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MATRICES
Definition
A matrix is a rectangular array of numbers or functionsarranged in rows and columns usually designated by acapital letter and enclosed by brackets, parentheses ordouble bars. A matrix may be denoted by:
mnmm
n
n
aaa
aaa
aaa
A
...
:::
...
...
21
22221
11211
Unless stated, we assume that all our matrices arecomposed of real numbers.
The horizontal groups of elements are called the rowsof thematrix. The ith row of A is
miaaa inii 1...21
The vertical groups of elements are called the columns ofthe matrix. The jth column of A is
nj
a
a
a
j
j
j
1:
3
2
1
The size of a matrixis denoted by m x n (m by n) where mis the number of rows and n is the number of columns.
We refer to aijas the entryor the elementin the ith row andjth column of the matrix.
We may often write a given matrix as
A = [aij].
SOME SPECIAL TYPES OF MATRICES
A. Row Matrix or Row Vector is a matrix consistsof only one row.
Example: B = [b1 b2 . . . bj . . . bn]
B. Column Matrix or Column Vector is a matrixconsists of only one column.
Example:
m
i
c
c
c
c
C
.
.
.
.
..
2
1
C. Square Matrix is a matrix in which the no. ofrows equals the no. of columns.
Order of a Square Matrix is the number of rowsor columns of the matrix. Thus, we can just referto a 3x3 matrix as a square matrix of order 3.
Principal Diagonal or Main Diagonal of a SquareMatrixconsists of the elements a11, a22, a33,
ann.
The Trace of a Square Matrixis the sum of theelements on the main diagonal of the matrix.
D. Upper Triangular Matrix a square matrix allelements of which below the principal diagonalare zero (aij= 0 for i>j).
Example:
33
2322
131211
00
0
u
uu
uuu
U
E. Lower Triangular Matrix a square matrix allelements of which above the principal diagonalare zero (aij= 0 for i
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EQUALITY OF MATRICES
Two matrices A = (aij) and B = (b ij) are equal if and only ifthe following conditions are satisfied:
a) They have equal number of rows.b) They have equal number of columns.c) All elements in A agree with the elements in
B. (aij=b
ij, for all i and j.)
Example: The matrices
143
221
112
3
21
21
c
bxBand
zb
ya
cx
A
are equal if and only if x = 2, y = 1, z = 2, a = 3,b = 4 and c = 1.
ELEMENTARY OPERATIONS ON MATRICES
A. MATRIX ADDITION AND SUBTRACTION
If A = (aij) and B = (bij) are matrices of the same size m x n,then the sum A + B is another m x n matrix C = [c ij] wherecij= aij+ bijfor i = 1 to m and j = 1 to n. Matrix addition isaccomplished by adding algebraically correspondingelements in A and B.
Example:
114
223
123
421BA
211
204
11)1(243
)2(42231
114
223
123
421BA
B. SCALAR MULTIPLICATION
If A = (aij) is an m x n matrix and k is a real number (or ascalar), then the scalar multiple of A by k is the m x n matrixC = [cij] where cij= kaij for all i and j. In other words, thematrix C is obtained by multiplying each element of thematrix by the scalar k.Examples: We obtain
156912
9306
5234
31023
28168
4820
1248
742
125
312
4
C. MATRIX SUBTRACTION
If A and B are m x n matrices, the difference between A andB denoted as AB is obtained from the addition of A and (-1)B.
AB = A + (-1)BMatrix subtraction is accomplished by subtracting from theelements of the first matrix the elements of the secondmatrix correspondingly.
Example:
383
412
475
243BA
718
651
3487)3(5
42)1(423
383
412
475
243BA
Note: We can only add or subtract matrices with the same
number of rows and columns.
D. MATRIX MULTIPLICATION
If A = (aij) is an m x n matrix and B = (bij) is an n xp matrix, then the product of A and B, AB = C =[cij] is an m x p matrix where
n
k
kjikij bac1
for i = 1 to m and j = 1 to p
The formula tells us that in order to get theelement cijof the matrix C, get the elements of the
ith row of A (the pre-multiplier) and the elementson the jth column of B (the post multiplier).Afterwards, obtain the sum of the products ofcorresponding elements on the two vectors.
Note: The product is defined only if thenumber of columns of the first factorA (pre-multiplier) is equal to thenumber of rows of the second factorB (post-multiplier). If this issatisfied, we say that the matricesare comformable in the order AB.
MATRIX EXPONENTIATION
The formula Anwill be defined as AAAA
Example:
13
42
43
21BA
20
1210
)4)(1()2(3)3)(1()1(3
)4)(4()2(2)3)(4()1(2
1618
68
)1(4)4(3)3(4)2(3
)1(2)4(1)3(2)2(1
BA
BA
AB
AB
Note: Although AB and BA are defined it is not necessarythat AB = BA.
