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Intermediate Algebra

4.6 Solve Systems of Equations Using Determinants

Intermediate Algebra4.6 Solve Systems of Equations Using Determinants

Learning Objectives

By the end of this section, you will be able to:
  • Evaluate the determinant of a 2×22×2 matrix
  • Evaluate the determinant of a 3×33×3 matrix
  • Use Cramer’s Rule to solve systems of equations
  • Solve applications using determinants

Be Prepared 4.6

Before you get started, take this readiness quiz.

  1. Simplify: 5(−2)(−4)(1).5(−2)(−4)(1).
    If you missed this problem, review Example 1.20.
  2. Simplify: −3(810)+(−2)(63)4(−3(−4)).−3(810)+(−2)(63)4(−3(−4)).
    If you missed this problem, review Example 1.19.
  3. Simplify: −12−8.−12−8.
    If you missed this problem, review Example 1.18.

In this section we will learn of another method to solve systems of linear equations called Cramer’s rule. Before we can begin to use the rule, we need to learn some new definitions and notation.

Evaluate the Determinant of a 2×22×2 Matrix

If a matrix has the same number of rows and columns, we call it a square matrix. Each square matrix has a real number associated with it called its determinant. To find the determinant of the square matrix [abcd],[abcd], we first write it as |abcd|.|abcd|. To get the real number value of the determinate we subtract the products of the diagonals, as shown.

A 2 by 2 determinant is show, with its first row being a, b and second one being c, d. These values are written between two vertical lines instead of brackets as in the case of matrices. Two arrows are shown, one from a to d, the other from c to b. This determinant is equal to ad minus bc.

Determinant

The determinant of any square matrix [abcd],[abcd], where a, b, c, and d are real numbers, is

|abcd|=adbc|abcd|=adbc

Example 4.45

Evaluate the determinate of [4−23−1][4−23−1] [−3−4−20].[−3−4−20].

Try It 4.89

Evaluate the determinate of [5−32−4][5−32−4] [−4−607].[−4−607].

Try It 4.90

Evaluate the determinate of [−13−24][−13−24] [−7−3−50].[−7−3−50].

Evaluate the Determinant of a 3×33×3 Matrix

To evaluate the determinant of a 3×33×3 matrix, we have to be able to evaluate the minor of an entry in the determinant. The minor of an entry is the 2×22×2 determinant found by eliminating the row and column in the 3×33×3 determinant that contains the entry.

Minor of an entry in 3 × 3 3 × 3 a Determinant

The minor of an entry in a 3×33×3 determinant is the 2×22×2 determinant found by eliminating the row and column in the 3×33×3 determinant that contains the entry.

To find the minor of entry a1,a1, we eliminate the row and column which contain it. So we eliminate the first row and first column. Then we write the 2×22×2 determinant that remains.

The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. a1 is highlighted. Lines strike out the first row and the first column. What remains is called minor of a1. It is shown as a separate determinant whose first row is b2, c2 and second row is b3, c3.

To find the minor of entry b2,b2, we eliminate the row and column that contain it. So we eliminate the 2nd row and 2nd column. Then we write the 2×22×2 determinant that remains.

The first row of the 3 by 3 determinant is a1, b1, c1. Row 2 is a2, b2, c2. Row 3 is a3, b3, c3. b2 is highlighted. Lines strike out the second row and second column. What remains is minor of b2. It is written as a separate determinant whose first row is a1, c1 and second row is a3, c3.

Example 4.46

For the determinant |4−2310−3−2−42|,|4−2310−3−2−42|, find and then evaluate the minor of a1a1 b3b3 c2.c2.

Try It 4.91

For the determinant |1−1402−1−2−33|,|1−1402−1−2−33|, find and then evaluate the minor of a1a1 b2b2 c3.c3.

Try It 4.92

For the determinant |−2−1030−1−1−23|,|−2−1030−1−1−23|, find and then evaluate the minor of a2a2 b3b3 c2.c2.

We are now ready to evaluate a 3×33×3 determinant. To do this we expand by minors, which allows us to evaluate the 3×33×3 determinant using 2×22×2 determinants—which we already know how to evaluate!

