[PDF] [PDF] Topic 3: MATRICES

Content • Adding, Subtracting and Multiplying Matrices • Matrix Inversion • Example: c12 = (2x2) + (3x3) + (4x4) = 29 Calculate the determinant of this as:



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[PDF] Topic 3: MATRICES

Content • Adding, Subtracting and Multiplying Matrices • Matrix Inversion • Example: c12 = (2x2) + (3x3) + (4x4) = 29 Calculate the determinant of this as:



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Topic 3: MATRICES

Jacques (3rd Edition):

Chapter 7.1- 7.2

Content

•Adding, Subtracting and Multiplying

Matrices

•Matrix Inversion •Example: Model of National Income

A Vector: list of numbers arran

g ed in a row or column e. g . consumption of 10 units X and 6 units of Y g ives a consumption vecto r (X,Y) of (10,6) (6,10)

A Matrix: a two-dimensional arra

y o f numbers arran g ed in rows an d columns e.g. A =

232221131211

aaaaaa a 2 X 3 matrix with 2 rows and 3 columns component a ij in the matrix is in the i th row and the j th column e.g. let a ij be amount g ood j consumed by individual i - columns1-3:represent goods X, Y& Z - rows 1-2:represent individuals 1 & 2

Matrix of consumption

C =

232221131211

cccccc

6045100

Individual 1 consumes 0 of X, 10 of Y

and 5 of Z

Individual 2 consumes 4 of X, 0 of Y an

d

6 of Z

NOTE

Row Vector is a matrix with only 1 row :

A = [5 4 3] 1 X 3 matrix

Column Vectoris a matrix with only 1

column : A = 3 45

3 X 1 matrix

Transposing Matrices

A =

232221131211

aaaaaa

2 X 3 matrix

Then A T

231322122111

aaaaaa

3X2 matrix

the transpose of a matrix replaces rowsb y columns. A=

6045100

then A T

6501040

Adding and Subtracting Matrices

Matrices must have same number o

f rows and columns, m X n

Just add (subtract) the corresponding

elements.....

A + B + C = D i.e. a

ij + b ij + c ij = d ij 56215
2331
6125
1439

A - B = E i.e. a

ij - b ij = e ij 5554
6125
1439

Multiplying Matrices To multiply A and B,

N o. Columns in A = No. Rows in B

Then A x B = C

(1x 3) (3x 2) = (1x 2) 1211

323122211211

131211

cc bbbbbb .aaa c 11 = (a 11 .b 11 )+ (a 12 .b 21
) +(a 13 .b 31
c 12 = (a 11 .b 12 )+ (a 12 .b 22
) +(a 13 .b 32
2925

423521

432
c 11 = (2x1) + (3x5) + (4x2) = 25 c 12 = (2x2) + (3x3) + (4x4) = 29

2423222114131211

114521011213

401012

cccccccc.

5617234527

c 11 = (2x3) + (1x1) + (0x5) = 7 c 12 = (2x1) + (1x0) + (0x4) = 2 c 13 = (2x2) + (1x1) + (0x1) = 5 c 14 = (2x1) + (1x2) + (0x1) = 4 c 21
= (1x3) + (0x1) + (4x5) = 23 c 22
= (1x1) + (0x0) + (4x4) = 17 c 23
= (1x2) + (0x1) + (4x1) = 6 c 24
= (1x1) + (0x2) + (4x1) = 5

SCALAR MULTIPLICATION

If A =

22211211

aaaa

Then 3A =

22211211

3333
aaaa A = 1234
then 2A = 2468

And 3A =

36912

Practice Transposing, Adding, Subtracting and

Multiplying Matrices using examples from any Text

Book - or simply by writing down some simple

matrices yourself....

Determinant of a Matrix If A =

dcba aaaa

22211211

Now we can find the determinant......

Multiply elements in any one row or any

one column by corresponding co-factors, and sum.....

Select row 1.... |A| = a

11 .C 11 + a 12 .C 12 ad - bc

Select column 2

|A| = a 12 .C 12 + a 22
.C 22
= b(-c)+da

MATRIX INVERSION

Square matrix: no. rows = no. columns

Identity Matrix I: AI = A and IA = A

I = 1001
(for 2 X 2 matrix)

Inverse Matrix A

-1 : A.A -1 = I A -1 .A= I T

O INVERT

2 X 2 M ATRIX

If A =

dcba

1) Get Cofactor Matrix:

abcd

2) Transpose Cofactor Matrix:

acbd 1) multiply matrix by |A|1 so acbd bcad1 (i.e. divide each element by ad- bc) If |A| =0 then there is no inverse......(matrix is singular)

Example....find the inverse of matrix A

A = 4321
|A| = ad-bc = (1.4)-(2.3) = -2 (non-singular) A -1 1324
21
21
23
12

Check : A.A

-1 = I = 1001

Example....find the inverse of matrix B

B = 10542
|B| = ad - bc = (2.10) - (4.5) = 0 therefore, matrix is singular and inverse does not exist

Example Expenditure model of national

income

Y = Income

C = Consumption

I = Investment

G = Government expenditure

Y = C+I+G (1)

The consumption function is

C = a + bY (2)

N ote C and Y are endogenous. I and G are exogenous.

How to solve for values of endogenous

variables Y and C?

Method 1

Solve the above equations directly,

substituting expression for C in eq. (2) into eq. (1)

Thus, Y = a + bY+I+G

Solve for Y as:

Y - bY = a + I + G

Y(1 - b) = a + I + G

Thus, bGIaY 1

Substitute this value for Y into eq. (2) an

d solve for C: babGI bGIabaC 1)( 1

Method 2 N

ow solve the same problem using matrix algebra:

Rewrite (1) and (2) with endogenous

variables, C and Y, on left hand side

From eq. 1: Y - C = I + G

From eq. 2: -bY + C = a

Now write this in matrix notation:

aGI CY b111 or A.X = B

We can solve for the endogenous variables

X, by calculating the inverse of the A

matrix and multiplying by B:

Since AX=B

X=A -1 B

To invert the 2 X 2 A matrix, recall the

steps from earlier in the lecture

If A =

dcba , then A -1 acbd bcad1

In this case, where

111
bA the determinant of A is : |A| = 1.1 - [- 1.- b] = 1 - b

Cofactor Matrix:

111b

Transpose Cofactor Matrix:

111
b

The inverse is :

bbbbb b bAquotesdbs_dbs14.pdfusesText_20