other side.*1508

*We would have solved for this.*1514

*So, if we can somehow get X alone, we will be done; we will have figured out what x, y, and z are equal to.*1515

*How can we do that, though--how can we get rid of A? Through inverse matrices!*1521

*That is no surprise, since this thing is titled Inverse Matrices.*1525

*We cancel out A; if A is invertible, then there exists some A ^{-1} that we can multiply that by that will cancel out A.*1527

*So, we started with Ax = B; we can multiply by A ^{-1} on the left side on both sides.*1537

*Remember: if you multiply by the left and the right, for matrix equations that doesn't work.*1542

*You have to always multiply both from the left or both from the right.*1548

*You are not allowed to do them on opposite sides; they have to both be coming from the same side when you multiply.*1553

*We multiply by A ^{-1} on the left side on both cases; the A^{-1} here and the A cancel out, and we are left with just X = A^{-1}B.*1558

*So, if we can compute A ^{-1}, and then we can compute what is A^{-1} times B, we will have solved our system.*1566

*We will have what our system is equal to.*1573

*Just make sure that you multiply from the same side for your inverse on both sides of the equation.*1575

*You have to multiply both from the left; otherwise it won't work out.*1579

*Finally, if A is not invertible (if the determinant of A is equal to 0, then you can't invert it),*1582

*then that means that the system has either no solutions or infinitely many solutions.*1588

*All right, let's try putting these things to use.*1593

*Oh, sorry; before we get into using them, the mighty graphing calculator:*1596

*all of these methods work great for solving linear systems: augmented matrices with Gauss-Jordan elimination,*1601

*Cramer's Rule, inverse matrices--they are all great ways to solve linear systems.*1606

*But they all have the downside of being really tedious; they take so much arithmetic to use.*1611

*We can work through it; we can see that we can do this stuff; but it is going to take us forever to actually work through this stuff by hand.*1617

*I have great news; it turns out that, if you have a graphing calculator, you can already do this right now, really fast, really quickly, and really easily.*1623

*Almost all graphing calculators have the ability to do matrix and vector operations.*1631

*You can enter matrices into your calculator, and then you can multiply them; you can take determinants of the matrices;*1636

*you can find inverses; or you can put them in reduced row echelon form.*1641

*Look on your graphing calculator, if you have a graphing calculator, for something that talks about where...*1645

*just look for a button about matrices; look for something like that.*1649

*And it will probably have more information about how to create a matrix, and then how to do things with it.*1652

*Inverse is probably just raising the matrix to the -1, and it will give out the value.*1657

*Each graphing calculator will end up being a little bit different for how it handles inputting the matrices.*1661

*But they will almost all have this ability, for sure.*1665

*If you have real difficulty figuring out how to do it on your calculator, just do a quick Internet search for "[name of your calculator] put in matrices."*1669

*Use "matrices," and you will be able to figure out an easy way to do it very quickly.*1675

*Someone has a guide up somewhere.*1678

*Also, if you don't have a graphing calculator, don't despair; it is still possible to get this stuff done really easily and really quickly.*1680

*There are a lot of websites out there where you can do these things for free.*1687

*Just try doing a quick Internet search for matrix calculator; just simply search the words matrix and calculator,*1690

*and the first 5 hits or so will all be matrix calculators, where you can plug in matrices,*1698

*and you can normally multiply them, or you can take their determinants, or you can get their inverses.*1704

*Or you can do other things that you don't even know you can do with matrices yet.*1708

*But just look for the things that you are looking for; there are lots of things you can do with it.*1711

*Do a quick Internet search for the words matrix and calculator, and you will be able to find all sorts of stuff for those.*1714

*So, even if you don't have a graphing calculator, there are lots of things out there.*1721

*If you are watching this video right now, you can go and find websites that will let you do this for free.*1724

*Finally, while this is great that we can do all of this stuff with a calculator,*1729

*and the calculator will do the work for us, I still want to point out that it is important to be able to do this stuff without a calculator.*1732

*So, it makes it so much easier to be able to use a calculator; but we still have to understand what is going on underneath the hood.*1738

*We don't have to constantly be using it, but we have to have some sense of what is going on under the hood*1744

*if we are going to be able to understand more, higher, complex-level stuff in later classes.*1749

*So, you want to be able to understand this stuff, just because you want to be able to understand things,*1754

*if you are going to be able to make sense of things that come later.*1757

*And also, you usually need to show your work on your tests.*1759

*Your teacher is not going to be very happy if you are taking a test, and you just say, "My calculator said it!"*1762

*You are not going to get any points for that.*1767

*So, you can't just get away with it all the time.*1769

*That said, it can be a great help for checking your work, so you can work through the thing by hand,*1771

*and then just do a quick check on your calculator to tell you that you got the problem right; that is really useful on tests.*1775

*Or if you are dealing with really huge matrices, where it is 4 x 4, 5 x 5, 6 x 6, or even larger,*1780

*where you can't reasonably be able to do that by hand, you just use a calculator, and that is perfectly fine.*1786

*All right, now let's go on to the examples.*1791

*The first example: Using Gauss-Jordan elimination, solve 2x + 5y = -3, 4x + 7y = 3.*1793

*Our very first thing to do is: we need to convert it into an augmented matrix.*1799

*We have 2 as the first coefficient on the x, and then 5 as the first coefficient on the y, and that equals -3.*1803

*So, there is our bar there; 4, 7, 3; we have converted it into an augmented matrix.*1809

*Our coefficients are on the left part of the matrix, and our constant terms are on the right part of the matrix.*1818

*All right, at this point, we just start working through it.*1825

*The very first thing that we need to do is to get that to turn into a 1--get the top left corner to turn into a 1.*1828

*So, we will do that by multiplying the first row: 1/2 times row 1.*1834

*All right, that is what we will do there: 2 times 1/2 becomes 1; 5 times 1/2 becomes 5/2; -3 times 1/2 becomes -3/2.*1839

*4, 7, 3; the bottom row didn't get touched, so it just stays there.*1849

*The next thing to do: we want to get this to turn into a 0.*1853

*So, we will subtract the top row; the top row is not going to end up doing anything on this step,*1858

*but we will subtract the top row 4 times, because we have 4 here; so - 4R _{1} + our second row.*1866

*-4 times 1 plus 4 gets us 0; -4 times 5/2 gets us -10; -10 + 7 gets us -3; -3/2 times -4 gets us +6; we got +6 out of that, so that gets us +9.*1877

*Our next step: we want to get this to turn into a 1; we will bring this whole thing up here.*1900

*The next step is to get the second row to turn into a 1: 1, 5/2, -3/2.*1908

*We multiply the bottom part by -1/3 times row 2; so 0 times -1/3 is still 0; -3 times -1/3 becomes +1; 9 times -1/3 becomes -3.*1916

*At this point, we can now turn this into a 0; we don't need to do anything to our bottom row; it is still 0, 1, -3.*1930

*But we will add -5/2 of row 2 to row 1; so 0 times anything is still going to be 0; so added there...it is still 1 there.*1940

*Then, -5/2 on 1 + 5/2 becomes 0; -5/2 times -3 becomes +15/2, and then still -3/2.*1953

*Let's simplify that: we have 1, 0, 0, 1...15/2 - 3/2 becomes 12/2; 12/2 is 6...6, -3.*1963

*So, at this point, we can convert that into answers; x = 6; y = -3.*1975

*There are our answers; however, we did have to do a whole lot of calculation to get to this point.*1983

*And it could be even more if we were working on a larger augmented matrix; they get big really fast.*1987

*So, it might be a good idea to do a quick check; let's just check our work and make sure it is correct.*1992

*Let's plug it into the first one: 2 times 6, plus 5 times -3; what does that come out to be?*1998

*We hope it will come out to be -3: 12 + -15 = -3; indeed, that is true.*2004

*We could check it again with this equation, as well, if we want to be really, really extra careful.*2011

*4 times 6, plus 7 times -3, equals positive 3; 24 - 21 = 3; that is true.*2015

*So, both of our checks worked out; we know that x = 6, y = -3; that is definitely a solution--great.*2025

*All right, the second example: let's see Cramer's Rule in action.*2031

*The first thing we want to do, if we are going to use Cramer's Rule, is: we need to get a coefficient matrix going.*2034

*A =...what are our coefficients here? We have a 2 x 2: 2, 5, 4, 7.*2039

*There are the coefficients; our next step is...we want A _{x}--what is A_{x} going to be?*2048

*Here is our x column; we are going to swap that out for the constants here.*2055

*-3 and 3 replaces what had been our x column here.*2062

*And then, the rest of it is just like normal; so we replace that one column, but everything else is just the same.*2067

*A _{y}: what will A_{y} be?*2073

*The same sort of thing, except now we are replacing the y column--what is the y column going to turn into?*2075

*It is going to also become -3, 3; so 2, 4 is just as it was before; the first column is still the same, because that is the x column.*2081

*But now we are swapping out the y column, so it becomes the constants, -3 and 3.*2088

*All right, so we were told that Cramer's Rule says that x is equal to the determinant of its special, swapped-out matrix, A _{x},*2093

*divided by the determinant of the normal coefficient matrix.*2103

*The determinant of -5, 3, 5, 7, divided by the determinant of 2, 5, 4, 7:*2107

*-3 times 7 gets us -21; minus 3 times 5 (is 15), over 2 times 7 (is 14), minus 2 times 5 (is 20);*2118

*we have -36/-6 =...that comes out to be positive 6, so we now have x = +6.*2130

*That checks out with what we just did in the previous one.*2140

*If you didn't notice, these equations are the same as what they were in the previous example,*2142

*so we are just seeing two different ways to do the same problem*2146

*(well, at least the part where we are trying to solve for x and y).*2149

*So, that checks out, because we checked it in the previous problem.*2153

*Now, what about y? y is going to be the same thing; y is equal to the determinant...same structure, at least...*2155

*of A _{y}, its special matrix, divided by the determinant of A, once again.*2162

*We could calculate the determinant of A, but we already calculated the determinant of A.*2168

*We figured out that it comes out to be -6, so we just drop in -6 here.*2173

*You only have to do it once; it is not going to change--the determinant of A will stay the determinant of A, as long as A doesn't change.*2177

*Then, the determinant of A _{y}: we don't know what A_{y}, that special matrix, is, yet.*2183

*2, -3, 4, 3: 2 times 3 is 6, minus 4 times -3 (-12); that cancels out; so we have 6 + 12,*2187

*divided by -6; 18/-6 equals -3; so y comes out to be -3.*2197

*Once again, that is the same answer as we had on the previous example, so we know that this checks out.*2205

*If you had just done this for the very first time, I would recommend doing a check,*2210

*because once again, you have to do a lot of arithmetic to get to this point.*2213

*And it is going to be even more if you are doing a larger Cramer's Rule, like, say, this one.*2216

*All right, so if we are working on this one, we only have to solve for the value of y.*2220

*There is one slight downside to only solving for one variable.*2224

*It means you can't check your work, because we can't plug in y and be sure that it works out to be true.*2227

*But we can at least get out what it should be.*2231

*All right, using Cramer's Rule, the first thing we need to do is figure out what our coefficient matrix A is.*2234

*A =...all of our first variables is w, so 2w, 3w...there is no w there at all, so it must be 0w; -2w.*2240

*Next, our x's: there are no x's in the first equation, so it is 0 x's; -2, -3, +5.*2250

*Next, 4y, 1y, 2y, 0y, -1z, 4z, 0z, 1z.*2258

*Great; now, we are looking to figure out what y is going to be.*2270

*We figure out that A _{y} is going to be the same thing, except it is going to have its y column swapped.*2274

*Notice that the y column is the third column in; so we are going to swap out the third column for the constants.*2281

*Other than that, it is going to look like our normal coefficient one.*2287

*So, we can copy over what we had in the previous one, except for that one column: 2, 3, 0, -2, 0, -2, -3, 5.*2289

*Now, this is the third column--this one right here--so we are swapping out for the column of constants.*2301

*That is 5, 16, 0, 17; and then back to copying the rest of it: -1, 4, 0, 1.*2309

*So, at this point, we have A _{y}; we have A; so we are going to need to figure out determinants.*2317

*First, let's figure out what the determinant of A is.*2322

*The determinant of A: if we want to figure out this here, remember: if we are going to be figuring out determinants,*2324

*we are going to be using cofactor expansion; so the very first thing we want to do is make a little +/- field: + - + -, - + - +, + - + -, - + - +.*2331

*We can use that as a reference point.*2346

*Which one would be the best one if we are looking to get the determinant of A?*2348

*If we are looking to get the determinant of A, which would be the best row or column to expand on?*2354

*I see two 0's on this column, so let's work off of that one.*2358

*The first 0 just disappears, because it is 0 times its cofactor; blow out that cofactor, because it is 0.*2362

*The next one is -3; so we are on the third row, second column, so that corresponds to that symbol, a negative.*2368

*It is negative, and then -3, what we have for what we are expanding around; negative -3 is minus -3;*2376

*and then times...what happens if we cut out everything on a line with that -3?*2384

*We have 2, 3, -2, 4, 1, 0, -1, 4, 1.*2390

*We have to keep going; now we are on the 2; let's swap to a new color.*2400

*2 here; we are on a + now, so it is +2 times...cut out what is on a line with that 2; so we have 2, 3, -2, 0, -2, 5, -1, 4, 1.*2406

*All right, at this point, we want to figure out what are the easiest rows or columns to expand on for these two matrices.*2426

*I notice that there is a 0 here and a 0 here; I personally find it easier to do expanding*2433

* based on a row than based on a column, so I will just choose to do rows.*2440

*- -3; these cancel to +3; so +3 times...expand on -2 first, so...*2445

*oh, and we are on a 3 x 3 now, so we are on that + there; so it is still positive...*2452

*it is 3 times...now we are figuring out the determinant of that matrix: -2 times...*2457

*cross out what is on a line with that: 4, -1, 1, 4; minus 0...but - 0 cancels out, so what is next after that?*2462

*Another plus: + 1 times...cross out what is on a line with that 1 here; 2, 4, 3, 1.*2471

*All right, let's work on our other half, the other determinant.*2485

*+2, times whatever the determinant is inside of this matrix: 2 here; 2 is a positive here,*2489

*because it is in the top left: so 2 times...whatever is on a line with that gets cancelled out.*2495

*So, we are left with -2, 4, 5, 1.*2504

*Then, 0 next: the 0 here we don't have to worry about.*2508

*And then, we are finally onto a +; but it is a negative 1, so + -1 times...cross out what is on a line with the -1; we are left with 3, -2, -2, 5.*2511

*OK, at this point, we just have a lot of arithmetic to work through.*2525

*3, -2...take the determinant of this matrix; we have 4 times 4; that is 16; 16 - -1 gets us +17.*2530

*Plus 1 times...forget about the 1...2 times 1 is 2, minus 3 times 4 is 12; so 2 - 12 is -10.*2540

*Plus 2 times...-2 times 1 is -2; minus 5 times 4 (is 20); so -2 - 20 is -22.*2550

*Plus -1...let's make it a negative...times...3 times 5 is 15; minus...-2 times -2 is +4; so 3 times 5 is 15, plus 4 is 19.*2560

*If you have difficulty doing that in your head, just write out the 2 x 2, as well.*2577

*Keep working through this: 3 times...-2 times 17 comes out to be -34; -34 - 10 + 2 times -22 (is -44), minus 19;*2582

*3 times -44 + 2(-44) - 19 becomes...oops, a mistake was made...oh, here it is.*2602

*I just caught my mistake--see how easy it is to make mistakes here?*2622

*That should be an important point: be really, really careful with this; it is really easy to make mistakes.*2625

*3 times 5 is 15, minus...-2 times -2 (let's work this one out carefully)...3 times 5, minus -2 times -2...*2629

*well, these cancel out, and we are left with +4; so 15 - 4 becomes 11.*2640

*So, this shouldn't be 19; it should be 11.*2644

*This shouldn't be a 19 here, either; it should be 11.*2649

*So, -44 - 11 becomes -55; see how easy it is to make mistakes?*2652

*I make mistakes; it is really easy to make mistakes; be very careful with this sort of stuff.*2658

*It is really, really a sad way to end up missing things, when you understand what is going on, but it is just one little, tiny arithmetic error.*2662

*All right, let's finish this one out.*2670

*3 times -44 becomes -132; plus 2 times -55 becomes -110; we combine those together, and we get -242.*2671

*-242 is the determinant of A; it is equal to -242; it takes a while to work through, doesn't it?*2691

*All right, the next one: We figured out the determinant of A; that comes out to be -242.*2701

*So, to use Cramer's Rule, we know that y is going to be equal to the determinant of A _{y}, over the determinant of A.*2706

*We now need to figure out what A _{y} comes out to be.*2716

*The determinant of A _{y}...let's figure this out.*2720

*We work through this one; I notice this nice row right here--we have three 0's on it.*2727

*That is going to make it easy to work through; if we are doing a cofactor expansion, we want to make our sign table.*2733

*OK, we can work along with that.*2744

*Our first one is a + on 0, but that doesn't matter; the next one is a - on -3.*2745

*- -3 on...we cut out what is on a line with that -3; we have 2, 3, -2, 5, 16, 17, -1, 4, 1.*2752

*Next is 0; once again, we don't have to worry about that.*2772

*Next is 0; once again, we don't have to worry about that.*2774

*All right, so we see that these cancel out, and we have 3 times...now we need to choose what we are going to expand along--which row or column.*2777

*Personally, I like the top row; I like expanding along rows, and 2, 5, -1 does at least have some kind of small numbers.*2786

*So, I will expand across that, just because I feel like it.*2793

*2, 5, -1: we will do it in three different colors here.*2797

*2 corresponds to this, so it is a positive 2, times what cuts along this...we are left with 16, 17, 4, 1.*2801

*The next one (do it with green): that corresponds to a negative there, so that is minus 5;*2814

*what does it cut out? We are left with 3, -2, 4, 1.*2822

*And then finally, go back to red; -1 corresponds to a positive, so + -1 times...what does it cross out? We have 3, -2, 16, 17 left.*2831

*All right, let's work this out: we have 3 times all of this stuff; 2 times...16 times 1 is 16; minus 17 times 4...*2850

*17 times 4 comes out to be -68; the next one: minus 5 times...3 times 1 is 3*2860

*(after that mistake last time, let's be careful) minus...-2 times 4 is -8; so that will cancel out to plus.*2874

*Finally, we turn this to a minus, since it was times -1; 3 times 17 comes to 51; minus -2 times 16 becomes -32.*2882

*OK, keep working this out: 3 times 2 times 16 - 68 is -52,*2899

*minus 5 times 3 + 8 is 11, minus 51 - -32 becomes 51 + 32; 51 + 32 is - 83.*2911

*So, 3 times...2 times -52 becomes -104; minus 5 times 11 becomes -55; and still, minus 83.*2929

*We combine all of those together, and that gets us 3 times -242.*2941

*Now, you could go through and multiply this together, and you would get a number out of it.*2948

*But notice: we have -242 here, and later on, in just a few moments, we are about to divide by that previous detA at -242.*2953

*So, why don't we just leave this as 3 times -242; that is equal to the determinant of A _{y}, our special matrix for A_{y}.*2962

*At this point, we know from Cramer's Rule that y equals (cut out a little space for it)*2976

*the determinant of its special matrix, A _{y}, divided by the determinant of the coefficient matrix.*2983

*We figured out that the determinant of our special matrix is 3(-242), so 3(-242) divided by the determinant*2991

*of our coefficient matrix--that is also -242; -242/-242--those parts cancel out, and we are left with 3; so y = 3.*3002

*Sadly, there is no good way to check it at this point, if we are going to have to work through the whole thing,*3013

*because we would have to solve for each one of them, w and x and z.*3018

*On the bright side, solving for w, x, and z is only having to figure out the determinant of A _{w}, A_{x}, and A_{z},*3023

*because we have already figured out the determinant of A.*3030

*But still, it clearly takes some effort to take the determinants of even just a size 4 x 4, so it is pretty difficult.*3031

*However, if you have a graphing calculator, it would be pretty easy to go through and enter the matrix,*3036

*and then enter an augmented matrix, including the constants, and then get the reduced row echelon form*3042

*and see if y = 3 pops out as the answer that you would have from it.*3049

*It would be the case that that is what you would get out of it.*3052

*Or you could use Cramer's Rule and do determinants: figure out what A _{x} is; figure out what A_{w} is;*3055

*figure out what A _{z} is; and then, be able to plug them all in and check afterwards.*3061

*Or you could also go through and do it with inverse matrices and see if y comes out to be 3, once you have figured out that on your calculator.*3066

*You can do this stuff by hand if you have to do it on a test;*3073

*but then, you can also, if you are allowed to just use the calculator (you just have to show your work)...*3075

*you can check your work in a second, different way to make sure that your work did come out to be true.*3079

*So, you can definitely get the problem right.*3083

*All right, the final example: do you remember that monster from solving systems of linear equations?*3086

*It is back, and we are going to solve it: we are going to knock out this thing that was way too difficult for us then.*3090

*It is going to be really easy for us now, because we have access to how inverse matrices work.*3095

*We can use calculators to be able to calculate an inverse matrix very quickly; this thing is going to be easy.*3099

*Our plan: remember, the idea was that we have the coefficient matrix A, times the column of the variables, is equal to the column of the constants.*3104

*So, AX = B: if we can figure out what A ^{-1} is, we can multiply by A^{-1} on the left side on both cases.*3117

*A ^{-1} cancels out there, and we are left with X = A^{-1}B.*3125

*We already know what B is: B is this thing right here, so that part is pretty easy.*3134

*Can we figure out what A ^{-1} is?*3139

*Well, this is A; I am assuming that we have access to a graphing calculator or some way to do matrix calculations.*3141

*Once again, matrix calculations are easy to do, but really tedious.*3149

*They take all of this time; it is easy to make a mistake, because just doing 100 calculations, you tend to make a mistake somewhere.*3153

*But that is what calculators and computers are for; that is why humans invented those sorts of things--*3159

*to be able to make tedious calculations like that go away, where we can trust the calculator*3164

*to do the number-crunching part, and we can trust us to do the thinking part (hopefully).*3168

*We figure what A is; it is going to be a big one: our u's first: 1u, -4u, 1u, -2, 1/5u, 2u;*3174

*next, our v's: 2v, 2v, 1v, 1/2v, -3v, +4v; 7w, 1w, 0w (because it didn't show up), 3w, -1w, -1w;*3188

*-3x, 1/3x, 0x, 0x, 2x, -3x; 4y, 2y, 1y, 2y, -1y, 5y; 2z, 1z, 1z, 4z, 4z, 0z.*3207

*So, what you do is: you take A, and you enter that into your graphing calculator.*3225

*You put that into a graphing calculator; you put that into some sort of matrix calculator.*3230

*You enter this into a calculator, or a computer, or something that is able to work with matrices.*3233

*Lots of programs are, because matrices are very useful.*3243

*Once again, we aren't even beginning to scratch the surface of how useful they are; we are just getting some sense with this one problem.*3245

*So, we enter this whole thing into a calculator; then you tell the calculator to take the inverse.*3251

*So, we do that; and I want to point out, before we actually go on to talk about the inverse:*3257

*you tell the calculator to take the inverse; before you do that, double-check that you entered the matrix correctly.*3261

*If you entered this wrong--if you entered this A, 6 x 6...that is 36 numbers that you just put into your calculator.*3268

*Chances are that you might have accidentally entered one of them wrong.*3274

*If you enter one of them wrong, your entire answer is going to be wrong.*3277

*Chances are it will end up being this awful decimal number, so you will think,*3280

*"Well, my teacher probably didn't give me something that would come out to be an awful decimal number."*3283

*But if you are working with something like physics, where you don't already know what the answer is going to be,*3286

*it is up to you to make sure that you get it in correctly the first time.*3290

*So, double-check: if you are entering a very large matrix, make certain that you entered that matrix correctly.*3294

*We have the entire matrix set up in our calculator, and we have double-checked that it is correct.*3300

*Now, we punch out A ^{-1}: on most calculators, that is going to end up being: take the matrix and raise it to the -1.*3304

*What does it come out to be; it comes out to be really ugly--it is awful.*3310

*For example, the very first term is going to be 1780/14131; the first row, second column, would be 45/14131; the third...this is awful.*3315

*So, what are we going to end up doing?*3337

*Do we have to write the whole thing down?*3339

*No, we don't have to write the whole thing down--it is in our calculator.*3340

*We just tell the calculator A ^{-1}, and then we don't have to worry about A^{-1} at all.*3342

*We don't have to figure it out and write the whole thing down on paper; there is no need for it.*3348

*The calculator will keep track of what the numbers for A ^{-1} are,*3352

*because all we are concerned about is taking A ^{-1} and applying it against B.*3356

*We leave it in the calculator; we know that X is going to be equal to A ^{-1} times B.*3362

*All right, that is what we just figured out from our plan of thinking about this.*3370

*So, we have, in our calculator, that A ^{-1} is in there.*3374

*We have it in the calculator; we don't have to actually see what the whole thing is, because it is already there.*3378

*What is our B? We enter in the column matrix, 41, 39, 4, 23, -30, 44; we make sure that our A ^{-1} is multiplying from the left side.*3383

*Otherwise, it won't work at all.*3396

*And what does this end up coming out to be?*3398

*This comes out to be the deliciously simple -5, 4, 1, -3, 6, -1.*3400

*So, we just figured out that our X (all of our variables at once) is equal to...what were all of our variables?*3410

*It was u, and then we put in v, and then we put in w, and then we put in x, y, z.*3420

*So, they go in that order in our column: u, v, w, x, y, z = this thing that we just punched out, -5, 4, 1, -3, 6, -1.*3425

*So, u = -5; v = 4; w = 1; x = -3; y = 6; z = -1.*3443

*If we really wanted to at this point, we could check it; we could plug each one of these into any one of these equations;*3450

*and if it came out right, chances are that we probably got the entire thing right.*3455

*So, it might not be a bad idea to check at that point.*3458

*But also, as long as we were really careful with entering in our A, and careful with entering in our column of constants, our B,*3460

*everything should have worked out fine there; otherwise there is some other error that cropped up.*3467

*So, it becomes really, really easy, with just a little bit of thinking, and this calculator*3470

*(to take care of the awful manual work of the numbers, of just having to work through that many numbers)--*3475

*as long as we have the calculator to be able to do that part, so that it is quick and easy,*3482

*and we can trust that it came out right, and we are able to do the thought of what is going on,*3485

*we see that A ^{-1}, our coefficient matrix inverted, times what the equations come out to be,*3489

*our constant column matrix, just comes out to be the answers for each one of them.*3496

*It is really cool, really fast, and really easy; any time you have a large linear system,*3500

*or even a small linear system, and you just want to check it, you can have it done like that,*3506

*if you have access to a matrix calculator--pretty cool.*3509

*All right, we will see you at Educator.com later--goodbye!*3512

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