231

Section 5.3 – Systems of Linear Equations: Determinants

In this section, we will explore another technique for solving systems called Cramer's Rule. Cramer's rule can only be used if the number of equations matches the number of variables. Also, if the problem consists of a system of linear dependent equations, then Cramer's Rule only tells us that our equations are dependent. Cramer's Rule works by using what are called determinants. Objective 1: Evaluate 2x2 determinants. If a matrix has the same number of rows and columns, it is called a square matrix. Every square matrix has a real number associated with it called its determinant. The determinant is defined as follows for 2×2 matrices: 2×2 Determinants

The determinant for 2×2 matrix is denoted as

€

a1 b1a2 b2

and is

defined as:

€

a1 b1a2 b2

= a1b2 – a2b1.

It is almost like cross multiplication except we are subtracting instead of setting the products equal. Evaluate the following:

Ex. 1

€

5 −34 1

Solution:

€

5 −34 1

= 5(1) – 4(– 3) = 5 + 12 = 17.

Ex. 2

€

3 22 −3

Solution:

€

3 22 −3

= 3(– 3) – 2(2) = – 9 – 4 = – 13.

232

Objective 2: Cramer’s Rule for Solving a System of Two Linear Equations in Two Variables

Cramer’s Rule for 2x2 Systems For the system: ax + by = s (the equations must be in standard form) cx + dy = t We can first calculate the following determinants:

D =

€

a bc d

Dx =

€

s bt d

Dy =

€

a sc t

Then, the solution to the system is

x =

€

DxD

and y =

€

DyD

If D = 0, and if either Dx or Dy or both are not zero, then system is inconsistent. If D, Dx, and Dy are all zero, then the equations are dependent. We must then use another technique to solve the system. To find Dx, replace the coefficients for x in the determinant by the constant terms. To find Dy, replace the coefficients for y in the determinant by the constant terms. To see where Cramer’s Rule comes from, let us solve the following system using the elimination method.

Solve using elimination: Ex. 3 ax + by = s cx + dy = t Solution: Let’s first eliminate x. – c•eqn. #1 + a•eqn. #2: – c•1) – acx – cby = – cs a•2) acx + ady = at ady – cby = at – cs (factor out y) (ad – cb)y = at – cs (solve for y) y =

€

at−csad−cb

But at – cs =

€

a sc t

= Dy and ad – cb =

€

a bc d

= D.So, y =

€

DyD .

Now, let’s eliminate y. d•eqn. #1 + – b•eqn. #2: d•1) adx + bdy = ds – b•2) – cbx – bdy = – bt

233

adx – cbx = ds – bt (factor out x) (ad – cb)x = ds – bt (solve for x) x =

€

ds−btad−cb

But ds – bt =

€

s bt d

= Dx and ad – cb =

€

a bc d

= D. So, x =

€

DxD .

Solve using Cramer’s Rule: Ex. 4 5x – 3y = 13 4x + y = 7 Solution:

D =

€

5 −34 1

= 5(1) – 4(– 3) = 5 + 12 = 17.

Dx =

€

13 −37 1

= 13(1) – 7(– 3) = 13 + 21 = 34.

Dy =

€

5 134 7

= 5(7) – 4(13) = 35 – 52 = – 17.

x =

€

DxD =

€

3417

= 2 and y =

€

DyD =

€

−1717

= – 1. So, the solution is (2, – 1). Ex. 5 6x – 4y = 12 – 9x + 6y = – 18 Solution:

D =

€

6 −4−9 6

= 6(6) – (– 9)(– 4) = 36 – 36 = 0.

Dx =

€

12 −4−18 6

= 12(6) – (– 18)(– 4) = 72 – 72 = 0

Dy =

€

6 12−9 −18

= 6(– 18) – (– 9)(12) = – 108 + 108 = 0

The system is dependent. Solve for y: 6x – 4y = 12 ⇒ – 4y = – 6x + 12 ⇒ y = 1.5x – 3 So, the solution is { (x, y) | y = 1.5x – 3, x is any real number}.

234

Objective 3: Understanding and evaluating 3×3 determinants.

