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Algebra 2 Solving Systems of Equations Using Matrices

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Section 4: Matrices: Lecture 7 | 22:32 min

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Lecture Comments (13)

	0 answers Post by Qing Xu on March 24, 2020 are x and x the same?
	1 answer Last reply by: Dr Carleen Eaton Fri May 24, 2013 9:29 PM Post by Kavita Agrawal on May 23, 2013 Which method would you say is faster - Cramer's rule or matrix multiplication (as shown above)?
	1 answer Last reply by: richard lawlor Tue Sep 4, 2012 6:55 AM Post by richard lawlor on September 4, 2012 What about dimension requirements for matrix multiplication i.e. number of rows must equal number of columns?This is not the case in the above example?
	1 answer Last reply by: Dr Carleen Eaton Sun Jan 29, 2012 4:42 PM Post by Edmund Mercado on January 27, 2012 Dr. Eaton, thank you for a very detailed explanation of a very difficult procedure.
	3 answers Last reply by: Carroll Fields Sat Sep 14, 2013 2:20 PM Post by Lemel Covington on November 10, 2011 example 3 was doe incorrectly as well. you did not flip the -7 and 1 in the matrix. you flipped the a and c.
	1 answer Last reply by: Dr Carleen Eaton Mon Nov 14, 2011 10:59 PM Post by Jeff Mitchell on March 7, 2011 In example 2 (approx. 10:22 into lecture) the equation for A inverse was incorrectly written as 1 / ab-bc and should be 1 / ad-bc. Jeff

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Solving Systems of Equations Using Matrices

We can write a system of equations in matrix form. The result is a matrix equation.
A system of equations can be solved using matrices. The solution is the product of the inverse of the matrix of coefficients and the matrix of constants.
The system has a unique solution if and only if the matrix of coefficients has an inverse.
If the matrix of coefficients does not have an inverse, then either there is no solution or there are an infinite number of solutions.

Solving Systems of Equations Using Matrices

Write the system as a matrix equation.
x + 6y = − 12
− 7x − 3y = 6

Recall that a matrix equation looks like this Ax = B
Find A, x, and B
A = [
1
6
− 7
− 3
]
x = [
x
y
]
B = [
− 12
6
]
Ax = B
[
1
6
− 7
− 3
]*[
x
y
] = [
− 12
6
]

[

− 7

− 3

]*[

] = [

− 12

]

Write the system as a matrix equation.
x − 3y = − 20
5x − y = 12

Recall that a matrix equation looks like this Ax = B
Find A, x, and B
A = [
1
− 3
5
− 1
]
x = [
x
y
]
B = [
− 20
12
]
Ax = B
[
1
− 3
5
− 1
]*[
x
y
] = [
− 20
12
]

[

− 3

− 1

]*[

] = [

− 20

]

Write the system as a matrix equation.
5x − 4y = − 2
x + 2y = 22

Recall that a matrix equation looks like this Ax = B
Find A, x, and B
A = [
5
− 4
1
2
]
x = [
x
y
]
B = [
− 2
22
]
Ax = B
[
5
− 4
1
2
]*[
x
y
] = [
− 2
22
]

[

− 4

]*[

] = [

− 2

]

Write the system as a matrix equation.
x + 3y = − 8
− 3x − 6y = 15

Recall that a matrix equation looks like this Ax = B
Find A, x, and B
A = [
1
3
− 3
− 6
]
x = [
x
y
]
B = [
− 8
15
]
Ax = B
[
1
3
− 3
− 6
]*[
x
y
] = [
− 8
15
]

[

− 3

− 6

]*[

] = [

− 8

]

Solve using a matrix equation
5x − 3y = 19
− 3x + y = − 17

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
5
− 3
− 3
1
]
B = [
19
− 17
]
x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
1
3
3
5
] = [
− [1/4]
− [3/4]
− [3/4]
− [5/4]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
− [1/4]
− [3/4]
− [3/4]
− [5/4]
]*[
19
− 17
] = [
8
7
]

Solution (8,7)

Solve using a matrix equation
x − 3y = − 20
5x − y = 12

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
1
− 3
5
− 1
]
B = [
− 20
12
]
x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
− 1
3
− 5
1
] = [
− [1/14]
[3/14]
− [5/14]
[1/14]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
− [1/14]
[3/14]
− [5/14]
[1/14]
]*[
− 20
12
] = [
4
8
]

Solution (4,8)

Solve using a matrix equation
5x − 4y = − 2
x + 2y = 22

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
5
− 4
1
2
]
B = [
− 2
22
]
x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
2
4
− 1
5
] = [
[1/7]
[4/14]
− [1/14]
[5/14]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
[1/7]
[4/14]
− [1/14]
[5/14]
]*[
− 2
22
] = [
6
8
]

Solution (6,8)

Solve using a matrix equation
x + 3y = − 8
− 3x − 6y = 15

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
1
3
− 3
− 6
]
B = [
− 8
15
]
x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
− 6
− 3
3
1
] = [
− 2
− 1
1
[1/3]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
− 2
− 1
1
[1/3]
]*[
− 8
15
] = [
1
− 3
]

Solution (1, − 3)

Solve using a matrix equation
− 4x + 6y = 0
7x + y = 23

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
− 4
6
7
1
]
B = [
0
23
]
x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
1
− 6
− 7
− 4
] = [
− [1/46]
[3/23]
[7/47]
[2/23]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
− [1/46]
[3/23]
[7/47]
[2/23]
]*[
0
23
] = [
3
2
]

Solution (3,2)

Solve using a matrix equation
4x − 3y = 6
x − 8y = 16

In order to solve this system of equations using matrices, it is assumed that you know how to
1) Find Determinant of a 2x2 matrix
2) Find the Inverse of a matrix
3) Multiply Matrices
Recall that to find the solution, you must isolate
x = A^{− 1}B
A = [
4
− 3
1
− 8
] B = [
6
16
] x = [
x
y
]
Step 1: Find the Inverse A^{− 1}
A^{− 1} = [1/detA][
− 8
3
− 1
4
] = [
[8/29]
− [3/29]
[1/29]
− [4/29]
]
Step 2: Multiply A^{− 1}B
[
x
y
] = [
[8/29]
− [3/29]
[1/29]
− [4/29]
]*[
6
16
] = [
0
− 2
]

Solution (0, − 2)

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Answer

Solving Systems of Equations Using Matrices

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

Matrix Equations

Example: System of Equations

Solving Systems of Equations

Example 1: Write as Matrix Equation

Example 2: Use Matrix Equations

Example 3: Use Matrix Equations

Example 4: Use Matrix Equations

Intro 0:00
Matrix Equations 0:11

Example: System of Equations

Solving Systems of Equations 4:01

Isolate x
Example: Using Numbers
Multiplicative Inverse

Example 1: Write as Matrix Equation 7:18
Example 2: Use Matrix Equations 9:12
Example 3: Use Matrix Equations 15:06
Example 4: Use Matrix Equations 19:35

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Solving Systems of Equations Using Matrices

Welcome to Educator.com.0000

In today's lesson, we are going to apply what we have learned with matrices so far,0002

in order to solve systems of equations using matrices.0006

OK, first: a system of equations can be written as a single matrix equation, and that is the first step to solving the system of equations.0011

So, let's use an example here: now, recall that, in previous lessons, we talked about a lot of other methods to solve equations,0022

such as substitution and elimination; and it depends on the situation which is the easiest method.0034

But this gives you another option, and it can be the best method in certain situations.0040

So, first, you are going to write one matrix using the coefficients of x and the coefficients of y; and I am going to call that matrix A.0044

In the first column is the coefficients of x; in the second column, I am going to have coefficients of y.0056