Example:
12
31
541
326
64
35
12
CBA
1. A x B is a 3 x 3 matrix while B x A isa 2 x 2 matrix.
2. A x C is a 3 x 2 matrix but C x A is notdefined.
3. B x C is not defined but C x B isdefined (2 x 3).
E. MATRIX TRANSPOSITION
If A = [aij] is an m x n matrix, then the transpose of A,denoted by A
T=[aij] is an n x m matrix defined by a ij=aji.
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The transpose of A is obtained by interchanging the rowsand the columns of A.
Example:
2342
3132
5211
235
312
431
221
TAthenAIf
Note: The transpose of a symmetric matrix is equal to itself.PROPERTIES AND THEOREMS ON MATRIXOPERATIONS:
MATRIX ADDITION
A + O = A Existence of Additive IdentityA + (-A) = O Existence of Additive InverseA + B = B + A Commutative Property(A + B) + C = A + (B + C) Associative Property
SCALAR MULTIPLICATION
0 x A = O1 x A = Ok l (A) = k (l A) = l (k A)(k + l) A = k A + l Ak(A + B) = k A + k B
MATRIX MULTIPLICATION
A(BC) = (AB)C Associative PropertyA(B + C) = AB + AC Left Distributive Property(A + B)C = AC + BC Right Distributive PropertyAI = IA = A Existence of Multiplicative
Identitykl (AB) = (k A)(l B) = (l A)(k B)
Note: In general, Matrix Multiplication is not
commutative. That is, AB BA.
MATRIX TRANSPOSITION(A
T)T= A
(A + B)T= A
T+ B
T
(k A)T= k A
T
(AB)T= B
TA
T
In general (A1A2A3An-1An)T= An
TAn-1
TA3
TA2
TA1
T
DETERMINANTS
Another very important number associated with a squarematrix A is the determinant of Awhich we will now define.This unique number associated to a matrix A is useful in thesolutions of linear equation.
Permutation:
Let S={1, 2, 3, , n} be the set of integers from 1 to n,arranged in increasing order. A rearrangement a1a2a3anofthe elements in S is called a permutation of S.
By the Fundamental Principle of Counting we can put anyone of the n elements of S in the first position, any one ofthe remaining (n-1) elements in the second position, anyone of the remaining (n-2) elements in the third position,and so on until the nth position. Thus there are n(n-1)(n-2)3*2*1 = n! permutations of S. We refer to the set of allpermutations of S by Sn.
Examples:
If S = {1, 2, 3} then S3= {123, 132, 213, 231, 312, 321}
If S = {1, 2, 3, 4} then there are 4! = 24 elements of S4.
Odd and Even Permutations
A permutation a1a2a3an is said to have an inversion if alarger number precedes a smaller one. If the total number ofinversion in the permutation is even, then we say that thepermutation is even, otherwise it is odd.Examples: ODD and EVEN Permutation
S1has only one permutation; that is 1, which is even sincethere are no inversions.
In the permutation 35241, 3 precedes 2 and 1, 5 precedes2, 4 and 1, 2 precedes 1 and 4 precedes 1. There is a totalof 7 inversions, thus the permutation is odd.
S3has 3! = 6 permutations: 123, 231and 312 are even while132, 213, and 321 are odd.
S4has 4! = 24 permutations: 1234, 1243, 1324, 1342, 1423,1432, 2134, 2143, 2314, 2341, 2413, 2431, 3124, 3142,3214, 3241, 3412, 3421, 4123, 4132, 4213, 4231, 4312,4321.
For any Sn, where n>1 it contains n!/2 even permutationsand n!/2 odd permutations.
DEFINITION: DETERMINANT
Let A = [aij] be a square matrix of order n. The determinant
of A denoted by det(A) or Ais defined by
nnjjj
aaaAA ...det21 21
where the summation is over all permutations j1j2jnof theset S = {1,2,,n}. The sign is taken as (+) if the permutationis even and () if the permutation is odd.
Examples:
If A = [a11] is a 1 x 1 matrix then det(A) or A= a11.
If A =
2221
1211
aa
aa , then to get Awe write down the terms a1-
a2-b and replace the dashes with the all-possiblepermutations of S = {1, 2}, namely 12 (even) and 21 (odd).
Thus A= a11a22 - a12a21.
If A =
333231
232221
131211
aaa
aaa
aaa , then to compute the A we write
down the six terms a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-, a1-a2-a3-. Replace the dashes with all the elements of S3,affix a (+) or (-) sign and get the sum of the six terms.
