Bba i-bm-u-2- matrix -

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MATRIX ALGEBRA Course BBA Subject: Business Mathematics Unit-2 1 2

Transcript of Bba i-bm-u-2- matrix -

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MATRIX ALGEBRA

Course BBA

Subject: Business Mathematics

Unit-2

1

2

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The notation is used for the algebraic

expression ad – bc.

It is called a determinant of order two.

a, b, c ,d are called the elements of the determinant.

ad – bc is called the expansion or the value of the

determinant.

a b

c d

3.

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a, b is called the first row and c,d is called the

second row of the determinant.

a, c is called first column and b, d is called the

second column of the determinant.

a, d is called the principal diagonal of the

determinant.

a b

c d

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a b c

d e f

g h i

is called a determinant of

order three.

Its expansion is as follows:

:

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= a(ei - fh) - b(di - fg) + c(dh - eg)

4.

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(1) Find the value of

Answer:

= 1 - 2 + (-3)

= 1 [(4*5) – (6*2)] – 2[(5* 5) – (2 * -8)] – 3 [(5 * 6) – (4 *-8)]= 1 [20 - 12] – 2 [25 + 16] – 3 [30 + 32]

= 1[8] – 2 [41] – 3 [62]= 8 – 82 – 186 = -260

1 2 3

5 4 2

8 6 5

4 2

6 5

5 2

8 5

5 4

8 6

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Suppose we are given the following equations.

a1x + b1y + c1 = 0………………. (1)

a2x + b2y + c2 = 0………………..(2)

To solve this equations, multiply equation (1) by b2 and

equation (2) by b1,

a1b2x + b1b2y + c1b2 = 0…………… (3)

a2b1x + b1b2y + c2b1 = 0…………….(4)

Subtracting equation (4) from (3), we get

(a1b2 – a2b1)x + c1b2 – c2b1 = 0

(a1b2 – a2b1)x = c2b1 – c1b2.

If a1b2 – a2b1 ≠ 0 then

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1 2 2 1 1 2 2 1

1x

b c b c a b a b

……(5)

Similarly, eliminating y, we get

2 1 1 2 1 2 2 1

1y

a c a c a b a b

……………. (6)

From (5) and (6), we get the following formula for the value of x and y.

1 2 2 1 2 1 1 2 1 2 2 1

1x y

b c b c a c a c a b a b

Expressing the denominators in the form of determinants, we get

1 1 1 1 1 1

2 2 2 2 2 2

1x y

b c c a a b

b c c a a b

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11 12 13 14

21 22 23 24

31 32 33 34

mn mn mn mn

a a a a

a a a a

a a a a

a a a a

Row 1

Row 2

Row 3

Row m

Column 2Column 1 Column n

5.

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5 3 1

1.) 11

2

2 X 3

3 8 9

2.) 2 5

6 7 1

3 X 3

83.)

7

2 X 1

4.) 2 3

1 X 2

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(1) Row matrix:

A 1 Χ n matrix of the type [ a11, a12, …….a1n] is called

a row matrix.

This matrix has only one row and n columns.

(2) Column matrix:

A m Χ 1 matrix of type is called a column matrix.

This matrix has m rows and only one column.

11

21

1m

a

a

a

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(3) Square matrix:

A matrix of order m Χ m is called a square matrix.

In a square matrix the number of rows equals the number of

columns.

For example, A = B =

2 2

1 2

3 4

3 3

a b c

d e f

g h i

(4) Diagonal matrix:

A square matrix, in which each element except the diagonal

elements is zero, is called a diagonal matrix.

For example, A = is a diagonal matrix. 4 0 0

0 2 0

0 0 5

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(5) Identity matrix:

A square matrix in which all the elements in the principal diagonal are equal to unity (1) and all other elements are 0 is called a unit matrix or an identity matrix.

A unit matrix of order n Χ n is denoted by In. For example.

I2 = I3 =

(6) Scalar matrix:

A diagonal matrix, in which all the elements in the principal diagonal are equal to a scalar k, is called a scalar matrix.

For example,

A =

1 0

0 1

1 0 0

0 1 0

0 0 1

3 0 0

0 3 0

0 0 3

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(7) Null matrix:

A matrix having all its elements as zero is called a zero or a null

matrix.

Null matrix can be square or rectangular.

It is usually denoted by 0mΧn or just 0.

For example 01Χ3 =

(8) Transpose of a matrix:

The matrix obtained from any given matrix A by changing its rows

into corresponding columns is called the transpose of A and it is

denoted by A’ or AT.

For example A = [1 2 3] then AT =

0 0 0

1

2

3

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(9) Symmetric matrix:

If for a square matrix A = [aij], A’ = A, then A is called a

symmetric matrix.

In a symmetric matrix, aij = aji for each pair (i, j).

For example A = B =

(10) Skew symmetric matrix:

If for a square matrix A = [aij], A’ = -A, then A is called

a skew symmetric matrix.

In a skew symmetric matrix, aij = -aji for each pair (i, j).

For example A =

1 3

3 2

3 6 7

6 5 4

7 4 8

0 3 2

3 0 5

2 5 0

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You can add or subtract matrices if they have the

same dimensions (same number of rows and

columns).

To do this, you add (or subtract) the corresponding

numbers (numbers in the same positions).

Example:

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A + B

5 6 0 3

4 2 1 3

1 3

6 4

5 0 6 3;

4 1 2 3A B

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When a zero matrix is added to another matrix of the

same dimension (i.e. number of rows and columns

should be same), that same matrix is obtained.

2 1 3 0 0 02.)

1 0 1 0 0 0

2 1 3

1 0 1

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1 2 1 1

3.) 2 0 1 3

3 1 2 3

1 1 2 ( 1)

2 1 0 3

3 2 1 3

0 3

3 3

5 4

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• If A is an m × n matrix and s is a scalar, then we let kA

denote the matrix obtained by multiplying every

element of A by k.

• This procedure is called scalar multiplication.

310

221A

930

663

331303

2323133A

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To multiply matrices A and B look at their dimensions

pnnm

MUST BE SAME

SIZE OF PRODUCT

If the number of columns of A does not equal the

number of rows of B then the product AB is

undefined.

6.

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•The multiplication of matrices is easier shown than put into

words.

•You multiply the rows of the first matrix with the columns of the

second adding products

140

123A

13

31

42

B

Find AB

First we multiply across the first row and down the first column

adding products. We put the answer in the first row, first column of

the answer.

23 1223 5311223

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140

123A

13

31

42

B

Find AB

We multiplied across first row and down first column so we put the

answer in the first row, first column.

5AB

Now we multiply across the first row and down the second column and we’ll put

the answer in the first row, second column.

43 3243 7113243

75AB

Now we multiply across the second row and down the first column and we’ll put

the answer in the second row, first column.

20 1420 1311420

1

75AB

Now we multiply across the second row and down the second column and we’ll

put the answer in the second row, second column.

40 3440 11113440

111

75AB

Notice the sizes of A and B and the size of the product AB.

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The matrix formed by taking the transpose of the cofactor

matrix of a given original matrix.

The adjoint of matrix A is often written adj A.

Example:

Find the adjoint of the following matrix:

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First find the cofactor of each element.

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As a result the cofactor

matrix of A is

Finally the adjoint of A is the transpose of the

cofactor matrix:

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If for a given square matrix A, there exists a matrix B

such that AB = I = BA, then the matrix B is called an

inverse of A.

It is denoted by A-1

Note: Inverse does not exist if product is not an identity .

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1. Find determinant of matrix. If det. is not equal to 0,

then inverse of matrix exist.

2. Find adjoint of a matrix.

a. find cofactor

b. change sign. (if odd.. –ve sign & if even.. +ve

sign)

c. transpose.

3. Find inverse. i.e. A-1 = (adjoint of A)1

A

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7.

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8.

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1 2 3

0 4 5

1 0 6

A

Find inverse of

Solution:

Step 1

= 1 (24-0) – 2 (0-5) + 3 ( 0 – 4)

= 1 (24) -2 (-5) + 3 (-4)

= 24 + 10 -12 = 22

4 5 0 5 0 41 2 3

0 6 1 6 1 0

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Step-2

Find Cofactor of

matrix

1 2 3

0 4 5

1 0 6

A

The cofactor for each element of matrix A:

11

4 524

0 6A 12

0 55

1 6A 13

0 44

1 0A

21

2 312

0 6A 22

1 33

1 6A 23

1 22

1 0A

31

2 32

4 5A 32

1 35

0 5A 33

1 24

0 4A

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Cofactor matrix of is then given by:

1 2 3

0 4 5

1 0 6

A

24 5 4

12 3 2

2 5 4

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Inverse matrix of is given by:1 2 3

0 4 5

1 0 6

A

1

24 5 4 24 12 21 1

12 3 2 5 3 522

2 5 4 4 2 4

T

AA

12 11 6 11 1 11

5 22 3 22 5 22

2 11 1 11 2 11

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Given AX = B

we can multiply both sides by the inverse of A, provided this exists, to give

A−1AX = A−1B

But A−1A = I, the identity matrix.

Furthermore, IX = X, because multiplying any matrix by an identity matrix of the appropriate size leaves the matrix unaltered.

So X = A−1B

This result gives us a method for solving simultaneous equations

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Example:

Solve x + y + z = 6

2y + 5z = -4

2x + 5y - z = 27

We can call the matrices "A", "X" and "B" and the

equation becomes:

AX = B

9.

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Where

A is the 3x3 matrix of x, y and z coefficients

X is x, y and z, and

B is 6, -4 and 27

Then the solution is this:

X = A-1B

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Multiply A-1 by B

The solution is:

x = 5, y = 3 and z = -2

10

.

11

.

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(A) Online Sources (for images):

1.https://encryptedtbn3.gstatic.com/images?q=tbn:ANd9GcTJQw7J0Ih3lz1g0FFGBevFpAra3nOrin6yt3PS6x_Qy--4E4U

2.https://encryptedtbn1.gstatic.com/images?q=tbn:ANd9GcR63sF2Y25j6cI_FzdrRa_seWbKVuVv0qRTWuxtpv4YboHiZEW

3.https://lh6.ggpht.com/mpWaWXPYIoGjnGJd_Tb80X8QKKwROA7zCI0wKkJ627NBG6WYnSqwfF_cwbV5QJ8H_5o7=s126

4. http://www.mathsisfun.com/algebra/images/matrix-3x3-det-c.gif

5.https://lh5.ggpht.com/n28coju14RsDs9cZL6_yJ3izr1tZ7FPnoJnMidvzy0sOIyBiOJw_hY4XBoCJG9m1OtE=s101

6.https://lh3.ggpht.com/CjbpIkLEbKk7bG4KhaQy5BXvU41TH8ey7N6KjNI2ZQB6vLurlpd5BQhJU1P89LOOb3W4=s170

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5N4zgacY4a9RExc01eu3bUXqlQWqwmXfEMdZoiz3A=s1

70

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DXtM5ICrpJpW9olN67Ch_fXggb9imB6w2ei7GGt8=s136

9. https://lh4.ggpht.com/_6VfrBUCjgni_ontV3K9dQVr-

KSQn1EPWMeqfgUpVKTGNzB4aOcVl3IEaLzRKKZ0CG

3T=s170

10.https://lh6.ggpht.com/iKzqvK9OQDvi_iFvgTl2J4UOyJhet_

ZTBsZ6XEbXYv75G2cCV9FmOkGxrpIA7JH2P=s170

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Z85pi2y2r_zGIE5avCQpoEixutD3L3bO_Mv5efyONQ=s17

0

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1.www.mathxtc.com/Downloads/NumberAlg/files/Matri

x%20Algebra.ppt

2.www.dgelman.com/powerpoints/algebra/alg2/spitz/4.1

%20Matrices.ppt

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1. Business Mathematics by A.G.Patel and G.C.Patel-

Atul Prakashan.