For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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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 leftside column gives a variable's coefficients,
 If a variable does not appear, it has coefficient 0,
 The vertical line represents `=',
 The rightside 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.
 We can use row operations to put a matrix in reduced rowechelon 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 topleft 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 rowechelon form, we will have solved its associated linear system.
 GaussJordan elimination is a method we can follow to produce reduced rowechelon 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 rowechelon 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 rowechelon 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 lefthand 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 singlecolumn matrix of the variables,
 B is a singlecolumn matrix of the constants.
 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 reducedrow 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 stepbystep.
Using Matrices to Solve Systems of Linear Equations

 An augmented matrix represents an entire linear system. The left side of the matrix has an entry for every coefficient attached to the variables, while the right side gives the constant terms.
 As long as the order of variables is the same for each equation, we can just pull out each number based on its location, then slot them into a matrix:
⎡
⎢
⎢
⎣3 5 −7 4 −3 10 ⎤
⎥
⎥
⎦


 




 An augmented matrix represents an entire linear system. The left side of the matrix has an entry for every coefficient attached to the variables, while the right side gives the constant terms.
 To convert a linear system into an augmented matrix, each of the variables must appear in the same order for each of the equations. This includes variables that do not appear in a given equation. Since they do not appear in the equation, they have a 0 in front of them (because none of that variable is there). Put the linear system into such an order. Let's put the variables into the order a, b, c, d. This is not the only possible choice, but it seems natural since it's an alphabetic ordering.
[Note: This is not the only possible ordering for the linear system. We could have put each equation into any row. We put them toptobottom in order of lefttoright appearance in the problem, but any vertical ordering would have been acceptable as long as we didn't split up the equations. Similarly, any ordering of the variables within the equations would be fine. There are many possible ways to convert the linear system into an augmented matrix, this is just (probably) the most convenient.]3a +2b −4c +0d = 6 −1a +0b +0c +5d = 13 0a +1b +1c +1d = 0 12a +3b +0c −5d = 3  Once the linear system is properly ordered and it is clear what the coefficient to each variable is, we can just pull out each number based on its location, then slot them into a matrix:
⎡
⎢
⎢
⎢
⎢
⎢
⎣3 2 −4 0 6 −1 0 0 5 13 0 1 1 1 0 12 3 0 −5 3 ⎤
⎥
⎥
⎥
⎥
⎥
⎦




 




 




 






 The method of GaussJordan elimination allows us to solve a linear system through augmented matrices and row operations. Begin by turning the linear system into an augmented matrix:
⎡
⎢
⎢
⎣4 7 6 1 3 −1 ⎤
⎥
⎥
⎦  Row operations allow us to do the following three things: add rows together, multiply an entire row by a number, and swap the location of two rows. Often we will combine the first two operations into a single action. Through the use of row operations, we turn the left half of the augmented matrix into an identity matrix. GaussJordan elimination tells us to begin by making the main diagonal nothing but 1's, and everything below the diagonal 0's. For this problem, we can get a 1 in the topleft simply by swapping the rows, so we do that first:
Then turn the number below the topleft 1 to a 0:⎡
⎢
⎢
⎣4 7 6 1 3 −1 ⎤
⎥
⎥
⎦R_{2} ↔ R_{1} → R_{2} ↔ R_{1} → ⎡
⎢
⎢
⎣1 3 −1 4 7 6 ⎤
⎥
⎥
⎦
Finally, we turn the bottomright number into a 1 as well to complete the main diagonal:−4R_{1} +R_{2} → ⎡
⎢
⎢
⎣1 3 −1 0 −5 10 ⎤
⎥
⎥
⎦− 1 5R_{2} → ⎡
⎢
⎢
⎣1 3 −1 0 1 −2 ⎤
⎥
⎥
⎦  Now that the main diagonal is 1's and there are 0's below, we reverse and work back up the main diagonal, turning all the numbers above the main diagonal to 0's as well.
At this point the GaussJordan elimination process has given us a matrix in reduced rowechelon form, and we can turn the augmented matrix back into a linear system:−3R_{2}+R_{1} ⎡
⎢
⎢
⎣1 0 5 0 1 −2 ⎤
⎥
⎥
⎦
This clearly gives us x=5 and y=−2.1x +0y = 5 0x +1y = −2