The 3×3 determinant will have the form:

€

a11 a12 a13a21 a22 a23a31 a32 a33

Before we can find the determinant for an n×n matrix with n being 3 or higher, we need to define the minor and cofactors of an entry of the determinant. Minors For any n×n determinant, the minor of Mij of entry aij is the determinant of the matrix with the ith row and jth column deleted. Hence, minor M23 can be found be crossing out the second row and third column where as the minor M31 can be found by crossing out the third row and first column.

M23 =

€

a11 a12 a13a21 a22 a23a31 a32 a33

=

€

a11 a12a31 a32

M31 =

€

a11 a12 a13a21 a22 a23a31 a32 a33

=

€

a12 a13a22 a23

Cofactor For any n×n determinant, the cofactor of entry aij, denoted by Aij, is given by: Aij = (– 1)i + j Mij where Mij is the minor of aij.

Find the following cofactors of the given matrix on the right:

€

3 −2 4−1 5 7−4 8 2

Solution:

a) A12 = (– 1)1 + 2

€

3 −2 4−1 5 7−4 8 2

= (– 1)3

€

−1 7−4 2 = – 1(– 2 + 28) = – 26

b) A31 = (– 1)3 + 1

€

3 −2 4−1 5 7−4 8 2

= (– 1)4

€

−2 45 7 = 1(– 14 – 20) = – 34

Ex. 6a A12 Ex. 6b A31

235

Evaluating an n×n determinant To calculate the value of an n×n determinant, multiply each entry in any column or row by its corresponding cofactor and add the results. The process is called expanding across a column or row.

Evaluate the following:

Ex. 7

€

2 −1 11 2 −13 4 −3

Solution: Expand across column #1:

€

2 −1 11 2 −13 4 −3

= 2(–1)2

€

2 −14 −3

+ 1(– 1)3

€

−1 14 −3

+ 3(– 1)4

€

−1 12 −1

= 2[2(– 3) – 4(– 1)] – 1[(– 1)(– 3) – 4(1)] + 3[(– 1)(– 1) – 2(1)] = 2[– 2] – 1[– 1] + 3[– 1] = – 4 + 1 – 3 = – 6.

Ex. 8

€

2 4 3−2 1 11 3 2

Solution: Expand across row #2:

€

2 4 3−2 1 11 3 2

= – 2(– 1)3

€

4 33 2

+ 1(– 1)4

€

2 31 2

+ 1(– 1)5

€

2 41 3

= 2[4(2) – 3(3)] + 1[(2)(2) – 1(3)] – 1[(2)(3) – 1(4)] = 2[– 1] + 1[1] – 1[2] = – 2 + 1 – 2 = – 3. Objective 4: Cramer’s Rule for solving a system of three linear equations in three variables. We will now state Cramer’s Rule for solving a system of three linear equations in three variables. The proof of Cramer’s Rule is quite a bit more complicated than the proof of Cramer’s Rule for 2x2 systems and it will be omitted. The proof does follow the same spirit as for the 2x2 systems.

236

Cramer’s Rule for 3x3 Systems For the system: a11x + a12y + a13z = c1 a21x + a22y + a23z = c2 a31x + a32y + a33z = c3 We can first calculate the following determinants:

D =

€

a11 a12 a13a21 a22 a23a31 a32 a33

Dx =

€

c1 a12 a13c2 a22 a23c3 a32 a33

Dy =

€

a11 c1 a13a21 c2 a23a31 c3 a33

Dz =

€

a11 a12 c1a21 a22 c2a31 a32 c3

Then, the solution to the system is

x =

€

DxD

, y =

€

DyD

, and z =

€

DzD

If D, Dx, Dy, and Dz are all zero, then the system is dependent and there are an infinite number of solutions. If D is zero and if at least one of Dx, Dy, and Dz is not zero, then the system inconsistent and has no solution. To find Dx, replace the coefficients for x in the determinant by the constant term. To find Dy, replace the coefficients for y in the determinant by the constant term. To find Dz, replace the coefficients for z in the determinant by the constant term. Solve the following systems using Cramer’s Rule: Ex. 9 3x + 2y – z = 6 – x + 3y + z = 4 2x – 5y + 6z = 9 Solution: Expand across column #1:

D =

€

3 2 −1−1 3 12 −5 6

= 3(– 1)2

€

3 1−5 6

– 1(– 1)3

€

2 −1−5 6

+ 2(– 1)4

€

2 −13 1

= 3(18 – (– 5)) + 1(12 – 5) + 2(2 – (– 3)) = 69 + 7 + 10 = 86.

237

Expand across column #1:

DX =

€

6 2 −14 3 19 −5 6

= 6(– 1)2

€

3 1−5 6

+ 4(– 1)3

€

2 −1−5 6

+ 9(– 1)4

€

2 −13 1

= 6(18 – (– 5)) – 4(12 – 5) + 9(2 – (– 3)) = 138 – 28 + 45 = 155. Expand across column #1:

Dy =

€

3 6 −1−1 4 12 9 6

= 3(– 1)2

€

4 19 6

– 1(– 1)3

€

6 −19 6

+ 2(– 1)4

€

6 −14 1

= 3(24 – 9) + 1(36 – (– 9)) + 2(6 – (– 4)) = 45 + 45 + 20 = 110. Expand across column #1:

Dz =

€

3 2 6−1 3 42 −5 9

= 3(– 1)2

€

3 4−5 9

– 1(– 1)3

€

2 6−5 9

+ 2(– 1)4

€

2 63 4

= 3(27 – (– 20)) + 1(18 – (– 30)) + 2(8 – 18) = 141 + 48 – 20 = 169.

Thus, x =

€

DXD

=

€

15586

, y =

€

DyD

=

€

11086

=

€

5543

, z =

€

DzD

=

€

16986

So, the solution is (

€

15586

,

€

5543

,

€

16986 ).

Ex. 10 2y + 6z = 7 – x – y + 2z = 0 x + 2y + z = – 2 Solution: Expand across column #1:

D =

€

0 2 6−1 −1 21 2 1

= 0(– 1)2

€

−1 22 1

– 1(– 1)3

€

2 62 1

+ 1(– 1)4

€

2 6−1 2

= 0(– 1 – 4) + 1(2 – 12) + 1(4 + 6) = 0 – 10 + 10 = 0

Expand across row #2:

DX =

€

7 2 60 −1 2−2 2 1

= 0(– 1)3

€

2 62 1

– 1(– 1)4

€

7 6−2 1

+ 2(– 1)5

€

7 2−2 2

= 0(2 – 12) – 1(7 + 12) – 2(14 + 4) = 0 – 19 – 36 = – 55. Since D = 0, but Dx ≠ 0, then the system is inconsistent. There is no solution.

238

Ex. 11 x + y + z = 1 – x + 2y + z = 5 2x – y = – 4 Solution: Expand across row #3:

D =

€

1 1 1−1 2 12 −1 0

= 2(– 1)4

€

1 12 1

– 1(– 1)5

€

1 1−1 1

+ 0(– 1)6

€

1 1−1 2

= 2(1 – 2) + 1(1 + 1) + 0(2 + 1) = – 2 + 2 + 0 = 0

Expand across column #3:

DX =

€

1 1 15 2 1

−4 −1 0 = 1(– 1)4

€

5 2−4 −1

+ 1(– 1)5

€

1 1−4 −1

+ 0(– 1)6

€

1 15 2

= 1(– 5 + 8) – 1(– 1 + 4) + 0(2 – 5) = 3 – 3 + 0 = 0

Expand across column #3:

Dy =

€

1 1 1−1 5 12 −4 0

= 1(– 1)4

€

−1 52 −4

+ 1(– 1)5

€

1 12 −4

+ 0(– 1)6

€

1 1−1 5

= 1(4 – 10) – 1(– 4 – 2) + 0(5 + 1) = – 6 + 6 + 0 = 0

Expand across column #1:

Dz =

€

1 1 1−1 2 5

2 −1 −4 = 1(– 1)2

€

2 5−1 −4

– 1(– 1)3

€

1 1−1 −4

+ 2(– 1)4

€

1 12 5

= 1(– 8 + 5) + 1(– 4 + 1) + 2(5 – 2) = – 3 + (– 3) + 6 = 0

Since D, Dx, Dy, Dz are all zero, the equations are dependent. We will have to use one of our other techniques to find the solution. Write as an augmented matrix:

€

1−1

2

⎡

⎣

⎢ ⎢ ⎢

12−1

110

15

−4

⎤

⎦

⎥ ⎥ ⎥

Use the elementary row operations to get in row echelon form:

239

€

1−1

2

⎡

⎣

⎢ ⎢ ⎢

12−1

110

15

−4

⎤

⎦

⎥ ⎥ ⎥

€

100

⎡

⎣

⎢ ⎢ ⎢

13−3

12−2

16−6

⎤

⎦

⎥ ⎥ ⎥

€

100

⎡

⎣

⎢ ⎢ ⎢

13−3

12−2

16−6

⎤

⎦

⎥ ⎥ ⎥ R3 = r3 + r2

€

100

⎡

⎣

⎢ ⎢ ⎢

130

120

160

⎤

⎦

⎥ ⎥ ⎥

€

100

⎡

⎣

⎢ ⎢ ⎢

130

120

160

⎤

⎦

⎥ ⎥ ⎥ R2 = r2 ÷ 3

€

100

⎡

⎣

⎢ ⎢ ⎢

110

1230

120

⎤

⎦

⎥ ⎥ ⎥

€

100

⎡

⎣

⎢ ⎢ ⎢

110

1230

120

⎤

⎦

⎥ ⎥ ⎥ R1 = r1 – r2

€

1 0 13

−1

0 1 23

20 0 0 0

⎡

⎣

⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥

Thus, our system is: x +

€

13

z = – 1 ⇒ x = –

€

13

z – 1

y +

€

23

z = 2 ⇒ y = –

€

23

z + 2 Hence, the solution is: {(x, y, z) | x = –

€

13

z – 1, y = –

€

23

z + 2, z is any real number}.

Objective 5: Properties of Determinants.

Property #1: The value of the determinant changes signs if you interchange any two rows or interchange any two columns. To illustrate, let's interchange column #1 and #2 and compare the determinants:

D1 =

€

5 −34 1

= 5(1) – 4(– 3) = 5 + 12 = 17.

D2 =

€

−3 51 4

= – 3(4) – 1(5) = – 12 – 5 = – 17.

R2 = r2 + r1 R3 = r3 – 2r1

240

Property #2: If all the entries of any row or column are zeros, the determinant is zero.

To prove this, expand the determinant across the row or column that contains all zeros for entries. Thus, all the cofactor will be multiplied by 0 which will yield a determinant of 0. Property #3: If any two rows or any two columns of a determinant have corresponding entries that are equal, the value of the determinant is zero. To illustrate this, let row #2 and row #3 be the same.

€

1 1 1−1 2 5−1 2 5

= 1(– 1)2

€

2 52 5

– 1(– 1)3

€

1 12 5

– 1(– 1)4

€

1 12 5

= 1(10 – 10) + 1(5 – 2) – 1(5 – 2) = 0 + 3 – 3 = 0 Property #4: If any row or column of a determinant is multiplied by a nonzero number k, the value of the determinant is changed by a factor of k. To illustrate this, multiply every entry in row #2 by 2 and compare to the original determinant:

D1 =

€

5 −34 1

= 5(1) – 4(– 3) = 5 + 12 = 17.

D2 =

€

5 −38 2

= 5(2) – 8(– 3) = 10 + 24 = 34 = 2•17.

Property #5: If the entries of any row or any column of a determinant are multiplied by a nonzero constant k and added to the corresponding entries of another row or column, the value of the determinant is unchanged. To illustrate this, add 2r1 to row #2 and compare to the original determinant:

D1 =

€

5 −34 1

= 5(1) – 4(– 3) = 5 + 12 = 17.

D2 =

€

5 −314 −5

= 5(– 5) – 14(– 3) = – 25 + 42 = 17.