Похожие презентации:

# Rank of a matrix. Theorem of Kronecker-Capelli

## 1.

Rank of a matrix. Theorem of Kronecker-CapelliConsider a system of n linear equations with n variables

(x1, x2, …, xn):

a x a x ... a x b (*)

11 1

12

2

1n

n

1

a x a x ... a x b

21 1

22 2

2n n

2

..........................................

a n1 x1 a n 2 x 2 ... a nn x n bn

Consider

a11

a

A(n; n) 21

...

a

n1

a12

a 22

...

an2

... a1n

... a 2 n

... ...

... a nn

x1

x

X ( n;1) 2

x

n

b1

b

B (n;1) 2 .

b

n

## 2.

Then the system (*) is written in a matrixrepresentation: A(n; n) X(n; 1) = B(n; 1)

X = A-1 B.

A matrix A(n; n) is called regular if its

determinant is not equal to zero, i.e.

a11

a12

... a1n

a 21

a 22

... a 2 n

...

...

...

a n1

an2

... a nn

...

0.

A matrix A-1(n; n) is called inverse to a matrix

A(n; n) if the product

A(n; n) A-1(n; n)= A-1(n; n) A(n; n) = E(n; n),

## 3.

A111 A12

1

A

...

A

1n

A21

A22

...

A2 n

An1

... An 2

.

... ...

... Ann

...

Example . Find the inverse matrix to the matrix

3 1 0

A 2 1 1 .

2 1 4

3

1) 2

2

1 0

1

1 12 2 0 0 8 3 5 0.

1 4

2) А11 = 5; A12 = 10; A13 = 0; A21 = 4; A22 = 12; A23 = 1; A31 = -1;

A32 = -3; A33 = 1.

5 10 0

5 4 1

5 4 1

1

T

1

3) A* 4 12 1 . 4) A * 10 12 3 . 5) A 10 12 3 .

5

1 3 1

0 1 1

0 1 1

## 4.

3 1 0 5 4 15 0 0 1 0 0

1

1

1

A A 2 1 1 10 12 3 0 5 0 0 1 0 .

5

0 1 1 5 0 0 5 0 0 1

2

1

4

Rank of a matrix.

Consider a matrix of the dimension m n :

a11

a

A( m; n) 21

...

a

m1

a12

a 22

...

am2

... a1n

... a 2 n

... ...

... a mn

The rank of a matrix A is the greatest order of its nonequal to zero minors. The rank of a matrix is denoted by

Rank A or r(A).

Theorem. If there is a non-equal to zero minor of the rth order in a matrix A and all its bordering minors of the

r+1-th order are equal to zero then the rank of A is

equal to r, i.e. r(A)=r.

## 5.

Theorem. The rank of a matrix doesn’t change if:a) All the rows are replaced by the corresponding

columns and vice versa;

b) Replace two arbitrary rows (columns);

c) Multiply (divide) each element of a row

(column) on the same non-zero number;

d) Add to (subtract from) elements of a row

(column) the corresponding elements of any other

row (column) multiplied on the same non-zero

number.

## 6.

Theorem of Kronecker-Capelli. A system oflinear equations is consistent if the rank of the

basic matrix A equals the rank of the extended

matrix C, i.e. Rank A = Rank C. Moreover:

•1) If Rank A = Rank C = n (where n is the

number of variables in the system) then the

system has a unique solution.

•2) If Rank A = Rank C < n then the system has

infinitely many solutions.

## 7.

Solving a system of linear equationsby the Gauss method

a11 x a12 y a13 z a14u a15

a x a y a z a u a

21

22

23

24

25

a31 x a32 y a33 z a34u a35

a 41 x a 42 y a 43 z a 44u a 45

(a)

(b)

(c )

(d )

Suppose that а11 0 (if а11 = 0 then we change the order of

equations by choosing as the first equation such an equation in

which the coefficient of х is not equal to zero).

I step: divide the equation (а) on а11, then multiply the obtained

equation on а21 and subtract from (b); further multiply (а)/a11 on

а31 and subtract from (с); at last, miltiply (а)/a11 on а41 and

subtract from (d).

## 8.

x b12 y b13 z b14u b15 (e)b22 y b23 z b24u b25 ( f )

b32 y b33 z b34u b35 ( g )

b42 y b43 z b44u b45 (i )

where bij are obtained from aij by the following formulas:

b1j = a1j / a11 (j = 2, 3, 4, 5);

bij = aij – ai1 b1j (i = 2, 3, 4; j = 2, 3, 4, 5).

II step: do the same actions with (f), (g), (i) (as with (a), (b),

(c), (d)) and etc.

As a final result the initial system will be transformed to a socalled step form:

x b12 y b13 z b14 u b15

y c 23 z c 24 u c 25

z d 34 u d 35

u e45

## 9.

Example 1.3 x 2 y z 5

x y z 0

4 x y 5 z 3

Interchange the first and the second equations of the system:

x y z 0

3 x 2 y z 5

4 x y 5 z 3

Subtract from the second equation the first equation multiplied on

3; also subtract from the third equation the first equation

multiplied on 4. We obtain:

x y z 0

y 4z 5

5 y 9z 3

## 10.

Further subtract from the third equation the second equationmultiplied on 5:

x y z 0

y 4z 5

11z 22

Multiply the second equation on (-2), and the third – divide on

(-11):

x y z 0

y 4 z 5

z 2

The system of equations has a triangular form, and consequently

it has a unique decision. From the last equation we have z = 2;

substituting this value in the second equation, we receive у = 3

and, at last from the first equation we find х = -1.

## 11.

• Maricopa's Success scholarship fund receivesa gift of $85000. The money is invested in

stocks, bonds, and CDs. CDs pay 3.25%

interest, bonds pay 4.1% interest, and stocks

pay 7.7% interest. Maricopa Success invests

$15000 more in bonds than in CDs. If the

annual income from the investments is

$3992.5. How much was invested in each

account?