Matrix Inversion Formula

7/30/2019 Matrix Inversion Formula

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Matrix Inversion Formula

EE 521 Analysis of Power Systems

Chen-Ching LiuWashington State University


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1. Sherman-Morrison-Woodbury Formula:

Let ML(Rn) be invertible and let U, V be nm

(mn) matrices. Then (M+UVT) is invertible if

and only if (I+VTM-1U) is invertible, and:

To prove (1), we simply pre-multiply and post-

multiply (M+UVT) with its inverse matrix to checkwhether the result is an identity matrix.

LetH=(I+VT

M-1

U)-1

, then

)1(MVU)MV(IUM-M)UV(M -1T-1

H

-1T-1-1-1T

+=+


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a) Pre-multiplication

b) Post-multiplication

[ ]( )[ ]

IMVHHIUI

MVHUMVIIUI

MVUHMVHIUI

MUHVMUVMUHVMMMUVMM

T

I

TT

TT

TTT

I

T

=

+=

++=

+=

+=

++

11

11

11

111111

1-T1-

H

1-T1-1-T MVU)MV(IUM-M)UV(M

[ ]( )[ ]

[ ] IVHHIUMI

VUMVIHIUMI

VUMHVHIUMI

UVMUHVMMMUHVMUVMMM

T

TT

TT

TTTT

=+=

++=

+=

+=

+

+

11

11

11

111111

T1-T1-

H

1-T1-1- )UV(MMVU)MV(IUM-M


4/13

2. Sherman-Morrison Formula:

In eq. (1), if m=1, i.e., U and V are column

vectors, let u and v (u,v Rn) represent these two

vectors, then let equation (1) can be simplified:

3. Special Case:

Suppose M is symmetrical and M-1 is known; and

in (2) we have uvT=aaT, where is a scalar and

a is an n-vector, then (2) becomes:

where:

( ) ( ) )2(11 11

111

+

=+ MuvMuMv

MuvMT

T

T

)3(1

T

rank

TbbMaaM =

+

( ) 111

+=

=

ba

aMb

T


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Example 1: Given

Let

Find M1-1

Solution: Here =2 and a=[0 1 0]T, So

b=M-1a=[-1 2 1]T =(-1+aTb)-1=0.4

Use(3)

The result can be verified by direct

multiplication.

=

=

321

221

111

M

110

121

012

1-M

[ ]

Ta

MM 010

0

1

0

2

110

141

012

+=

=

[ ]

=

=

4.14.02.0

4.04.02.02.02.06.0

221

2

21

4.0

321

221111

1

1M


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4. Comments:

(M+UVT) is a modification of M. Since M-1 is

assumed known, the inverse of this modifiedmatrix can be performed without inverting annn matrix again. Instead, the necessarymatrix inversion is only for an mm matrix.

When m=1 (Eqs. (2-3)), the matrix inversionrequires only matrix multiplications and matrixadditions.

The admittance matrix is symmetrical, and itschange can be easily represented by findinga vector and a scalar . The corresponding

computational efforts is lower.

a


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5. Inverting Admittance Matrix When LineAdmittances Are Changed

Suppose that we have Ybus and its inverse Zbus.

Suppose that the admittance of the k-th branchconnecting node i and node j is changed from yk

To yk+ yk. Let to be the new admittance

matrix. We need to find its inverse, .

First, we try to find a vector and scalar k.

There are 2 cases to consider:

Case 1: Neither node i nor node j is the referencenode. Then we get the new admittance matrix:

i line k j

Ref.

new

busZ

new

busY

ka


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Thus in we can put 1 in the i-th row and 1 in the

j-th row. All other entries are zero. Also k=yk.

[ ]

T

kkbus

bus

bus

new

bus

aaY

Y

YY

k

k

kk

kk

y

1-1

1

1

y

yy-

y-y

+=

+=

+=

ka


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Case 2: The k-th branch connects the i-th node to the

reference node. Similar to case 1, we can put a

1 in the i-th row of , and put all other entrieszero. Also we have k=yk.

i line k Ref

Now we consider calculation of . We already haveZbus=[Ybus

]-1. By eq. (3) we obtain:

where

ka

newbusZ

(4))()( 11 TkkkbusT

kkkbus

new

bus

new

bus bbZaayYYZ =+==

(5)1

+

=

=

k

T

k

k

k

kbusk

bay

aZb


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Let the i-th and the j-th columns of Zbus be Zi and Zj,

respectively, and the ii-th, jj-th, and ij-th elements be

Zii, Zjj, and Zij, respectively. According to the two

cases above, Eq. (5) can be simplified:

Case 1: by Eq. (5):

(6)

Case 2: j is the reference node:

(7)

[ ]

[ ] [ ]1

21

11

1

1

++

=

=

=

=

ijjjiik

k

jik

T

k

jijik

zzzy

ZZTba

ZZZZb

1

+

=

=

iik

k

ik

zy

Zb


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Example 2:Suppose that for a given network we already have

The network is shown in the figure.

(all values are line admittances p.u.)

( to are branch number)

Note that at the beginning there are no connections between node 1

and node 2. Now we add a line (branch number 5)

between node 1 and node 2. The line admittance is j10.

Find:

=

=

2.02.01.0

2.05.01.0

1.01.02.0

3and

301010

10100

10020j

ZjY busbus

-j10-j10

-j10

-j10 2

3Ref.

1

12

5

43

1 5

new

busZ


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Solution: The change of the admittance of branch 5

y5=-j10-j0=-j10

Since neither node 1 nor node 2 is the reference,

we use Eq. (6).

75.3))2.05.02.0(310

1(

)2

1

(

1.0

4.01.0

32.0

5.01.0

1.0

1.02.0

3

1

1

122211

5

5

215

jj

j

zzzy

jjzzb

=++

=

+=

=

==

Th b i i (4)


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Then by using equation (4):

We can also get directly from the figure:

To check the result, one can simply multiply with

[ ]

=

=

+=

=

1875.01500.01125.0

1500.03000.01500.0

1125.01500.01875.0

3

01.004.001.0

04.016.004.0

01.004.001.0

3

25.1

2.02.01.0

2.05.01.0

1.01.02.0

3

1.04.01.03

1.0

4.0

1.0

375.3

j

jj

jjjZ

bbZZ

bus

T

kkkbus

new

bus

new

busZnew

busY

new

busY

=

301010

102010101030

jYnew

bus

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Matrix Inversion Formula