To evaluate a 3×33×3 determinant by expanding by minors along the first row, we use the following pattern:

A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

Remember, to find the minor of an entry we eliminate the row and column that contains the entry.

Expanding by Minors along the First Row to Evaluate a 3 × 3 3 × 3 Determinant

To evaluate a 3×33×3 determinant by expanding by minors along the first row, the following pattern:

A 3 by 3 determinant is equal to a1 times minor of a1 minus b1 times minor of b1 plus c1 times minor of c1.

Example 4.47

Evaluate the determinant |2−3−1320−1−1−2||2−3−1320−1−1−2| by expanding by minors along the first row.

Try It 4.93

Evaluate the determinant |3−240−1−223−1|,|3−240−1−223−1|, by expanding by minors along the first row.

Try It 4.94

Evaluate the determinant |3−2−22−14−10−3|,|3−2−22−14−10−3|, by expanding by minors along the first row.

To evaluate a 3×33×3 determinant we can expand by minors using any row or column. Choosing a row or column other than the first row sometimes makes the work easier.

When we expand by any row or column, we must be careful about the sign of the terms in the expansion. To determine the sign of the terms, we use the following sign pattern chart.

|+++++||+++++|

Sign Pattern

When expanding by minors using a row or column, the sign of the terms in the expansion follow the following pattern.

|+++++||+++++|

Notice that the sign pattern in the first row matches the signs between the terms in the expansion by the first row.

A 3 by 3 determinant has row 1: plus, minus, plus, row 2: minus, plus, minus and row 3: plus, minus, plus. The three signs in the first row each point to a minor determinant in the expansion of a 3 by 3 determinant. Plus points to minor of a1, minus to the minor of b1 and plus to the minor of c1.

Since we can expand by any row or column, how do we decide which row or column to use? Usually we try to pick a row or column that will make our calculation easier. If the determinant contains a 0, using the row or column that contains the 0 will make the calculations easier.

Example 4.48

Evaluate the determinant |4−1−33025−4−3||4−1−33025−4−3| by expanding by minors.

Try It 4.95

Evaluate the determinant |2−1−303−43−4−3||2−1−303−43−4−3| by expanding by minors.

Try It 4.96

Evaluate the determinant |−2−1−3−1224−40||−2−1−3−1224−40| by expanding by minors.

Use Cramer’s Rule to Solve Systems of Equations

Cramer’s Rule is a method of solving systems of equations using determinants. It can be derived by solving the general form of the systems of equations by elimination. Here we will demonstrate the rule for both systems of two equations with two variables and for systems of three equations with three variables.

Let’s start with the systems of two equations with two variables.

Cramer’s Rule for Solving a System of Two Equations

For the system of equations {a1x+b1y=k1a2x+b2y=k2,{a1x+b1y=k1a2x+b2y=k2, the solution (x,y)(x,y) can be determined by

x is Dx upon D and y is Dy upon D where D is determinant with row 1: a1, b1 and row 2 a2, b2, use coefficients of the variables; Dx is determinant with row 1: k1, b1 and row 2: k2, b2, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1 and row 2: a2, k2, replace the y coefficients with constants

Notice that to form the determinant D, we use take the coefficients of the variables.

The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is D with row 1: a1, b1 and row 2: a2, b2. Column 1 has coefficients of x and column 2 has coefficients of

Notice that to form the determinant DxDx and Dy,Dy, we substitute the constants for the coefficients of the variable we are finding.

The equations are a1x plus b1y equals k1 and a2x plus b2y equals k2. Here, a1, a2, b1, b2 are coefficients. The determinant is Dx has row 1: k1, b1 and row 2: k2, b2. Here columns 1 and 2 re constants and coefficients of y respectively. Determinant Dy has row 1: a1, k1 and row 2: a2, k2. Here, columns 1 and 2 are coefficients of x and constants respectively.

Example 4.49

How to Solve a System of Equations Using Cramer’s Rule

Solve using Cramer’s Rule: {2x+y=−43x2y=−6.{2x+y=−43x2y=−6.

Try It 4.97

Solve using Cramer’s rule: {3x+y=−32x+3y=6.{3x+y=−32x+3y=6.

Try It 4.98

Solve using Cramer’s rule: {x+y=22x+y=−4.{x+y=22x+y=−4.

How To

Solve a system of two equations using Cramer’s rule.

  1. Step 1. Evaluate the determinant D, using the coefficients of the variables.
  2. Step 2. Evaluate the determinant Dx.Dx. Use the constants in place of the x coefficients.
  3. Step 3. Evaluate the determinant Dy.Dy. Use the constants in place of the y coefficients.
  4. Step 4. Find x and y. x=DxD,x=DxD, y=DyDy=DyD
  5. Step 5. Write the solution as an ordered pair.
  6. Step 6. Check that the ordered pair is a solution to both original equations.

To solve a system of three equations with three variables with Cramer’s Rule, we basically do what we did for a system of two equations. However, we now have to solve for three variables to get the solution. The determinants are also going to be 3×33×3 which will make our work more interesting!

Cramer’s Rule for Solving a System of Three Equations

For the system of equations {a1x+b1y+c1z=k1a2x+b2y+c2z=k2a3x+b3y+c3z=k3,{a1x+b1y+c1z=k1a2x+b2y+c2z=k2a3x+b3y+c3z=k3, the solution (x,y,z)(x,y,z) can be determined by

x is Dx upon D, y is Dy upon D and z is Dz upon D, where D is determinant with row 1: a1, b1, c1, row 2: a2, b2, c2, row 3: a3, b3, c3, use coefficients of the variables; Dx is determinant with row 1: k1, b1, c1, row 2: k2, b2, c2 and rwo 3: k3, b3, c3, replace the x coefficients with the consonants; Dy is determinant with row 1: a1, k1, c1, row 2: a2, k2, c2 and row 3: a3, k3, c3, replace the y coefficients with constants; Dz is determinant with row 1: a1, b1, k1; row 2: a2, b2, k2, row 3: a3, b3, k3; replace the z coefficients with constants.

Example 4.50

Solve the system of equations using Cramer’s Rule: {3x5y+4z=55x+2y+z=02x+3y2z=3.{3x5y+4z=55x+2y+z=02x+3y2z=3.

Try It 4.99

Solve the system of equations using Cramer’s Rule: {3x+8y+2z=−52x+5y3z=0x+2y2z=−1.{3x+8y+2z=−52x+5y3z=0x+2y2z=−1.

Try It 4.100

Solve the system of equations using Cramer’s Rule: {3x+y6z=−32x+6y+3z=03x+2y3z=−6.{3x+y6z=−32x+6y+3z=03x+2y3z=−6.

Cramer’s rule does not work when the value of the D determinant is 0, as this would mean we would be dividing by 0. But when D=0,D=0, the system is either inconsistent or dependent.

When the value of D=0D=0 and Dx,DyDx,Dy and DzDz are all zero, the system is consistent and dependent and there are infinitely many solutions.

When the value of D=0D=0 and Dx,DyDx,Dy and DzDz are not all zero, the system is inconsistent and there is no solution.

Dependent and Inconsistent Systems of Equations

For any system of equations, where the value of the determinant D=0,D=0,
Value of determinantsType of systemSolutionD=0andDx,DyandDzare all zeroconsistent and dependentinfinitely many solutionsD=0andDx,DyandDzare not all zeroinconsistentno solutionValue of determinantsType of systemSolutionD=0andDx,DyandDzare all zeroconsistent and dependentinfinitely many solutionsD=0andDx,DyandDzare not all zeroinconsistentno solution

In the next example, we will use the values of the determinants to find the solution of the system.

Example 4.51

Solve the system of equations using Cramer’s rule : {x+3y=4−2x6y=3.{x+3y=4−2x6y=3.

Try It 4.101

Solve the system of equations using Cramer’s rule: {4x3y=88x6y=14.{4x3y=88x6y=14.

Try It 4.102

Solve the system of equations using Cramer’s rule: {x=−3y+42x+6y=8.{x=−3y+42x+6y=8.

Solve Applications using Determinants

An interesting application of determinants allows us to test if points are collinear. Three points (x1,y1),(x1,y1), (x2,y2)(x2,y2) and (x3,y3)(x3,y3) are collinear if and only if the determinant below is zero.

|x1y11x2y21x3y31|=0|x1y11x2y21x3y31|=0

Test for Collinear Points

Three points (x1,y1),(x1,y1), (x2,y2)(x2,y2) and (x3,y3)(x3,y3) are collinear if and only if

|x1y11x2y21x3y31|=0|x1y11x2y21x3y31|=0

We will use this property in the next example.

Example 4.52

Determine whether the points (5,−5),(5,−5), (4,−3),(4,−3), and (3,−1)(3,−1) are collinear.

Try It 4.103

Determine whether the points (3,−2),(3,−2), (5,−3),(5,−3), and (1,−1)(1,−1) are collinear.

Try It 4.104

Determine whether the points (−4,−1),(−4,−1), (−6,2),(−6,2), and (−2,−4)(−2,−4) are collinear.

Media

Access these online resources for additional instruction and practice with solving systems of linear inequalities by graphing.

Section 4.6 Exercises

Practice Makes Perfect

Evaluate the Determinant of a 2 × 2 Matrix

In the following exercises, evaluate the determinate of each square matrix.

232.

[ 6 −2 3 −1 ] [ 6 −2 3 −1 ]

233.

[ −4 8 −3 5 ] [ −4 8 −3 5 ]

234.

[ −3 5 0 −4 ] [ −3 5 0 −4 ]

235.

[ −2 0 7 −5 ] [ −2 0 7 −5 ]

Evaluate the Determinant of a 3 × 3 Matrix

In the following exercises, find and then evaluate the indicated minors.

236.

|3−14−10−2−415||3−14−10−2−415|
Find the minor a1a1 b2b2 c3c3

237.

|−1−324−2−1−20−3||−1−324−2−1−20−3|
Find the minor a1a1 b1b1 c2c2

238.

|2−3−4−12−30−1−2||2−3−4−12−30−1−2|
Find the minor a2a2 b2b2 c2c2

239.

|−2−231−30−23−2||−2−231−30−23−2|
Find the minor a3a3 b3b3 c3c3

In the following exercises, evaluate each determinant by expanding by minors along the first row.

240.

| −2 3 −1 −1 2 −2 3 1 −3 | | −2 3 −1 −1 2 −2 3 1 −3 |

241.

| 4 −1 −2 −3 −2 1 −2 −5 7 | | 4 −1 −2 −3 −2 1 −2 −5 7 |

242.

| −2 −3 −4 5 −6 7 −1 2 0 | | −2 −3 −4 5 −6 7 −1 2 0 |

243.

| 1 3 −2 5 −6 4 0 −2 −1 | | 1 3 −2 5 −6 4 0 −2 −1 |

In the following exercises, evaluate each determinant by expanding by minors.

244.

| −5 −1 −4 4 0 −3 2 −2 6 | | −5 −1 −4 4 0 −3 2 −2 6 |

245.

| 4 −1 3 3 −2 2 −1 0 4 | | 4 −1 3 3 −2 2 −1 0 4 |

246.

| 3 5 4 −1 3 0 −2 6 1 | | 3 5 4 −1 3 0 −2 6 1 |

247.

| 2 −4 −3 5 −1 −4 3 2 0 | | 2 −4 −3 5 −1 −4 3 2 0 |

Use Cramer’s Rule to Solve Systems of Equations

In the following exercises, solve each system of equations using Cramer’s Rule.

248.

{ −2 x + 3 y = 3 x + 3 y = 12 { −2 x + 3 y = 3 x + 3 y = 12

249.

{ x 2 y = −5 2 x 3 y = −4 { x 2 y = −5 2 x 3 y = −4

250.

{ x 3 y = −9 2 x + 5 y = 4 { x 3 y = −9 2 x + 5 y = 4

251.

{ 2 x + y = −4 3 x 2 y = −6 { 2 x + y = −4 3 x 2 y = −6

252.

{ x 2 y = −5 2 x 3 y = −4 { x 2 y = −5 2 x 3 y = −4

253.

{ x 3 y = −9 2 x + 5 y = 4 { x 3 y = −9 2 x + 5 y = 4

254.

{ 5 x 3 y = −1 2 x y = 2 { 5 x 3 y = −1 2 x y = 2

255.

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

256.

{ 6 x 5 y + 2 z = 3 2 x + y 4 z = 5 3 x 3 y + z = −1 { 6 x 5 y + 2 z = 3 2 x + y 4 z = 5 3 x 3 y + z = −1

257.

{ 4 x 3 y + z = 7 2 x 5 y 4 z = 3 3 x 2 y 2 z = −7 { 4 x 3 y + z = 7 2 x 5 y 4 z = 3 3 x 2 y 2 z = −7

258.

{ 2 x 5 y + 3 z = 8 3 x y + 4 z = 7 x + 3 y + 2 z = −3 { 2 x 5 y + 3 z = 8 3 x y + 4 z = 7 x + 3 y + 2 z = −3

259.

{ 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3 { 11 x + 9 y + 2 z = −9 7 x + 5 y + 3 z = −7 4 x + 3 y + z = −3

260.

{ x + 2 z = 0 4 y + 3 z = −2 2 x 5 y = 3 { x + 2 z = 0 4 y + 3 z = −2 2 x 5 y = 3

261.

{ 2 x + 5 y = 4 3 y z = 3 4 x + 3 z = −3 { 2 x + 5 y = 4 3 y z = 3 4 x + 3 z = −3

262.

{ 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1 { 2 y + 3 z = −1 5 x + 3 y = −6 7 x + z = 1

263.

{ 3 x z = −3 5 y + 2 z = −6 4 x + 3 y = −8 { 3 x z = −3 5 y + 2 z = −6 4 x + 3 y = −8

264.

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

265.

{ x 4 y = −1 −3 x + 12 y = 3 { x 4 y = −1 −3 x + 12 y = 3

266.

{ −3 x y = 4 6 x + 2 y = −16 { −3 x y = 4 6 x + 2 y = −16

267.

{ 4 x + 3 y = 2 20 x + 15 y = 5 { 4 x + 3 y = 2 20 x + 15 y = 5

268.

{ x + y 3 z = −1 y z = 0 x + 2 y = 1 { x + y 3 z = −1 y z = 0 x + 2 y = 1

269.

{ 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20 { 2 x + 3 y + z = 12 x + y + z = 9 3 x + 4 y + 2 z = 20

270.

{ 3 x + 4 y 3 z = −2 2 x + 3 y z = −12 x + y 2 z = 6 { 3 x + 4 y 3 z = −2 2 x + 3 y z = −12 x + y 2 z = 6

271.

{ x 2 y + 3 z = 1 x + y 3 z = 7 3 x 4 y + 5 z = 7 { x 2 y + 3 z = 1 x + y 3 z = 7 3 x 4 y + 5 z = 7

Solve Applications Using Determinants

In the following exercises, determine whether the given points are collinear.

272.

(0,1),(0,1), (2,0),(2,0), and (−2,2).(−2,2).

273.

(0,−5),(0,−5), (−2,−2),(−2,−2), and (2,−8).(2,−8).

274.

(4,−3),(4,−3), (6,−4),(6,−4), and (2,−2).(2,−2).

275.

(−2,1),(−2,1), (−4,4),(−4,4), and (0,−2).(0,−2).

Writing Exercises

276.

Explain the difference between a square matrix and its determinant. Give an example of each.

277.

Explain what is meant by the minor of an entry in a square matrix.

278.

Explain how to decide which row or column you will use to expand a 3×33×3 determinant.

279.

Explain the steps for solving a system of equations using Cramer’s rule.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 4 rows and a header row. The header row labels each column: I ca, confidently, with some help and no, I don’t get it. The first column has the following statements: Evaluate the Determinant of a 2 by 2 Matrix, Evaluate the Determinant of a 3 by 3 Matrix, Use Cramer’s Rule to Solve Systems of Equations, Solve Applications Using Determinants. The remaining columns are blank.

After reviewing this checklist, what will you do to become confident for all objectives?

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