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Algebra 2 Cramer's Rule

Name: Algebra 2: Cramer's Rule
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Section 4: Matrices: Lecture 5 | 28:25 min

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Lecture Comments (21)

	0 answers Post by Jinjin Chai on February 21, 2021 I just checked the work for example 1 at 3:34 because I was confused - I noticed that the denominator for x was -9, and 4 was the value of x. I solved for Dr.Eaton's determinant and got -18/4, or -4.5. However, when I plugged in the value for x into the first equation, I got that y was 2, not -4.5. Furthermore, if the denominator was -9, which was what the denominator was for x, the solution for the determinant for y would've been 2, which is correct. In short, Dr.Eaton made a small error for the first example at 3:34 - for any people confused. the answer is 2, and the equation for the determinant is \|3 2\| \|-3 -8\| -18 _______ = __________ = 2! -9 -9
	0 answers Post by Qing Xu on March 20, 2020 THANKS SO MUCH! You can explain this to a person really well. I'm a 5th grader, and I understood
	1 answer Last reply by: Joshua Chen Tue Jun 16, 2020 10:08 AM Post by DJ Sai on September 10, 2018 Why are your outros different every time
	1 answer Last reply by: Dr Carleen Eaton Sun Apr 8, 2018 2:48 PM Post by John Stedge on March 29, 2018 Never mind you caught your mistake! Great job!
	1 answer Last reply by: DJ Sai Mon Sep 10, 2018 11:30 PM Post by John Stedge on March 29, 2018 At about 22:24 you miss-multiplied the last group of #'s (12-2) and you put (-2) in the denominator instead of (-4) as it should be therefore the final answer is incorrect.
	1 answer Last reply by: Dr Carleen Eaton Sat Oct 3, 2015 7:15 PM Post by Peter Ke on September 30, 2015 For example 4, I used the expansion b minors, and I used (2, -3 . 1) for my denominator. And when I solved it I got 12 + 18 - 18 which wasn't equal to 0. What did I do wrong?
	3 answers Last reply by: Dr Carleen Eaton Wed Jan 1, 2014 12:52 AM Post by Rob Escalera on May 21, 2012 Yes, I think the determinant (D) should be -9, and x = 4 while y = 2.
	4 answers Last reply by: Miranda Winther Tue Oct 29, 2013 9:19 PM Post by aloosh aloosh on February 17, 2011 to solve for y isnt d=9 not 4 ????
	1 answer Last reply by: Dr Carleen Eaton Thu Feb 3, 2011 1:48 PM Post by Santhini Dheenathayalan on February 1, 2011 Great video, Dr. Eaton! Thank you!

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Cramer's Rule

Cramer’s rule allows us to solve a system of equations using only determinants. There is no graphing or algebraic work involved.
Cramer’s rule indicates situations in which a system does not have a unique solution. It does not specify whether the system has no solution or an infinite number of solutions.

Cramer's Rule

Solve the system of equations using Cramer's Rule
2x + y = − 2
2x − 3y = − 10

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
− 2
− 10
and Dy is the determinant of D, but the second column replaced with
− 2
− 10
Find the three Determinants
D = |
2
1
2
− 3
| Dx = |
− 2
1
− 10
− 3
| Dy = |
2
− 2
2
− 10
|
D = − 8
Dx = 16
Dy = − 16
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [16/( − 8)] = − 2
y = [( − 16)/( − 8)] = 2

Solution = ( − 2,2)

Solve the system of equations using Cramer's Rule
2x − y = − 5
3x + y = 0

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
− 5
0
and Dy is the determinant of D, but the second column replaced with
− 5
0
Find the three Determinants
D = |
2
− 1
3
1
| Dx = |
− 5
− 1
0
1
| Dy = |
2
− 5
3
0
|
D = 5
Dx = − 5
Dy = 15
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [( − 5)/5] = − 1
y = [15/5] = 3

Solution = ( − 1,3)

Solve the system of equations using Cramer's Rule
3x + y = − 6
2x + 2y = − 8

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
− 6
− 8
and Dy is the determinant of D, but the second column replaced with
− 6
− 8
Find the three Determinants
D = |
3
1
2
2
| Dx = |
− 6
1
− 8
2
| Dy = |
3
− 6
2
− 8
|
D = 4
Dx = − 4
Dy = − 12
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [( − 4)/4] = − 1
y = [( − 12)/4] = − 3

Solution = ( − 1, − 3)