 The method of GaussJordan elimination allows us to solve a linear system through augmented matrices and row operations. Begin by turning the linear system into an augmented matrix:
⎡
⎢
⎢
⎢
⎣1 1 −2 3 2 3 −1 2 −1 2 4 −15 ⎤
⎥
⎥
⎥
⎦  Row operations allow us to do the following three things: add rows together, multiply an entire row by a number, and swap the location of two rows. Often we will combine the first two operations into a single action. Through the use of row operations, we turn the left half of the augmented matrix into an identity matrix. GaussJordan elimination tells us to begin by making the main diagonal nothing but 1's, and everything below the diagonal 0's.
⎡
⎢
⎢
⎢
⎣1 1 −2 3 2 3 −1 2 −1 2 4 −15 ⎤
⎥
⎥
⎥
⎦−2R_{1}+R_{2} → R_{1}+R_{3} → ⎡
⎢
⎢
⎢
⎣1 1 −2 3 0 1 3 −4 0 3 2 −12 ⎤
⎥
⎥
⎥
⎦−3R_{2}+R_{3} → ⎡
⎢
⎢
⎢
⎣1 1 −2 3 0 1 3 −4 0 0 −7 0 ⎤
⎥
⎥
⎥
⎦− 1 7R_{3}→ ⎡
⎢
⎢
⎢
⎣1 1 −2 3 0 1 3 −4 0 0 1 0 ⎤
⎥
⎥
⎥
⎦  Now that the main diagonal is 1's and there are 0's below, we reverse and work back up the main diagonal, turning all the numbers above the main diagonal to 0's as well.
⎡
⎢
⎢
⎢
⎣1 1 −2 3 0 1 3 −4 0 0 1 0 ⎤
⎥
⎥
⎥
⎦2R_{3}+R_{1}→ −3R_{3} + R_{2} → ⎡
⎢
⎢
⎢
⎣1 1 0 3 0 1 0 −4 0 0 1 0 ⎤
⎥
⎥
⎥
⎦−R_{2}+R_{1}→ ⎡
⎢
⎢
⎢
⎣1 0 0 7 0 1 0 −4 0 0 1 0 ⎤
⎥
⎥
⎥
⎦  At this point the GaussJordan elimination process has given us a matrix in reduced rowechelon form, and we can turn the augmented matrix back into a linear system:
1a + 0b +0c = 7 0a + 1b +0c = −4 0a + 0b +1c = 0

 It is possible to do this problem without using a graphing calculator or any sort of matrix calculating aid. That said, since it is such a large linear system, using one will make this problem much easier. If we are using such an aid, all we have to do is set it up properly and enter it in correctly. Begin by converting the linear system into an augmented matrix:
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣3 0 −2 1 2 21 1 −2 3 0 1 18 0 1 6 −3 −2 16 −3 3 4 −2 −1 3 1 1 1 1 1 14 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦  The GaussJordan elimination method allows us to turn the augmented matrix into its reduced rowechelon form (where the left side is an identity matrix). Most graphing calculators have some sort of function that allows you to convert it into that form. It is unlikely to be a button on the calculator, but is something you can find if you search the menus. [For example, on most Texas Instruments graphing calculators, you can find a menu that deals specifically with matrices, then choose the `rref' function and apply it to the matrix.] You may need to find specific instructions for how to use your calculator in such a way. You can probably find out how with a simple internet search of "[name/model of your graphing calculator] reduced row echelon matrix", or something similar. If you do not have a graphing calculator, you can also search the internet for "reduced row echelon matrix calculator"there are plenty you can use with just a web browser.
 Carefully enter the augmented matrix (it won't have a vertical bar before the sixth column) into your calculator and apply the reduced rowechelon function to it. Your calculator will churn out the following result:
We can then convert this into the value for each variable based on location.⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣1 0 0 0 0 5 0 1 0 0 0 2 0 0 1 0 0 3 0 0 0 1 0 −4 0 0 0 0 1 8 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦

 We can solve a linear system through Cramer's Rule. First, start off by converting the coefficients of the linear system into a coefficient matrix. (Notice that this does not include the constants on the right side of the equations.)
A = ⎡
⎢
⎢
⎣2 −7 −5 6 ⎤
⎥
⎥
⎦  Depending on which variable we're interested in solving for, we swap the column of constants for the corresponding column of coefficients from that variable:
To actually find the solution for a variable, Cramer's Rule says it is given by the determinant of the modified matrix for that variable divided by the determinant of the nonmodified coefficient matrix:A_{x} = ⎡
⎢
⎢
⎣−14 −7 −11 6 ⎤
⎥
⎥
⎦⎢
⎢A_{y} = ⎡
⎢
⎢
⎣2 −14 −5 −11 ⎤
⎥
⎥
⎦x = det A_{x} det A ⎢
⎢y = det A_{y} det A  Thus, to find the solutions, we need to calculate detA, detA_{x}, and detA_{y}. [If you are unfamiliar with calculating determinants, make sure to check out the previous lesson.]
det A = ⎢
⎢
⎢
⎢2 −7 −5 6 ⎢
⎢
⎢
⎢= 2 ·6 − (−7)·(−5) = −23 det A_{x} = ⎢
⎢
⎢
⎢−14 −7 −11 6 ⎢
⎢
⎢
⎢= −14·6 − (−7)·(−11) = −161
Thus, by Cramer's Rule, we havedet A_{y} = ⎢
⎢
⎢
⎢2 −14 −5 −11 ⎢
⎢
⎢
⎢= 2 ·(−11) − (−14)·(−5) = −92 x = det A_{x} det A = −161 −23= 7 ⎢
⎢y = det A_{y} det A = −92 −23= 4

 We can solve a linear system through Cramer's Rule. First, start off by converting the coefficients of the linear system into a coefficient matrix. (Notice that this does not include the constants on the right side of the equations.)
A = ⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣−4 4 2 −2 5 3 −1 −2 −5 5 −4 −3 −2 5 2 −1 −4 5 −3 −1 3 −2 −4 −1 1 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦  Depending on which variable we're interested in solving for, we swap the column of constants for the corresponding column of coefficients from that variable. In this case, we're interested in solving for d, so we swap in the column of constants for the column of coefficients that match up to d.
To actually find the solution for a variable, Cramer's Rule says it is given by the determinant of the modified matrix for that variable divided by the determinant of the nonmodified coefficient matrix:A_{d} = ⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣−4 4 2 −6 5 3 −1 −2 37 5 −4 −3 −2 16 2 −1 −4 5 −31 −1 3 −2 −4 32 1 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦d = det A_{d} det A  Thus, to find the solutions, we need to calculate detA and detA_{d}. Notice that it would take a lot of effort to find these determinants by hand since they are coming from 5 ×5 matrices. Luckily, the problem told us that we could (and should) use a graphing calculator or matrix calculator. If we've got some sort of computing device to do this process for us, it's a breeze. If you're using a graphing calculator, there will be some sort of det function in a menu about matrices. Choose that, then apply it to the matrix (which you will also have to enter). If you're using a matrix calculator on your computer (just do an internet search for "matrix calculator"), the process will likely be reversed: enter the matrix then tell it to take the determinant. However you're doing it, the results should be as follows:
Therefore, by Cramer's Rule, we havedet (A) = 2067 ⎢
⎢det (A_{d}) = 6201 d = det A_{d} det A = 6201 2067= 3

 Notice that, by the rules of matrix multiplication, we can rewrite the above linear system as the interaction of matrices:
Symbolically, we can express this as3x −3y = 9 −3x +5y = −7 ⇔ ⎡
⎢
⎢
⎣3 −3 −3 5 ⎤
⎥
⎥
⎦⎡
⎢
⎢
⎣x y ⎤
⎥
⎥
⎦= ⎡
⎢
⎢
⎣9 −7 ⎤
⎥
⎥
⎦
where A is the coefficient matrix from the system, X is a column matrix of the variables, and B is a column matrix of the constants on the right side of the equations. Therefore, if A has an inverse matrix A^{−1}, we can use algebra to apply that inverse to the equation and cancel out the A, leaving X isolated on one side. Thus, if we multiply by A^{−1} on both sides, we getAX = B,
Thus, if we find A^{−1}, we can easily solve for X, which will tell us the solutions to the linear system.A^{−1} AX = A^{−1} B ⇒ X = A^{−1} B  From the previous lesson, we saw that it was quite easy to find the inverse of a 2×2 matrix. For any matrix A, we have
Thus, we can find the inverse to the coefficient matrix by just plugging in:A = ⎡
⎢
⎢
⎣a b c d ⎤
⎥
⎥
⎦⇒ A^{−1} = 1 ad−bc⎡
⎢
⎢
⎣d −b −c a ⎤
⎥
⎥
⎦A^{−1} = 1 3·5−(−3)(−3)⎡
⎢
⎢
⎣5 3 3 3 ⎤
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎢
⎢
⎣5 61 21 21 2⎤
⎥
⎥
⎥
⎥
⎥
⎦  Finally, by the logic we talked about in the first step, we have
Therefore, since X represented a column matrix of the variables, we have shownX = A^{−1} B = ⎡
⎢
⎢
⎢
⎢
⎢
⎣5 61 21 21 2⎤
⎥
⎥
⎥
⎥
⎥
⎦⎡
⎢
⎢
⎣9 −7 ⎤
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎢
⎢
⎣15 2− 7 29 2− 7 2⎤
⎥
⎥
⎥
⎥
⎥
⎦= ⎡
⎢
⎢
⎣4 1 ⎤
⎥
⎥
⎦⎡
⎢
⎢
⎣x y ⎤
⎥
⎥
⎦= ⎡
⎢
⎢
⎣4 1 ⎤
⎥
⎥
⎦

 As in the previous problem, we can use matrix multiplication to rewrite the system:
Symbolically, we can express this as2x −2y −3z = −2 −3x +3y +5z = 4 2x −2y −5z = −14 ⇔ ⎡
⎢
⎢
⎢
⎣2 −2 −3 −3 3 5 2 −2 −5 ⎤
⎥
⎥
⎥
⎦⎡
⎢
⎢
⎢
⎣x y z ⎤
⎥
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎣−2 4 −14 ⎤
⎥
⎥
⎥
⎦
where A is the coefficient matrix from the system, X is a column matrix of the variables, and B is a column matrix of the constants on the right side of the equations. Therefore, if A has an inverse matrix A^{−1}, we can use algebra to solve for XAX = B,
From the above, we see that if we can find A^{−1}, we can easily find a unique solution for X, which means we have found a unique solution for the linear system.A^{−1} AX = A^{−1} B ⇒ X = A^{−1} B  However, if we can not find an inverse for A, then it is not possible for us to find a unique solution to the linear system. Therefore, to show that the linear system does not have a unique solution (which is what the problem told us to do), we only need to show that A is not invertible. From the previous lesson on determinants and inverses, we know that a matrix is not invertible when its determinant equals 0. Thus, we need to show that det(A) = 0.
 Take the determinant of A [If you're not sure how the below works, make sure to check out the previous lesson on determinants.]:
det (A) = ⎢
⎢
⎢
⎢
⎢2 −2 −3 −3 3 5 2 −2 −5 ⎢
⎢
⎢
⎢
⎢2 · ⎢
⎢
⎢
⎢3 5 −2 −5 ⎢
⎢
⎢
⎢−(−2) · ⎢
⎢
⎢
⎢−3 5 2 −5 ⎢
⎢
⎢
⎢+(−3) · ⎢
⎢
⎢
⎢−3 3 2 −2 ⎢
⎢
⎢
⎢2·(3·(−5) − 5·(−2)) + 2 ·((−3)·(−5) − 5·2) + (−3) ·((−3)·(−2) − 3 ·2)
Thus det(A) = 0, so we have shown that the linear system does not have a unique solution because we cannot invert A.2·(−5) + 2 ·(5) −3 ·(0) = 0

 Using matrix multiplication, we can rewrite the linear system:
Symbolically, we can express this as⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣2 3 5 4 −2 1 0 2 −3 1 −1 2 3 −2 1 0 3 −3 3 −3 1 2 4 −7 1 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣a b c d e ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣48 −14 −11 18 −34 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
where A is the coefficient matrix from the system, X is a column matrix of the variables, and B is a column matrix of the constants on the right side of the equations. Therefore, if A has an inverse matrix A^{−1}, we can use algebra to solve for XAX = B, A^{−1} AX = A^{−1} B ⇒ X = A^{−1} B  Thus, to solve the linear system, we only need to find A^{−1} B using a graphing calculator or matrix calculator. One possibility would be to find A^{−1} using the calculator, then enter that result in and multiply it by B. However, that will require a lot of work entering two different 5 ×5 matrices (one for A, then again for A^{−1} once we know it). Instead, we can usually speed up the process by telling the calculator to do both steps one after the other. Enter in to the calculator something along the lines of
Doing this will cause the calculator to both find the inverse and then immediately multiply that by the constant matrix B. [However, depending on the specific calculator you are using, you might have to enter in A^{−1} by hand. But it is likely you can find some method to avoid that extra work.]⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣2 3 5 4 −2 1 0 2 −3 1 −1 2 3 −2 1 0 3 −3 3 −3 1 2 4 −7 1 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦−1
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣48 −14 −11 18 −34 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦  However you wind up doing it, the calculator will give that
Thus, since X = A^{−1} B, we have that⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣2 3 5 4 −2 1 0 2 −3 1 −1 2 3 −2 1 0 3 −3 3 −3 1 2 4 −7 1 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦−1
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣48 −14 −11 18 −34 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣2 −1 3 6 −4 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣a b c d e ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦= ⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎣2 −1 3 6 −4 ⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Using Matrices to Solve Systems of Linear Equations
Lecture Slides are screencaptured 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!
 GaussJordan Elimination  Idea
 GaussJordan 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
 GaussJordan Elimination  Idea 8:04
 GaussJordan Elimination  Idea, cont.
 Reduced RowEchelon Form
 GaussJordan Elimination  Method 11:36
 Begin by Writing the System As An Augmented Matrix
 GaussJordan 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
Precalculus with Limits Online Course
Transcription: Using Matrices to Solve Systems of Linear Equations
Hiwelcome back to Educator.com.0000
Today, we are finally going to see why we have been studying matricesjust 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 leftside 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 rightside 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 sensethere 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 system0213
it is a way to portray a linear systemthen 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 nonzero number.0239
I can multiply an entire row by 2 or by 5 or by 10whatever I want to0243
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 sensehow 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 simpleswap; multiply; or add a multiplenot 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 sensesomething must have gone wrong0405
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, leftright, 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 ias 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
GaussJordan elimination: here is an idea: since all of our row operations make sense for solving a linear system0482
they all make sense for a way to do itwe 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 nonexistent, 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 matrixone the right side of the augmented matrix, past that vertical line that shows equality.0662
We will have solved its associated linear system.0669
GaussJordan 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 GaussJordan 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 multiplehow 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; greatwe 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 0we still have 0.0891
0 times 4, added to thiswe 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 manyif our linear system can't be solved,1007
or it has infinitely many solutionsthis 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 twovariable 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 solution1125
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 nvariable linear system.1165
Let A_{i} be the same as A, except the i^{th} column; the column that represents1169
the variable we are currently working withthe variable that we want to solve forthat 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: 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 thing1247
that we want to be able to understandthis 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, 1notice that that is just our normal coefficient matrix A showing up here1287
if we multiply it by the column matrix x, y, z equals our coefficient column here, that ends up being just the same1298
it is completely equivalent; these two ideas here are completely equivalentmultiplying 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 singlecolumn 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 singlecolumn 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, thoughhow 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 GaussJordan elimination,1601
Cramer's Rule, inverse matricesthey 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, complexlevel 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 GaussJordan 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 1get 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 solutiongreat.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 columnwhat 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, swappedout 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 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 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 changethe 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 columnthis one right hereso 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 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 mistakesee 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 herewe 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 alongwhich 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 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 determinant2991
of our coefficient matrixthat is also 242; 242/242those 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 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 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 things3159
to be able to make tedious calculations like that go away, where we can trust the calculator3164
to do the numbercrunching 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, doublecheck that you entered the matrix correctly.3261
If you entered this wrongif 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, doublecheck: 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 doublechecked 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 uglyit 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 downit 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 calculatorpretty cool.3509
All right, we will see you at Educator.com latergoodbye!3512
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