If A is a square matrix of order n, there will be n! terms inthe determinant of A with n!/2 positive terms and n!/2negative terms.
9
620098
)2)(1)(3()1)(1)(2()3)(2)(0()1)(1)(0()3)(1)(3()2)(2)(2(
210
123
312
METHODS IN GETTING THE DETERMINANT
A. DIAGONAL METHOD
This method is applicable to matrices with sizeless than or equal to 3.
1. 2 x 2 Matrices
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12212211
2221
1211aaaa
aa
aa s
2. 3 x 3 Matrices
B. METHOD OF COFACTORS
Complementary Minor, det(Mij) or MijThe complementary minor or simply minor of an element a ijof the matrix A is that determinant of the sub-matrix Mijobtained after eliminating the ith row and jth column of A.
Algebraic Complement or Cofactor, Aij
The algebraic complement or cofactor of an element a ij ofthe matrix A is that signed minor obtained from the formula
(-1)i+j
Mij
DETERMINANT USING THE COFACTOR METHOD
The determinant of a square matrix maybe obtained usingexpansion about a row or expansion about a column. Thefollowing formulas maybe used in getting the determinant:
n
k
ikikininiiii AaAaAaAaA1
2211 ...)det(
(expansion about the ith row)
and
n
k
kjkjnjnjjjjj AaAaAaAaA1
2211 ...)det(
(expansion about the ith row)
Note: We may choose any row or any column in getting thedeterminant of a given matrix.
Example: To evaluate
0130
0423
1412
0301
It is best to expand about the fourth row because it has themost numbers of zeros. The optimal course of action is toexpand about the row or column that has the largestnumber of zeros, because in that case the cofactors Aijof
those aijwhich are zero need not be evaluated since theproduct of aijAij= (0)Aij= 0.
THEOREMS ON DETERMINANTS
1. If a square matrix A = [aij] contains a row (or a column)
that has elements all equal to zero, then A= 0.
Example:
0
042
032
021
2. The determinant of a square matrix A = [aij] is equal to
the determinant of its transpose AT = [aij].(i.e.A=AT)
Example:
233
422
141
241
324
321
3. If a row (or column) of a square matrix A = [aij] ismultiplied by a constant k, then the determinant of theresulting matrix B = [bij] is equal to k times the
determinant of A (i.e. B= kA).
Example:
10)5(2
224
224
1
5
224
2
1
ba
yx
ba
then
ba
yx
ba
If
4. As a corollary to the third theorem, if A has a row (orcolumn) that has a common factor l, then this k may befactored out of the determinant of A, where a simplified
matrix B is formed. (i.e. .A= kB.
Example:
xy
yyx
xxy
xy
xy
yyx
xxy
xy
xyxy
yyyx
xxxy
218343
23
)2(22183862
23
22186864
232
2
2
5. If two row (or columns) of a square matrix A = [aij] were
interchanged to form a new matrix B = [b ij], then B=-A.
Example:
1
02
002
1
002
02
yzyx
xxy
x
yzyx
x
xxy
6. If two rows (or columns) of a matrix A = [aij] are
identical then A= 0.Example:
0
xyzxy3z4
xyzz5yx3x2
xz2x3yx2
xzyz5yx3x2
x3
xyzxy3z4
yzx3xz15xy3x9x6
xz2x3yx2
xyzz5yx3x2
22
222
7. As a corollary to the sixth theorem, if the elements in a
row (or column) of a square matrix A = [a ij] aremultiples of the corresponding elements of another row
or column of the matrix A, then A= 0.Example:
0
216
464
232
xyy
8. If B = [bij] is a square matrix of order n that is derivedfrom another square matrix A = [a ij] of order n, byadding correspondingly the elements of a row (orcolumn) to a multiple of the elements of another row
(or column), then B=A.Example:
333231
232221
331332123111
333231
232221
131211
aaa
aaa
kaakaakaa
aaa
aaa
aaa
9. If the elements of one row (or column) of a squarematrix A = [aij] of order n may be expressed as
binomials such that two square matrices B = [bij] and C= [cij] both of order n, are formed after splitting the
binomial elements, then A=B+C.Example:
3231
2221
1211
333231
232221
131211
aa
aa
aa
aaa
aaa
aaa
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333231
232221
131211
333231
232221
131211
333231
232322222121
131211
aaa
ccc
aaa
aaa
bbb
aaa
aaa
cbcbcb
aaa
10. The determinant of the product of two square matrices A= [aij] and B = [bij] of the same order n is equal to the
product of the determinant of A and the determinant of B.Example:
If 2
216
464
232
xyy
z
x
and
222
56
543
bbaa
ba
a
= 3
then
xyy
z
x
216
464
232
222
56
543
bbaa
ba
a
= 32 = 6 .
11. The determinant of a triangular matrix is equal to theproduct of the elements in its principal diagonal.
Example:
4
1
02
0022
x
yzyx
xxy
x
12. The determinant of an Identity Matrix is equal to 1.
ADJOINT OF A MATRIX
The Adjoint of a square matrix A=[aij] of order n is thatsquare matrix with the same order n denoted by adj(A)=[Aji]where Aijis the cofactor of the element aijof matrix A. Theadjoint of a matrix is the transpose of the matrix of cofactors
of the elements of A.
Input: Square MatrixOutput: Square Matrix (with the same size as the originalmatrix)
Notation: adj A, adj B
Step 1: Get the cofactors of all the elements in theoriginal matrix.
Recall: the cofactor of an element aij can be denoted as Aijand is defined by:
ijji
ij MA )1(
Step 2: Set up the adjoint matrix by taking the transposeof the matrix of cofactors.
TijAAadj
Example:
If A =
dc
bathen adj(A) =
ac
bd.
Inverse of a Matrix
The inverse of a square matrix A = [aij] of order n is that
matrix B = [bij] of the same order n such that AB = BA = In.We denote the inverse matrix of A by A-1. Thus, we define
the inverse of A as that matrix A-1
such that
A(A-1) = (A
-1)A = In.
Not all matrices has its inverse. However, if the inverse of amatrix exists, it is unique.
If the inverse of a matrix exists, we say that the matrix isinvertible or non-singular. Otherwise, we say that the matrixis non-invertible or singular.
Matrix Inversion Using the Adjoint
and the Inverse
Matrix Inversion applies only to square matrices and can beproduced using the adjoint matrix and thedeterminant.
Notation: A-1, B
-1
The proof of this needs the knowledge of the followingtheorem:
The sum of the products of the elements in one row (or
column) and the cofactors of the elements of another row(or column) of a given square matrix is zero.
From the above formula for inverse, it is highly suggestedthat the determinant be computed first. If it sohappened that the matrix is singular (i.e., thedeterminant is zero), then the inverse of thematrix is said to be non-existent.
Note that it is a waste of effort to still produce the adjoint if
the matrix is singular. Therefore, it is advised
that you first check for singularity.Example: Set up the Inverse of the given matrix.
Using the diagonal method to compute for the determinantof the given matrix:
Since matrix A is singular, as evidenced by its zerodeterminant, it can thus be concluded that theInverse of A (or A-1) does not exist.
Example 2: Set up the Inverse of the given matrix
324
321
111
A 73682126 A
Since the determinant is not zero, then matrix A is said to be non-singular. In this case, the inverse exists and there is a need to set upthe adjoint.
Getting the cofactors of all the elements in the original matrix.
ijji
ij MA )1(
1232
32)1( 211
A 934
31)1( 312 A
10
24
21)1( 413
A 1
32
11)1( 321
A
134
11)1( 422 A
224
11)1( 523
A
771
252
111
A
A
adjAA 1
01414514235 A
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532
11)1( 431
A 2
31
11)1( 532 A
321
11)1( 633
A
325
211
10912
ijA ,thus
Consequently,A
AadjA 1
3210
219
5112
7
11A
SOLUTION TO SYSTEM OF LINEAR EQUATIONS
In general, we can think of a system of linearequations as a set of m equations that contains nunknowns. There are several forms by which a system ofequations can be written.
We can have the equation form:
mnmnmmm
nn
nn
nn
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
bxaxaxaxa
332211
33333232131
22323222121
11313212111
Or we can transform that to the matrix form:
mnmnmmm
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaa
aaaa
3
2
1
3
2
1
321
3333231
2232221
1131211
Referring to the matrix form, we can actually rewrite thesystem of equations as a compact matrix operation:
AX = B.
Where:
A Coefficient Matrix
X
Column Matrix of Unknowns/VariablesB Column Matrix of Constants
SOLUTION TO SYSTEM OF n-LINEAR EQUATIONSWITH n UNKNOWNS
A. USING THE INVERSE METHOD
The Inverse Method maybe applied only to asystem of linear equations in which the number ofindependent equations is equal to the number of unknowns.If the number of equations is equal to the number ofunknowns, the equation AX = B will have a matrix ofcoefficients that is square.
If the matrix of coefficients A is non-singular, thesolution to the system is unique. On the other hand, if A issingular, either the system has a unique solution or nosolution at all.
Derivation of the Solution forxis:
BAX
BAXI
BAXAA
BAAXA
BAX
1
1
11
11
*
**)(
*
Example: Determine the values of x1, x2 and x3 in thefollowing system of equations.
5324
632
1
321
321
321
xxx
xxx
xxx
Solution:
The above system of equations can be written in matrixform:
56
1
324321
111
3
2
1
xx
x
We can write this in matrix form AX = B and let X = A-1B,
where:
324
321
111
A
Getting A-1
3210
219
5112
7
11A
To getx1,x2andx3, multiplyA-1to B:
5
6
1
*
3210
219
5112
7
1
3
2
1
x
x
x
X
Performing the operationA-1B will yield the solution
matrix:
1
1
1
3
2
1
x
x
x
X
Make it a habit to check if all the computed values of the
unknowns satisfy all the given equations. Checking is doneby substituting the valuesx1 = 1,x2= 1 andx3= 1 to theoriginal equations.
Equation11(1)1(1) + 1(1) =? 1Satisfied
Equation 21(1) + 2(1) + 3(1) =? 6Satisfied
Equation 34(1)2(1) + 3(1)=? 5Satisfied
Since all the equations were satisfied, then (x1,x2,x3) = (1,1, 1) is indeed the solution to the system.
SOLUTION TO SYSTEM OF EQUATIONS USINGCRAMER'S RULE
Recall that A system of equation n equations in nunknowns can be modeled as a matrix operation AX = B.
3210
219
5112
Aadj
where aijareconstantcoefficients of theunknowns xjand bI
Take note that thederivation assumesthat A
-1exists. If A
-1
does not exist, wecan not find thesolution to the
system AX = B.
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nnnnnnn
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaa
aaaa
3
2
1
3
2
1
321
3333231
2232221
1131211
Let:A
coefficient matrixxi ithvariable
B right hand side constantsA I matrix resultingfrom replacing the
ithcolumn ofAby the column
vector of constants B
The solution of the system of equations can be determinedby using the formula:
A
Ax
i
i
Notice that regardless of the variable ithat is computed, thedenominator of the above formula is fixed at |A|. Therefore,it is suggested that the determinant of the coefficient matrixbe the first to be computed.
Example:Using Cramer's Rule, determine the values ofx1,x2andx3that simultaneously satisfy the following system ofequations.
3
2
1
121
133
121
3
2
1
x
x
x
Solution:
Compute for the determinant ofA first:
6623623
121
133
121
A
A
Now, let us compute for the value ofx1by using the formula
A
Ax
11
The right hand side matrix B is
3
2
1
To set up the matrix A1, all you just have to do is to replacethe first column ofAby b. Doing what has just beendescribed will result in:
0429463
123132
121
1
1
A
A 06
011
A
A
x
Applying the same process to solvex2 and x3:
1
321
233
121
*6
11
131
123
111
*6
1 33
22
A
Ax
A
Ax
SOLUTION TO SYSTEM OF LINEAR EQUATIONS USINGL-U FACTORIZATION
Direct L-U Factorization:
In theory any square matrix A may be factored into aproduct of lower and upper triangular matrices.
Let us take the case of a 4thorder matrix:
1000
100
10
1
*0
00
000
34
2423
141312
44434241
333231
2221
11
44434241
34333231
24232221
14131211
u
uu
uuu
llll
lll
ll
l
aaaa
aaaa
aaaa
aaaa
Notice that the diagonal elements of the upper triangularmatrix have been set to values of 1 for reason of simplicity.(L-U Factorization is not unique.)
From matrix multiplication, we know that:
)0(0)0(0)0(0)1(1111 la or 1111 al
)0(0)0(0)0()1( 222121 lla or 2121 al
)0(0)0()0()1( 33323131 llla or 3131 al
)0()0()0()1( 4443424141 lllla or
4141 al
)0(0)0(0)1(0)( 121112 ula or 111212 / lau
)0(0)1(0)(0)( 23131113 uula or 111313 / lau
)1(0)(0)(0)( 3424141114 uuula or
111414 / lau
)0(0)0(0)1()( 22122122 lula or
)( 12212222 ulal
)0(0)0()1()( 3332123132 llula or
)( 12313232 ulal
)0()0()1()( 444342124142 lllula or
)( 12414242 ulal
)0(0)1(0)()( 2322132123 ulula or
2213212323 /)]([ lulau
)1(0)(0)()( 232422142124 uulula or
2214212424 /)]([ lulau
)0(0)1()()( 332332133133 lulula or
)()( 233213313333 ululal
)0()1()()( 44432342134143 llulula or
)()( 234213414343 ululal
)1(0)()()( 34332432143134 ululula or
33243214313434 /)]()([ lululau
)1()()()( 4434432442144144 lululula or
)()()( 3443244214414444 ulululal
How to get the solution to a system of equations usingL-U Decomposition Method?
Recall:A system of equations can be written as a compactmatrix operationAX = B
If we factor out the coefficient matrix A as L*U andsubstitute to AX = B, we can generate the equationL(UX)=B.
Momentarily define UX = Y which suggests LY = B. Fromthis transformation, we have actually decomposedAX = b totwo systems of equations.
Two-stage solution:
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ES 21 NotesPage 8 of 12
I. Solve for Y in the equation LY = B using forwardsubstitution.
II. Solve for X in the equation UX = Y using backsubstitution.
Example: Determine the values ofxi'sin :
8
4
3
121
153
124
3
2
1
x
x
x
Knowing that
121
153
124
A
, therefore
25251
0273
004
L
and
100211041
211
U
Stage 1: Forward substitution using LY = B
8
43
25
251
02
73004
3
2
1
y
yy
Note that the computed values of yi'shere are not yet thesolution since the original system of equations is in terms ofxi's.
Stage 2: Back substitution using UX = Y
32
14
3
1002
110
41
211
3
2
1
x
x
x
This time (x1,x2,x3) = (1, 2, 3) is the solution to the originalsystem of equations.
AUGMENTED MATRIX OF A AND B
If A is an m x n matrix and B is a p x n matrix, then theaugmented matrix of A and B denoted by [A : B] is thematrix formed by the elements of A and B separated bypipes.
Example:
If
833
617
521
A
and
47
20
12
Bthen A : B is
47
20
12
|
|
|
833
617
521
The augmented matrix associated to a system of linearequation AX=B is the matrix [A : B]. For example, we cannow rewrite the system of equation:
1
1
1
142
621
312
z
y
xas simply
1
1
1
|
|
|
142
621
312.
ECHELON FORM OF A MATRIX
An m x n matrix A is said to be in row echelon form if itsatisfies the following properties:
1. All rows whose elements are all zeros, ifexist, are at the bottom of the matrix.
2. If at least one element on a row is not equalto zero, the first non-zero element is 1, andthis is called the leading entry of the row.
3. If two successive rows of the matrix haveleading entries, the leading entry of the rowbelow the other row must appear to theright of the leading entry of the other row.
An m x n matrix A is said to be in reduced row echelon formif added to the first three properties it satisfies a fourthproperty:
4. If a column contains a leading entry ofsome row, then all the other entries mustbe zero.
Example:
The following matrices are not in row echelon form. (Whynot?)
0010
8100
7610
4221
0000
1000
0100
0210
0001
100000
000000
291000
531010
231021
CBA
The following matrices are in row echelon form but not inreduce row echelon form.
0000
0000
0100
3010
0801
00100
80010
26501
1000
2100
3210
4321
FED
The following matrices are in reduced row echelon form.(Hence, in row echelon form.)
000
000
000
010
301
01000
30100
20010
1000
0100
0010
0001
JHG
ELEMENTARY ROW (COLUMN) OPERATIONS ONMATRICES
An elementary row (column) operation on a matrix A is anyone of the following operations:
Type I. Interchange any two rows (columns).Type II. Multiply a row (column) by a non-zero
constant k.Type III. Add to elements of a row k times of the elementsof another row the correspondingly.
Example: Let
2282
4613
0201
A
Interchanging rows 1 and 3 of A (R1R3) obtain
0201
4613
2282
B
Multiplying row 3 by (R3R3), we obtain
34 1 y 43
1 y
42
7321
yy2
12y
82
52
5321
yyy 33y
33 x
21
21
32 xx 22 x
43
41
21
321 xxx 11x
33 x
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1141
4613
0201
C
Adding 3 times the elements in row 1 to the elements in row
2 (R2R2 + 3R1), we obtain
22824010
0201
A
ELEMENTARY ROW OPERATIONS AS APPLIED TOTHE A SYSTEM OF EQUATION A:B
As a applied to the augmented matrix [A : B] as a system ofequation, the three elementary row operation willcorrespond to the following:
TYPE I rearranging the order of the equationsTYPE II multiplying both side of the equation by
a constant
TYPE III working with two equations
From this observation, we could see that as applied to aoperations does not alter the solution of the system.
ROW (COLUMN) EQUIVALENT MATRICES
An m x n matrix A is row (column) equivalent to an m x nmatrix B if B can be obtained from A by applying a finitesequence of elementary row operations.
THEOREMS ON MATRIX EQUIVALENCE
1. Every nonzero m x n matrix A = [aij] is row (column)equivalent to a matrix in row (column) echelon form.
2. Every nonzero m x n matrix A = [aij] is row (column)equivalent to a matrix in reduced row (column) echelonform.
3. Let AX = B and CX = D be two systems of m linearequations in n unknowns. If the augmented matrices[A : B] and [C : D] are row equivalent, then the linearsystems are equivalent (i.e. they have exactly thesame solutions).
4. As a corollary to the third theorem, if A and B are rowequivalent matrices, then the homogeneous systemsAX = 0 and BX = 0 are equivalent.
SOLUTIONS TO A SYSTEM OF m EQUATIONS in nUNKNOWNS
In general a system of m equations inn unknowns maybe written in matrix form:
mnmnmmm
n
n
n
b
b
b
b
x
x
x
x
aaaa
aaaa
aaaa
aaaa
3
2
1
3
2
1
321
3333231
2232221
1131211
.
This system may now be represented by the augmentednotation:
mmnmmm
n
n
n
b
b
b
b
aaaa
aaaa
aaaa
aaaa
3
2
1
321
3333231
2232221
1131211
|
|
|
|
Applying the theorems on equivalent matrices we now havethe following methods of solution:
GAUSSIAN ELIMINATION METHOD
The objective of the Gaussian Elimination Method is totransform the augmented matrix [A : B] to the matrix [A* :B*] in row echelon form by applying a series of elementaryrow transformations. Getting the solution of the system [A* :B*] using back substitution will also give the solution to theoriginal system [A : B].
To reduce any matrix to row echelon form, apply thefollowing steps:
1. Find the leftmost non-zero column.2. If the 1
st row has a zero in the column of step 1,
interchange it with one that has a non-zero entry in
the same column.3. Obtain zeros below the leading entry by adding
suitable multiples of the top row and to the rowsbelow that.
4. Cover the top row and repeat the same processstarting with step 1 applied to the leftover submatrix.Repeat this process with the rest of the rows.
5. For each row obtain leading entry 1 by dividing eachrow by their corresponding leading entry.
Example: The linear system
33
82
932
zx
zyx
zyx
has the augmented matrix associated to the system
3|103
8|112
9|321
:BA
which can be transformed as a matrix in row echelon form
3|100
2|110
9|321
: ** BA
using back substitution we have
2)3(3)1(29932
1322
3
xzyx
yzy
z
thus we have the solution (x, y, z) = (2, -1, 3).
GAUSS-JORDAN REDUCTION METHOD
On the other hand a second method called the Gauss-Jordan Reduction Method gets rid of the back substitutionphase. The objective of the Gauss-Jordan ReductionMethod is to transform the augmented matrix [A : B] to thematrix [A* : B*] in reduced row echelon form by applying aseries of elementary row transformations. Doing this willautomatically give the solution of the system [A* : B*] whichalso provides the solution to the original system [A : B].
To reduce any matrix to reduced row echelon form, apply
the following steps (SINE):
1. Search search the ith column of the augmentedmatrix from the ith row to the nth row for the
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maximum pivot, i.e. element with the largest absolutevalue.
2. Interchange assuming the maximum pivot occursin thejthrow, interchange the ithrow and thejthrowso that the maximum pivot will now occur in thediagonal position.
3. Normalizenormalize the new ithrow by dividing itby the maximum pivot on the diagonal position.
4. Eliminateeliminate the ithcolumn from the first upto the nthequation, except in the ithequation itselfusing the transformations.
Example: The linear system
33
82
932
zx
zyx
zyx
has the augmented matrix associated to the system
3|103
8|112
9|321
: BA
which can be transformed as a matrix in row echelon form
3|100
1|010
2|001
: ** BA
thus we have the solution (x, y, z) = (2, -1, 3).
SUBMATRIX AND RANK
SUBMATRIXA submatrix of A=[aij] is any matrix obtained by eliminatingsome rows and/or columns of the matrix A.
Example: Let
275114
25961
28303
81032
A
The following are some submatrices of A:
)remaining
columnond(sec28303
)remaining
columnfourth(
7
5
8
1
)columnfifthand
third,firstremove(
2511
296
230
803
)columnfifthandthird
androwondsecremove(
7114
561
132
)rowthirdremove(
275114
28303
81032
RANK OF A MATRIX
The rank of a matrix A = [a ij] is the order of the largestsquare submatrix of A with a non-zero determinant. Wedenote the rank of A by rank(A) or simply r(A).
Example: What is the rank of A?
10503
3412
4321
A
Solution:
Checking out first the determinants of 3x3 submatrices:
060
1050
341
432
014
1053
342
431
0
1003
312
421
0
503
412
321
Since at least one 3x3 submatrix of A has a non-zerodeterminant, then r(A) = 3.
Example: What is the rank of B?
161284
12963
8642
4321
B
Solution:
The determinant of B is equal to zero (THEOREM:Proportional rows). And it can also be shown that 3x3submatrices of B will have determinants equal to zero.
0
1684
1263
842
.g.e
(Rows are proportional)
But at least one 2x2 submatrix has non-zero determinant.
02463
42.g.e
Therefore r(B) = 2.
THEOREMS ON RANKS
1. The rank of a matrix is not altered by anysequence of elementary row (column)transformations.
2. Let A = [aij] and B = [bij] be two mxn matrices, ifrank(A) = rank(B) then A and B are equivalent.
3. If A = [aij] and B = [bij] are mxn matrices, andrank(A) = rank(B) = n, then rank(AB) = rank(BA) =n.
Example: What is the rank of C?
98765
87654
76543
65432
54321
C
Solution:Operating on rows of matrix C, we obtain the equivalentmatrix C
455
344
233
122
RR'R
RR'R
RR'R
RR'R
11111
11111
11111
11111
54321
'C
We could easily see that all 5x5, 4x4 and 3x3 submatricesof C have determinants equal to zero (THEOREM: Identicalrows). But for at least one 2x2 submatrix of C has a non-zero determinant.
(e.g. 0111
21 )
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Consequently r(C) = 2. But C and C are equivalentmatrices and hence they have equal ranks. Therefore r(C) isalso equal to 2.
RANKS AND THE TYPES OF SOLUTION TO A SYSTEMOF EQUATION
Recall that for the system of m linear equations in nunknowns AX = B. We can associate the system ofequation to the augmented matrix of the system [A:B].
The type of solution may be classified as unique, non-unique or inconsistent. Applying the concept of rank to theaugmented matrix [A:B], we have the following propositions:
1. If r(A) = r([A:B]) = n then the solution to thesystem is unique.
Example:
3:100
1:010
3:001
1:111
27:139
1:111
2. If r(A) = r([A:B]) < n, then the solution to the
system is non-unique.
Example:
0:000
1:010
1:001
6:224
3:112
0:111
3. If r(A) < r([A:B]), then the system has no solutionor inconsistent.
Example:
3:000
23:01045:001
6:116
2:224
3:112
Example 1: Rank and the Type of Solution to a System
For what values of k will the system of equations havea) a unique solutionb) a non-unique solutionc) no solution
2
k8z5y
k5z4y2x
k2z3y3x2
Solution: In augmented matrix form, we have:
28|510
5|421
2|332
k
k
k
Performing Gaussian Elimination Method:
kk
k
kRRR
k
k
kRRR
k
k
kRR
k
k
k
88|000
8|510
5|421
8|510
8|510
5|4212
8|510
2|332
5|421
8|510
5|421
2|332
2
23
'
3
2
12
'
2
2
21
2
Therefore we have the following conclusions:
a) For a unique solution, r(A) = r[A:B] = n
There will be no value of k that will satisfy thissince r(A) = 2 < n =3.
b) For non-unique solutions, r(A) = r[A:B] < n.
This will be satisfied if r[A:B] is also equal to 2.This will happen when the last element in the third row ofthe augmented matrix is also equal to zero.
8k2
- 8k = 0 k = 0, 1.
c) For the system to be inconsistent, r(A) < r[A:B]. This willbe satisfied if r[A:B] = 3 > 2. This will happen when the lastelement in the third row of the augmented matrix is notequal to zero.
8k2- 8k 0 k 0, 1.
Example 2:
For what values of m will the system of equations haved) a unique solutione) a non-unique solutionf) no solution
4mc2mba
0ca)2m(2cba
Solution: In augmented matrix form, we have:
4|11
0|102
2|111
2mm
m
Performing Gaussian Elimination Method:
2|100
42|120
2|111)2(
4|11
0|102
2|111
213
'
3
12
'
2
2
mm
mmm
RRR
RmRR
mm
m
Therefore we have the following conclusions:
Therefore we have the following conclusions:
a) For a unique solution, r(A) = r[A:B] = n
This will be satisfied if m2-10 andm-20. Thuswe have a unique solution if m 1,2.
b) For non-unique solutions, r(A) = r[A:B] < n.
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This will happen when the last element in the thirdrow of matrix A and the augmented matrix are both equal tozero.
m2- 1 = 0 and m+2 = 0.
There is no value of m that will satisfy bothequation.
The other value of m to be checked is when m = -2.Substituting this to the system gives the system:
0|300
0|100
2|111
Clearly, we could see that the resulting systemgives a non-unique solution because, r(A) = r[A:B]=2 2. This willhappen when the last element in the third row of theaugmented matrix is not equal to zero but the last elementof the third row of A is equal to zero.
m21 = 0 and m+2 0.
Thus, the system is inconsistent when m = 1 .
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