Solve the system of equations using Cramer's Rule
2x + 3y = 13
3x − 3y = − 3

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
13
− 3
and Dy is the determinant of D, but the second column replaced with
13
− 3
Find the three Determinants
D = |
2
3
3
− 3
| Dx = |
13
3
− 3
− 3
| Dy = |
2
13
3
− 3
|
D = − 15
Dx = − 30
Dy = − 45
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [( − 30)/( − 15)] = 2
y = [( − 45)/( − 15)] = 3

Solution = (2,3)

Solve the system of equations using Cramer's Rule
− x + 2y = − 1
x + y = 4

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
− 1
4
and Dy is the determinant of D, but the second column replaced with
1
4
Find the three Determinants
D = |
− 1
2
1
1
| Dx = |
− 1
2
4
1
| Dy = |
− 1
− 1
1
4
|
D = − 3
Dx = − 9
Dy = − 3
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [( − 9)/( − 3)] = 3
y = [( − 3)/( − 3)] = 1

Solution = (3,1)

Solve the system of equations using Cramer's Rule
x + y = − 6
− 2x − 2y = 12

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
12
and Dy is the determinant of D, but the second column replaced with
− 6
12
Find the three Determinants
D = |
1
1
− 2
− 2
| Dx = |
− 6
1
12
− 2
| Dy = |
1
− 6
− 2
12
|
D = 0
Dx = 0
Dy = 0
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [0/0] = undefined
y = [0/0] = undefined

No Unique Solution

Solve the system of equations using Cramer's Rule
− 2x − 3y = 2
x − 5y = − 14

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute three Determinants as follows
x = [Dx/D] y = [Dy/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
2
− 14
and Dy is the determinant of D, but the second column replaced with
2
− 14
Find the three Determinants
D = |
− 2
− 3
1
− 5
| Dx = |
2
− 3
− 14
− 5
| Dy = |
− 2
2
1
− 14
|
D = 13
Dx = − 52
Dy = 26
Now Find x and y
x = [Dx/D] y = [Dy/D]
x = [( − 52)/13] = − 4
y = [26/13] = 2

Solution = ( - 4,2)

Solve the system of equations using Cramer's Rule
5x + 2y − 4z = 9
− 6x + y + 5z = 2
− 6x − z = 22

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute four Determinants as follows
x = [Dx/D] y = [Dy/D] z = [Dz/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
9
2
22
Dy is the determinant of D, but the second column replaced with
9
2
22
Dz is the determinant of D, but the third column replaced with
9
2
22
Find the four Determinants. Choose between Expansion by Minors or Diagonal Method
D = |
5
2
− 4
− 6
1
5
− 6
0
− 1
| Dx = |
9
2
− 4
2
1
5
22
0
− 1
| Dy = |
5
9
− 4
− 6
2
5
− 6
22
− 1
| Dz = |
5
2
9
− 6
1
2
− 6
0
22
|
D = − 101
Dx = 303
Dy = − 404
Dz = 404
Now Find x, y and z
x = [Dx/D] y = [Dy/D] z = [Dz/D]
x = [303/( − 101)] = − 3
y = [( − 404)/( − 101)] = 4
z = [404/( − 101)] = − 4

Solution = ( − 3,4, − 4)

Solve the system of equations using Cramer's Rule
− 4x − 2y + z = − 7
− 5x − 2y + 2z = − 11
6x + 3y = 3

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute four Determinants as follows
x = [Dx/D] y = [Dy/D] z = [Dz/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
− 7
− 11
3
Dy is the determinant of D, but the second column replaced with
− 7
− 11
3
Dz is the determinant of D, but the third column replaced with
− 7
− 11
3
Find the four Determinants. Choose between Expansion by Minors or Diagonal Method
D = |
− 4
− 2
1
− 5
− 2
2
6
3
0
| Dx = |
− 7
− 2
1
− 11
− 2
2
3
3
0
| Dy = |
− 4
− 7
1
− 5
− 11
2
6
3
0
| Dz = |
− 4
− 2
− 7
− 5
− 2
− 11
6
3
3
|
D = − 3
Dx = 3
Dy = − 9
Dz = 15
Now Find x, y and z
x = [Dx/D] y = [Dy/D] z = [Dz/D]
x = [3/( − 3)] = − 1
y = [( − 9)/( − 3)] = 3
z = [15/( − 3)] = − 5

Solution = ( − 1,3, − 5)