OK, that is the first matrix: the second matrix I will call B, and for this matrix,0080

you are going to use the constants that are on the right side of the equation: 1 and 3.0088

Finally, you are going to write a third matrix that I will call X; and that is going to consist of variables in the system of equations.0095

OK, so now I have these three matrices; but I want to write them as an equation.0106

And how I could write these is that I could say that matrix A times matrix X equals matrix B.0112

And let's look at why: if I have matrix A, that is 5, 6, 4, 3 for the elements, times matrix X (that is my variables), it is going to give me my constants.0124

Well, let's just look at what this is saying by using matrix multiplication.0140

This is my matrix equation for this system of equations: I have written this matrix equation right here.0147

All right, let's multiply this out and see what happens.0156

Well, with matrix multiplication, I am going to multiply this first row times the column (I only have one column).0159

And it is going to give me, for my product matrix, 5x + 4y; I can't go any farther.0165

OK, now for this second row, I am going to also multiply it by the column; and then, that is going to give me 6x + 3y =...0177

Recall the definition of equal matrices: equal matrices have corresponding elements that are equal.0193

So, this element here (row 1, column 1) and this element here (row 2, column 2) correspond to row 1, column 1, row 2, column 1, here and here.0200

OK, so that gives me 5x + 4y =...I look at the corresponding element in row 1, column 1.0213

Here, I look at row 2, column 1, and my corresponding element here; and I get my original equations back.0222

So, that shows that this system of equations is equivalent to this matrix equation.0229

And that is what allows us to write this system of equations as a matrix equation.0235

OK, now let's talk about solving systems of equations.0242

We found this matrix equation AX = B: however, if I want to solve, what I really want to find is...0246

I want to isolate X, because recall that the matrix X, we defined as containing the variables.0257

And I want to isolate that, the same way as, if I am solving a regular system of equations, I want to isolate the variable.0265

Well, we already talked about how this matrix equation, AX = B, can be written for a system of equations.0272

So now, let's talk about isolating X.0279

Recall that what we want to do is get rid of the A; so we can do that by multiplying it by its inverse.0284

I am thinking for a second about why we are doing that: we are taking it, and we are multiplying it by its multiplicative inverse--matrix inverse--A^-1.0296

Well, think about when we are working with just regular numbers.0308

If I have 3x + 4, what I am going to do is multiply 3 by its multiplicative inverse, 1/3.0310

If I have 3x = 4, and I want to isolate the x, I am going to divide by 3; or I could say I am going to multiply by 1/3.0319

So, if I multiply 3 by its multiplicative inverse, 1/3, this is going to cancel out, and I am going to end up with 1.0330

Well, what is going to happen here? Recall that, when we talked about identity matrices,0344

we said they function somewhat like the number 1 in the world of matrices.0348

So, if I am multiplying a matrix by its multiplicative inverse, then I am going to get the identity matrix.0353

In the same, when I multiplied a number by its inverse, I got the number 1.0362

So, A times A^-1, which we discussed in previous lessons on identity matrices, is the identity matrix.0367

So, this is going to give me the identity matrix, times X, equals A^-1 times B.0377

OK, now also recall that, if you multiply a matrix by its identity matrix, you get the original matrix back.0384

What that tells me is that, if I multiply X by its identity matrix, that is the same as the matrix X.0396

So, I can just write this as X = A^-1B.0405

OK, so this is what is important to know; but this shows you how we got there.0409

Now, in the last slide, we talked about writing the matrix equation AX = B.0414

Here, we talked about isolating X; so what you really want to work with is this, where X equals A^-1 times B,0419

because that will allow you to solve for this matrix, and therefore to solve for the variables in the system of equations.0429

For example, write the system as a matrix equation.0441

First, all this is asking is to write it as a matrix equation; we don't even need to solve.0446

So, I have 2x - 3y = -7; this is 3x + 2y = 8; and we talked earlier about writing a matrix equation, AX = B.0457

And that is going to consist of three matrices.0472

The first matrix, A, is going to have, in the first column, the coefficients of x.0474

In the second column, it is going to have the coefficients of y.0482

Then, I am going to write a second matrix, B, which is going to have elements that consist of the constants, -7 and 8.0487

Then, I am going to have a third matrix that contains the variables.0499

The next thing to do is to put these together into an equation: that is going to give me A, times X, equals B.0506

The matrix equation for this system of equations is shown here: and this is AX = B.0532

Again, I wasn't asked to solve it, so I have done what I have been asked, which is to write the system as a matrix equation.0540

And we can use that as the basis of solving systems of equations, just as we are going to do in this problem.0547

The first thing to do is to write the matrix equation, so I need my three matrices: the first one is A;0563

and the coefficient of x is 1; and the other is 2; the coefficients of y are -2 and 1.0574

The second matrix, B, contains the constants for elements.0581

And then, finally, we have X, with my variables x and y.0587

Now, from this I could write AX = B; and we saw, earlier on, how, in order to solve it, I need to rearrange things0590

and do some manipulation and end up with this, because I really want to solve for the matrix X, because it contains the variable.0601

So, this is the equation I want; and I have A, B, and X, so what I need is A^-1.0609

Recall from an earlier lesson that A^-1 is 1 over the determinant, times this matrix.0615

OK, so all I am doing right now is finding A^-1, if it exists.0630

This is ad, which is 1 times 1, minus -2 times 2.0636

And just looking over at this, this is going to be -4, so it is 1 minus -4, so that is not 0.0643

So, I am OK--the determinant does exist--so I am going to continue.0650

I am going to switch these two, but they are the same number, so it ends up 1 and 1.0654

Then, I am going to take -2 and reverse its sign; so that is going to be 2.0658

Here, I am going to take 2 and reverse its sign (because that is c), and I am going to make it -2.0664

This is going to give me 1 over 1 - -4, times this matrix.0672

A^-1 is 1 over 1 + 4, which is going to give me 1/5.0682

So, I don't even have to multiply this out, because I am going to be doing some multiplication over here.0692

So, leaving this as it is: now I want to have X = A^-1 times B.0696

So, X equals...here is A^-1, 1/5 times its matrix, 1, 2, -2, 1, times B; B is 3, 7.0703

Now, I am going to go ahead and do my matrix multiplication to find the product matrix.0718

So, the product matrix would be: for row 1, column 1, this is 1, 2, times 3, 7--so 1 times 3, plus 2 times 7, equals 3 + 14; that is 17.0727

So, 17 goes right here; now, in row 1, column 2, right here, this is going to be 1...actually, it only has one row,0747

so it is just row 2, column 1; that is all I need to find; row 2 right here is the next position, because I have no other column here.0764

So, it is going to be row 2, column 1; that is going to give me -2 times 3, plus 1 times 7; that is -6 plus 7, which is going to give me 1.0774

OK, I found that I have x = 1/5 times 17/1, which equals...I am going to multiply 1/5 by 17--that is 17/5.0792

And multiplying 1/5 by 1, that is 1/5.0814

Now, the matrix X equals this; recall that matrix X also equals xy; so these two are equal;0820

therefore, the corresponding elements are equal; so I can say that x = 17/5 and y = 1/5.0830

It is a little bit complicated, but you just have to take it one step at a time.0840

I was asked to solve this using a matrix equation.0843

I wanted to find this equation, because then I would have the matrix with the variables in it isolated.0847

I started out by writing three matrices--one containing the x and y coefficients, one containing the constants, and one containing the variables.0854

Then, I had A, but I needed to find A^-1; and I used my formula to find that.0864

I went through and found that this is A^-1.0870

With that, I could then insert it into this formula: A^-1, right here, times B.0874

I went ahead and did matrix multiplication to find that I got this matrix as the product.0884

And then, I multiplied it by 1/5, which I still had out here, to get the matrix X equals 17/5 and 1/5, which equals x and y.0892

So, that is my solution to this system of equations.0901

Example 3: again, solve using a matrix equation.0908

So, I am going to go ahead and find my three matrices that I will need to have,0913

because I eventually want to end up with X = A^-1B.0918

So, A is a matrix with coefficients of x 3 and -1, and coefficients of y 7 and -2.0923

Matrix B has elements that are the constants 4 and 3; and then, matrix x has the variables.0934

The next thing is: I actually don't need A; I need A^-1.0943

And I am going to recall my formula, that this equals 1 over the determinant, ad - bc, times this matrix.0948

So, A^-1 is 1 over the determinant, which is 3 times -2, minus 7 times -1,0962

times this matrix: a and d switch positions, so it is going to be -2, 3; b reverses its sign; and c reverses its sign.0976

Proceeding to find the determinant: 3 times -2 is -6...that is minus -7, times this matrix...0989

that is going to give me A^-1 is 1 over...well, this is -6 minus -7, or -6 + 7, so that is just 1.1006

OK, I have A^-1: I found A, B, and C matrices; now I have A^-1.1022

And this is what I want: I want that X = A^-1B.1027

So, I am going to write A^-1 up here; and I am just going to write this as 1/1, times B; B is right here.1033

I am doing some matrix multiplication: row 1, column 1 is going to be -2 times 4, plus 7 times 3.1050

That is going to be -8; let me see, this should actually be a -7.1069

This is going to be -2 times 4; and this is going to be 3, -7, times 3.1087

So, row 1, column 1: -2 times 4 and -7 times 3.1097

That is going to give me -8, minus 21, which is going to be -29 for row 1, column 1.1101

Row 2, column 1: row 2 here, column 1 here: that is going to be 1 times 4, plus 3 times 3, equals 4 + 9, equals 13.1112

It is actually -29; OK, so row 1, column 1 gave me -29; row 2, column 1 gave me 13.1130

So, x equals 1 times this; well, since this is just 1, I am going to end up with x equals -29, 13 as elements.1142

And I also know that x is this; corresponding elements in equal matrices are equal, so -29 equals x, and 13 equals y.1153

So, this is my solution to this system of equations, using matrices.1168

Again, solve using a matrix equation: and I am going to keep in mind that I want to find x = A^-1B,1177

where A is a matrix with the coefficients of x and the coefficients of y;1186

B is a matrix containing the constants 8 and 9; and x is a matrix containing variables--the variables in the system of equations.1196

I need A^-1: A^-1 is 1 over the determinant of A, ad - bc, times the matrix d, -b, -c, a.1209

Therefore, A^-1 is 4 times -6, minus -3 times 8, times this matrix.1225

We are reversing a and d; so I take a, and I put it in the d position; d goes in the a position.1238

For b, I am going to reverse the sign; I am going to make it 3.1245

For c, I am going to reverse the sign; and that is -8.1250

OK, here I have 4 times -6; that is -24, minus -3 times 8; that is -24; and you can probably already see what the issue is.1255

That A^-1 would equal 1 over -24 minus -24, times its matrix; and we have a problem,1271

because taking this one step further, this gives me 1/0.1284

And it really doesn't matter what this matrix is over here, because this is undefined.1289

A^-1 does not exist; I cannot use this method.1296

So, the situation is that there is no unique solution; there may be no solution at all; there may be an infinite number of solutions.1304

But this method only works if there is a unique solution--a unique solution for x, and a unique value for y, that satisfies the system of equations.1314

So, I started out trying to use my method and writing this matrix equation.1324

And I got my matrices OK; but once I got to this step, finding A^-1, then I discovered1328

that the determinant for this matrix A is 0, and in that case, A^-1 does not exist; I can't use this method.1335

OK, that concludes this lesson on solving systems of equations using matrices.1344

Thanks for visiting Educator.com!1350

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Dr. Carleen Eaton

Solving Systems of Equations Using Matrices

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro