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Using Matrices to Solve Systems of Linear Equations

    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.
  • We can represent an entire linear system with an an augmented matrix:
    + y
    + z


    • 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:
    1. Interchange the locations of two rows.
    2. Multiply (or divide) a row by a nonzero number.
    3. Add (or subtract) a multiple of one row to another.
    While row operations are a simple idea, working with them involves a 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.
    1. Write the linear system as an augmented matrix.
    2. 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.
    3. 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.
    4. 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 Ai be the same as A except the ith column is replaced with the column of constants from the linear system (the right side of the equations). Then, if det(A) ≠ 0, the ith variable xi is
    xi =



    If det(A) = 0, then the system has either no solutions or infinitely many solutions.
  • 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.
    Notice that if we could get X alone on one side of the equation, we will have solved for the variables.
  • 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−1B, 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.

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 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

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 it0019

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 matrix0085

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 10126

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 matrix0220

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 mistakes0372

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 R2.0440

If we wanted to talk about the ninth row, it would be R9; and so on, and so on.0444

So, if we are swapping row i and row j, we have Ri, little arrow going back and forth, Rj.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 kRi (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 operations0541

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 enough0577

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 constant0657

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 left0715

(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), a11x, a12y, a21x, a22y,1042

is equal to constants on the right side (b1 and b2 are also constants),1050

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

All of our coefficients, a11, a12, a21, a22...1063

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

Now, Ax...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 a1, a21 with this; and so, we have b1, b2 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 Ax is just like A, except that it has this constant column replacing the x column.1111

Similarly, Ay 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 Ax, 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 Ai be the same as A, except the ith column; the column that represents1169

the variable we are currently working with--the variable that we want to solve for--that column1177

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: Ai in this case,1191

if we are looking for the ith variable, up until we are looking for the ith one.1195

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

the b1, b2 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 ith variable, xi, 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 thing1247

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 matrices1313

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-1B.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 hood1744

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 - 4R1 + 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 Ax--what is Ax 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

Ay: what will Ay 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, Ax,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 problem2146

(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 Ay, 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 Ay: we don't know what Ay, 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 Ay 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 Ay; 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 expanding2433

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 Ay, over the determinant of A.2706

We now need to figure out what Ay comes out to be.2716

The determinant of Ay...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 32860

(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 Ay, our special matrix for Ay.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, Ay, 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 determinant2991

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 Aw, Ax, and Az,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 form3042

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 Ax is; figure out what Aw is;3055

figure out what Az 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-1B.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 calculator3164

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 calculator3470

(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