Solve the system of equations using Cramer's Rule
− 4x − 2z = 4
− 4x + 6y + z = − 26
− x + 4y − 5z = 7

To find the solution (x,y) for this problem, we're going to use Determinants, we need to compute four Determinants as follows
x = [Dx/D] y = [Dy/D] z = [Dz/D]
Where D is the determinant of the coefficients of the linear equations.
Where Dx is the determinant of D, but first column replaced with
4
− 26
7
Dy is the determinant of D, but the second column replaced with
4
− 26
7
Dz is the determinant of D, but the third column replaced with
4
− 26
7
Find the four Determinants. Choose between Expansion by Minors or Diagonal Method
D = |
− 4
0
− 2
− 4
6
1
− 1
4
− 5
| Dx = |
4
0
− 2
− 26
6
1
7
4
− 5
| Dy = |
− 4
4
− 2
− 4
− 26
1
− 1
7
− 5
| Dz = |
− 4
0
4
− 4
6
− 26
− 1
4
7
|
D = 156
Dx = 156
Dy = − 468
Dz = − 624
Now Find x, y and z
x = [Dx/D] y = [Dy/D] z = [Dz/D]
x = [156/156] = 1
y = [( − 468)/156] = − 3
z = [( − 624)/156] = − 4

Solution = (1, − 3, − 4)

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Answer

Cramer's Rule

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

System of Two Equations in Two Variables

System of Three Equations in Three Variables

Example 1: Two Equations

Example 2: Two Equations

Example 3: Three Equations

Example 4: Three Equations

Intro 0:00
System of Two Equations in Two Variables 0:16

One Variable
Determinant of Denominator
Determinants of Numerators
Example: System of Equations

System of Three Equations in Three Variables 7:06

Determinant of Denominator
Determinants of Numerators

Example 1: Two Equations 8:57
Example 2: Two Equations 13:21
Example 3: Three Equations 17:11
Example 4: Three Equations 23:43

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: Cramer's Rule

Welcome to Educator.com.0000

In previous lessons, we talked about determinants; today, we are going to apply determinants0002

in order to solve systems of equations and solve for variables.0007

And we are going to do that using what is called Cramer's Rule.0011

Cramer's Rule shows us a way to solve for systems of equations; and we can do that in two variables or three variables.0017

Today, we are going to start out with two variables; and then we will go on in a minute and talk about systems of equations in three variables.0026

Now, one thing to notice, before we even delve into this, is that this is broken down into this formula for x and another one for y.0033

And what makes this rule helpful is that you may be in a situation where you just need to find one variable.0043

You have two or three variables in your system, but you only need one variable.0050

So, instead of solving the entire system, you can just hone in on the variable that you are looking for.0054

OK, so looking at this: if we have a system of equations ax + by = e, and the second equation cx + dy = f; we can find x by using determinants.0059

Let's first look at the determinant in the denominator; let's look at x, and I am going to call this determinant d.0074

Looking at y, it is actually the same determinant; and this is comprised of the coefficients of x in this row and the coefficients of y in this row.0085

So, whether you are looking for x or y, it doesn't matter; the denominator is going to be the same,0116

and it is going to be a determinant that is comprised of, in this first column, coefficients of x, and then here coefficients of y in the second column.0122

OK, now let's look at the numerator; and these are different; and I am going to call the numerator for x...0142

d is the determinant for x, and here I am going to have d and the subscript y to indicate that it is looking for the y; OK.0149

Now, if you look here, you will see that the second columns, where I had the y coefficients, are the same.0158

But if I am looking for x, I am actually going to replace the column where I had the x coefficients,0166

the coefficients of x, with the constants that are on the right side of the equation.0171

And look over here at y: this determinant for y--I kept the x coefficients the same; they are in their place.0179

But where I had the y coefficients, in that column, I replaced those with the constants.0190

So, I just think of this first column as the x column, and the second column as the y column.0195

And if I am looking for x, I replace the x column with the constants; if I am looking for y, I replace the y column with the constants.0200

And then, I just need to solve: OK, so using an example...let's say I have a system of equations, 3x - 5y = 2 and -3x + 2y = -8.0209

And from previous lessons, you know other ways to solve this.0224

You know substitution; you could use elimination; but this is introducing another way that is useful in certain situations.0227

So, looking at this: if I would like to find x, first I am going to find x.0235

Now, this determinant in the denominator is going to consist of the coefficients of x in this first column,0245

3 and -3, and the coefficients of y in the second column, which are -5 and 2.0260

OK, so I have the denominator, and now I need to find the numerator.0271

Recall that, in the numerator, if I am looking for x, I am going to replace the x column with constants; here, that is 2 and -8.0275

The y column--I am just going to keep it as it is down there.0288

OK, once I get this far, I just use my usual methods for finding a determinant.0292

And remember that that is...if a determinant is a, b, c, d--a second-order determinant--I just need to find ad - bc; OK.0298

So, I am going to do that: so x = 2(2) - -8(-5), over 3(2) - -3(-5).0312

OK, therefore, x equals...that is 4 minus...-8 times -5 is 40; in the denominator, 3 times 2 is 6; minus -3 times -5...that gives me 15.0340

This gives me x = (4 - 40), which is -36, over (6 - 15), which is -9; so my negatives cancel out, and -36/-9 is simply 4, so x equals 4.0359

If I wanted to solve for y, I would already have the determinant in the denominator.0379

So, if I wanted to solve for y, it would actually be even simpler, because I already know that the denominator is the same.0387

So, I would just put 4 in here; then I would find the determinant that I need in the numerator;0394

and I would look over here and say, "OK, my x column is the same as in the denominator; it is 3 and -3."0400

y I would replace with the constants, 2 and -8.0409

And then, I would just go ahead and solve, using ad - bc for the determinant, and divide by 4.0415

Cramer's Rule can also be used to solve systems of equations in three variables.0426

And although this looks pretty complicated, it is actually the same idea as we just covered.0432

Looking in the denominator: this first column is the coefficients of x, just like we talked about with systems of two equations.0438

This second column is coefficients of y, and the third is the coefficients of z.0452

And that is exactly the same, whether you are looking for the variable x, y, or z.0466

In the numerator, here you see that this is the same as below--coefficients of x.0474

Here, this is what I think of as my x column; I think of this as my y column; and then here, my z column--0484

I am going to replace that with the constants, since I am looking for z.0497

If I am looking for y, I am going to replace this second column with the constants, j, k, and l.0506

If I am looking for x, in the numerator, I replace that x column with the constants.0516

So, the denominator is always the same; in the numerator, these two columns are the same,0523

but for x this one is different; for y this one is different; and for z this one is different.0529

OK, for example, solving this system of equations using Cramer's Rule: this is in the form ax + by = some constant (I am going to call it e).0536

Here, I am going to call this cx + dy = f; and then, just to remind you that, if I am looking for x or y, what I am going to do...0557

In the denominator, I am going to have a determinant that consists of ac (coefficients of x)0573

in the first columns, and bd (coefficients of y) in the second column.0580

And that is going to be the same for both.0584

In the numerator, I am going to replace this first column with the constants; the second column will remain the same (coefficients of y).0587

Here, the x column still will have the coefficients of x, but the second column, where the y coefficients were--I am going to replace that with the constants, ef.0597

Therefore, if I am solving for x, let's start out with the denominator.0606

The denominator is going to be...the coefficient of x is 1; the other coefficient of x is 3.0614

The coefficient of y is -2, and the other is 1.0623

In the numerator, I am going to keep the y section the same; I am going to replace x with the constants 1 and 2, the x coefficients.0627

Now, I am just going to recall that, in order to find the second-order determinant here, it is going to be...this is a, b, c, d; it is going to be ad - bc.0635

Therefore, x equals...this is going to be 1 times 1, minus -2 times 2, over 1 times 1, minus -2 times 3.0647

Therefore, x equals 1 minus...-2 times 2 is -4, over 1 minus...-2 times 3 is -6; x equals 1...a negative and a negative gives me a positive;0670

the same here--a negative and a negative gives me a positive; so x equals 5/7.0688

Solving for y is actually going to be easier, because recall that these determinants are the same: the denominator is the same for x and y.0698

Well, I already did all this work, finding the denominator, which is 7.0709

So, I am not going to repeat that; I am just going to get rid of this and write 7 here.0713

In the numerator, in this first column, I am going to have my coefficients of x, 1 and 3, just like I did in the denominator.0717

However, in this second column, I am going to replace it with the constants 1 and 2.0728

Again, I am using my formula ad - bc to find this determinant; that is going to give me 1(2) - 3(1), all over 7.0737

This is y = (2 - 3)/7, so y = -1/7.0754

So, x = 5/7; y = -1/7; and I was able to figure that out using Cramer's rule, where the denominator is the determinant0763

formed by the coefficients of x and the coefficients of y, and the numerator is...0774

the first column consists of the constants; the second is the coefficients of y.0779

Once I did that, y was easier to find, because I already had the denominator.0785

I just plugged that in right here; for the numerator, I had a determinant consisting of the coefficients of x, and then (in the second column) the constants.0790

OK, again, we are solving a system of equations with two variables for x.0802

OK, recall that in the denominator, this first column is going to be coefficients of x; the second is going to be the coefficients of y.0815

In the numerator, I am going to leave the y's alone; but I am going to replace x with 8 and 4, the constants.0825

Now, I am going to use my rule, ad - bc, assuming that this is the set-up, to find these determinants.0837

x equals a times d, minus b times c; therefore, x equals...8 times 3 is 24, minus -16, or x = 24...a negative and a negative...that is + 16.0847

So, 24 and 16 is 40; x equals 40 in the numerator, over this denominator.0870

Let's find the denominator; the denominator determinant is 3 times 3, so that is the numerator for x.0886

Now, to find the denominator (this is just the numerator), I have 3 times 3, minus -4 times 2.0899

3 times 3 is 9, and in the denominator, I have -4 times 2, so that is -8; and then, this is going give me 9.0920

A negative and a negative is a positive; 9 plus 8 equals 17.0933

This is my numerator right here, and I have my x denominator right here; and this is going to give me x = 40/17.0937

I found my determinants; this consists of the x coefficients and the y coefficients; this one is the constants and the y coefficients.0951

And then, I found the numerator right here; I found the denominator right here.0960

And then, I went ahead put those together, and then x equals 40/17.0968

Now, I am finding y: finding y is going to be easier, because here I have my denominator,0975

which is 17--the same denominator, so that takes a lot of work out of it.0981

In the numerator, in that first column, go the x coefficients.0986

I replace the second column with the constants; and then, I go ahead and find this determinant,0992

which is going to be 3 times 4, minus 8 times 2.1002

This is going to give me 12 minus 16 over 17, and that is going to be...12 - 16 is -4, over 17.1010

So, here I have...x is 40/17 y is -4/17--using Cramer's Rule.1022

Now, looking at a system of equations with three variables, I am being asked only to solve for x.1032

And this shows you how this rule can be useful, because I only have to focus in on that, and not solve the entire system.1038

You use the same general approach as with three variables.1046

So, I am going to take x; and in the denominator, I am going to say, "OK, the first column is the x coefficients."1052

The coefficient here is 1; 2; 1; y column: -2, 1...there is no y here, so the coefficient is just 0.1060

z--there is no z at this first one, so that is going to be 0, 1, 1.1075

All right, now I am looking for x; the first column represents the x coefficients--I am going to hold on to that thought for a second.1081

And I am just going to copy over the other two columns, because I am not messing with those.1090

I am going to replace this first column with the constants, 3, 1, 3.1095

Now that I have set this up, I have to find a third-order determinant in the numerator and the denominator.1104

And you recall that we went over a couple of methods to do that: expansion of the minors and the diagonal method.1110

And now, I am going to use the diagonal method.1116

The first step of the diagonal method is to copy over the first two columns--so, up here, 3, 1, 3, -2, 1, 0.1120

Here, it is 1, 2, 1, -2, 1, 0.1130

Once I have done that, then I need to start at the upper left and go along the diagonals.1137

And I am going to find, actually, right along this way...the product of that and add1145

the product of this diagonal to the product of the next diagonal to the product of the next diagonal.1163

OK, so x equals...the first product...this is 3 times 1 times 1, plus -2 times 1 times 3, plus 0 times 1 times 0.1171

OK, I have taken care of those diagonals; now I start at the lower left and go along those diagonals.1200

And I am going to subtract those products: so minus 3 times 1 times 0.1206

Next, 0, 1, 3--minus 0 times 1 times 3--minus the last diagonal: and that is 1, times 1, times -2.1215

OK, that takes care of the numerator; in the denominator, I am going to start at the upper left.1232

My first diagonal is 1 times 1 times 1; and I am going to add that to the next diagonal, -2 times 1 times 1.1238

I am adding that to this diagonal, 0 times 2 times 0.1253

Now, subtracting: we are starting down here: 1 times 1 times 0; we are subtracting 0 times 1 times 1,1258

and then finally subtracting 1 times 2 times -2.1274

I'm running out of room; OK.1285

1 times 2 times -2; OK, I am working this out...1287

Therefore, x equals 3, and this is -2 times 1 times 3, which gives me -6, plus 0, minus 0, minus 0.1295

