Spring 2013

Solving a System of Linear Equations

Matrix inverseSimultaneous equations

Cramer’s ruleSecond-order Conditions

Lecture 7

Spring 2013

1. Matrix Equation and Matrix Inversion

Market equilibrium model

Demand: Q + P = 10

Supply : Q – 4P = -3

In matrix form,

3

10

41

11

P

Q

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General equation system

(N equations, N unknowns)

NNNNNN

N

N

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

.11

NNNNbxA

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If we can find a matrix A -1 such that A -1 Ax = x, then we can solve the equations by observing that

A-1Ax = A-1b

x = A-1b

Existence of an inverse A-1:

• A is a square matrix.

• A is not a null matrix.

• Rows and columns are linearly independent.

• A-1A = AA-1 = IN.

Spring 2013

In the previous example,

3

10

41

11

P

QAx = b

Example 7.1: Consider the matrix

11

14

5

1B

10

01

41

11

11

14

5

1BA

62

47

3

10

11

14

5

1.

.* Bbx

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2. Determinants

2.1 Determinants of Order 2

|A| = a11a22 – a12a21.

|A| = 1·4 - 2·3 = -2.

2221

1211

aa

aaA

43

21AExample 7.2:

Spring 2013

333231

232221

131211

aaa

aaa

aaa

A

|A| = a11a22a33 + a12a23a31 + a13a21a32

- a11a23a32 – a12a21a33 – a13a22a31

2.2 Determinants of Order 3

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333231

232221

131211

aaa

aaa

aaa a11 a12

a21 a22

a31 a32

+ + +– – –

|A| = a11a22a33 + a12a23a31 + a13a21a32

- a11a23a32 – a12a21a33 – a13a22a31

Spring 2013

Example 7.3: Find the determinants of the following square matrices.

325

011

203

A

2

2

2

cc1

bb1

aa1

B

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|Mij| =

2.3 Expansion by Cofactors

1. The minor |Mij| of a matrix is the determinant of the matrix obtained by deleting the ith row and jth column.

11 1, 1 1 1, 1 1

21 2, 1 2 2, 1 2

1 , 1 , 1

1 , 1 , 1

j j j n

j j j n

i i j ij i j in

n n j nj n j nn

a a a a a

a a a a a

a a a a a

a a a a a

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|M11| = 4, |M12| = 3, |M21| = 2, |M22| = 1

|M11| = 5, |M12| = -17, |M13| = -14

43

21A

315

746

238

B

Example 7.4: Find all the minors of the matrices A and B.

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2. The cofactor |Cij| of a matrix = (-1)i+j|Mij|

Example 7.5: Find all the cofactors of the following matrices.

43

21A

315

746

238

B

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3. The determinant is the expansion along any row or column.

|A| = ai1Ci1 + ai2Ci2 + + aijCij + ainCin

= a1jC1j + a2jC2j + + aijCij + anjCnj

Spring 2013

1st row: |A| = a11C11 + a12C12 = 1·4 + 2·(-3) = -2

2nd row: |A| = a21C21 + a22C22 = 3·(-2) + 4·1 = -2

1st column: |A| = a11C11 + a21C21 = 1·4 + 3·(-2) = -2

2nd column: |A| = a12C12 + a22C22 = 2·(-3) + 4·1 = -2

43

21A

Example 7.6: Calculate the determinant of matrix A.

Spring 2013

Expansion along the 1st row

|B| = 8·5 + 3·17 + 2·(-14) = 63.

Expansion along the 3rd column

315

746

238

B

037

032

165

C

.27733237

321C

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2.4 Basic Rules for Determinatns

Let A be an n n matrix. Then:

• If all the elements in a row (or column) of A are zero, then |A| = 0.

• If two of the rows (or columns) of A are proportional, then |A| = 0.

Spring 2013

Let A and B be n n matrices. Then:

• If all the elements in a single row (or column) of A are multiplied by a number , the determinant is multiplied by .

• The value of the determinant of A is unchanged if a multiple of one row (or column) is added to a different row (or column) of A.

• |AB| = |A||B|.

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3. Matrix Inversion

Let A be n by n. If there is a n by n matrix X such that

AX = XA = In,

then X is called the inverse of A and is denoted by A-1.

Spring 2013

Example 7.7: Consider the following three matrices:

5 6 1/2 -3/10 1 0A , X and B .

5 10 -1/4 1/4 0 0

Show that A and X are inverses of each other and B has no inverse.

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Step 1: Check the singularity of A

|A| 0 Carry on matrix inversion

|A| = 0 Stop

Step 2: Find the cofactor matrix of A

NNNN

N

N

CCC

CCC

CCC

C

21

22221

11211

3.1 Finding the Inverse of a Matrix

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Step 3: Find the adjoint of A

adj(A) = CT.

Step 4: Compute the inverse

A-1 = adj(A) / |A|.

Example 7.8: Find the inverse of

43

21A

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Step 1: |A| = 1x4 – 3x2 = -2 < 0

Step 2: M11 = 4, M12 = 3, M21 = 2, M22 = 1

C11 = 4, C12 = -3, C21 = -2, M22 = 1

12

34C

Step 3:

13

24TCAadj

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Step 4:

5051

12

13

24

2

11

..A

AadjA

Example 7.9: Show that the matrix A has an inverse and find the inverse.

421

134

432

A

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3.2 Solving Equations

Matrix equation

b

N

x

N

A

NNNN

N

N

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

Ax = b A-1Ax = A-1b

x = A-1b.

Spring 2013

Example 7.10: Solve the following simultaneous equations.

3x1 – 2x2 = 11

2x1 + x2 = 12

Matrix equation

bxA

x

x

12

11

12

23

2

1

1. |A| = 31 – 2(-2) = 7 > 0

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32

212 C .

32

213 Aadj .

7372

7271

32

21

7

14 1-A .

2

5

12

11

7372

7271*x

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4. Cramer’s Rule

Matrix equation

Let

b

N

x

N

A

NNNN

N

N

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

A 2 ,,

NNNNN

N

N

NNNN

N

N

aaba

aaba

aaba

aab

aab

aab

A

31

223221

113111

2

2222

1121

1

Spring 2013

Cramer’s rule: If |A| 0, then for i = 1,...,N,

In the previous example,

x1* = A1 / |A| = 35/7 = 5

x2* = A2 / |A| = 14/7 = 2

.*

A

Ax ii

14122

11335

112

2111

2A and A

Spring 2013

Example 7.11: Solve the following system for the unknowns x1, x2 and x3 using Cramer’s rule.

2x1 + 4x2 – x3 = 15

x1 – 3x2 + 2x3 = -5

6x1 + 5x2 + x3 = 28

Spring 2013

Example 7.12: Solve the following system of equations:

x1 + x2 = 3

x1 + x3 = 2

x2 + x3 + x4 = 6

x2 + x4 = 1

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3 Possible Cases of an Equation System

Case 1: Unique solution

|A| 0

b

N

x

N

A

NNNN

N

N

b

b

b

x

x

x

aaa

aaa

aaa

2

1

2

1

21

22221

11211

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Case 2: No solution

|A| = 0 and equations are inconsistent.

Case 3: Multiple solutions

|A| = 0 and equations are consistent.