1 times 1 times -2...so that is minus -2.1308

In the denominator: the 1's multiply and give me 1; plus -2 times 1 is -2, times 1...-2 times 1 times 1 is -2;1313

plus...this is going to come out to 0; minus 1 times 1 times 0; minus 0 times 1 times 1 (that is another 0); minus 1 times 2 times -2.1326

OK, let's clean this up: this is 3 minus 6; get rid of all these 0's; and this is minus -2, and that is the same as plus 2.1340

In the denominator, I have 1 minus 2; get rid of all the 0's; and this is minus -2; actually, that is 2 times -2--this should be -4; minus -4 gives me plus 4.1352

Now, x equals...well, 3 minus 6 plus 2 is going to give me -1;1373

in the denominator, I have 1 minus 2 (that is -1), plus 4: that is going to give me 3.1382

So, x is -1/3; and as you can see, once you get it set up, then it is a matter of just keeping track of all of that multiplication and these signs.1388

So, I set this up by forming determinants with coefficients of x, y, and z in the denominator.1398

In the numerator, there are coefficients of y and z, and then the constants in the first row.1407

Then, I solve this by using the diagonal method; and I did that right here to come up with x = -1/3.1412

OK, again, we are asked to solve a system of three equations.1423

I am going to start out by looking for x, and looking at the denominator.1427

For the denominator, I am going to use the coefficients of x: 2, -6, and 4; the coefficients of y: -3, 9, -1; and the coefficients of z: 1, -3, 1.1437

OK, in the numerator, I am looking for x, so I am going to keep the y and z columns4 just as they were.1455

And I am going to go over here to the x area, and I am going to substitute the constants 4, 11, and 10.1465

Now, I am going to, again, use the diagonal method; and I am actually going to start with the denominator.1471

And you will see why in a minute--why you should always start with the denominator, and how it can save you work.1475

So, I need to find this third-order determinant; and I am going to use the diagonal method.1481

I am going to rewrite these first two columns: 2, -6, 4, -3, 9, -1.1486

After I have done that, I am just going ahead and rewriting this; I am not working with the numerator yet.1496

I am just going to leave that like this.1504

And I am only working at finding this determinant in the denominator.1510

Starting at the upper left, using the diagonal method, I am going to find the products of this first diagonal: that is 2 times 9 times 1,1513

plus this next number right here--next set of numbers: -3, -3, 4--that product, plus, coming down right here, 1, -6, -1.1531

Then, from that, I am going to start subtracting this other set of diagonals, starting down here, and going up:1558

the product of 4 times 9 times 1; continuing on: minus -1, -3, and 2, minus this last diagonal right here: 1, -6, -3.1566

So, x equals the same determinant; let's just work with the denominator right now.1591

2 times 9 times 1 is 18; plus -3 times -3 (is 9) times 4--that is 36; -6 and 1--that is 6.1598

OK, and now subtracting: 4 times 9 times 1--that is 36; minus...-1 and -3 is 3, times 2 is 6, minus 1 times -6 is -6, times -3 is 18.1610

Now, you can see what is happening here: x equals this determinant in the numerator.1628

But I have 18 and 18, so those cancel; that gives me 0.1634

I have 36 and 36, so those cancel: 0; 6 and 6--those cancel; so I have a 0 in the denominator, and that is not allowed.1640

Since that is not defined, what I have ended up with is a situation where Cramer's Rule does not work.1651

It doesn't mean that there is not necessarily a solution to this system of equations, but what it means that there is no unique solution.1657

It may be that there is an infinite number of solutions (it is a dependent system).1666

It may mean that there is no solution.1671

But I can't go any farther, and I can't use Cramer's Rule in this situation.1674

And that is why the first thing you should do, if you are using Cramer's Rule, is to check out the denominator and make sure that it is not 0.1678

If the determinant in the denominator, d, is 0, you can't use this method.1685

And that concludes this session of Educator.com on Cramer's Rule; I will see you next lesson.1699

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Algebra and Trigonometry: Structure and Method, Book 2

Author: McDougal Littell

ISBN: 395977258

Publisher: McDougal Littell

Year: 1999

This classic textbook does a great job of explaining concepts and includes lots of in-depth, worked out examples in every section as well as practice problems.

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Dr. Carleen Eaton

Cramer's Rule

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro