determinants. if is a square matrix of order 1, then |a| = | a 11 | = a 11 if is a square matrix of...
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Determinants
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Determinants
If is a square matrix of order 1,
then |A| = | a11 | = a11
ijA = a
If is a square matrix of order 2, then11 12
21 22
a aA =
a a
|A| = = a11a22 – a21a12
a a
a a
11 12
21 22
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Example
4 - 3Evaluate the determinant :
2 5
4 - 3Solution : = 4 ×5 - 2 × -3 = 20 + 6 = 26
2 5
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Solution
If A = is a square matrix of order 3, then11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
[Expanding along first row]
11 12 1322 23 21 23 21 22
21 22 23 11 12 1332 33 31 33 31 32
31 32 33
a a aa a a a a a
| A | = a a a = a - a + aa a a a a a
a a a
11 22 33 32 23 12 21 33 31 23 13 21 32 31 22= a a a - a a - a a a - a a + a a a - a a
11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 22a a a a a a a a a a a a a a a a a a
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Example2 3 - 5
Evaluate the determinant : 7 1 - 2
-3 4 1
2 3 - 5
1 - 2 7 - 2 7 17 1 - 2 = 2 - 3 + -5
4 1 -3 1 -3 4-3 4 1
= 2 1 + 8 - 3 7 - 6 - 5 28 + 3
= 18 - 3 - 155
= -140
[Expanding along first row]
Solution :
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Minors
-1 4If A = , then
2 3
21 21 22 22M = Minor of a = 4, M = Minor of a = -1
11 11 12 12M = Minor of a = 3, M = Minor of a = 2
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Minors4 7 8
If A = -9 0 0 , then
2 3 4
M11 = Minor of a11 = determinant of the order 2 × 2 square sub-matrix is obtained by leaving first row and first column of A
0 0= =0
3 4
Similarly, M23 = Minor of a23
4 7= =12- 14=-2
2 3
M32 = Minor of a32 etc.4 8
= =0+72=72-9 0
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Cofactors i+ j
ij ij ijC = Cofactor of a in A = -1 M ,
ij ijwhere M is minor of a in A
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Cofactors (Con.)
C11 = Cofactor of a11 = (–1)1 + 1 M11 = (–1)1 +1 0 0
=03 4
C23 = Cofactor of a23 = (–1)2 + 3 M23 = 4 7
22 3
C32 = Cofactor of a32 = (–1)3 + 2M32 = etc.4 8
- =- 72-9 0
4 7 8
A = -9 0 0
2 3 4
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Value of Determinant in Terms of Minors and Cofactors
11 12 13
21 22 23
31 32 33
a a a
If A = a a a , then
a a a
3 3
i jij ij ij ij
j 1 j 1
A 1 a M a C
i1 i1 i2 i2 i3 i3= a C +a C +a C , for i =1 or i = 2 or i = 3
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Properties
1. If each element of a row (or column) of a determinant is zero, then its value is zero.
2 2 2
3 3 3
0 0 0
a b c =0
a b c
Proof: Expanding the determinant along the row containing only zeros:
0
..21
0..002....2221
1...1211
nnanana
naaanaaa
.00det1
n
jijCA
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Properties 2. If all the elements of a row (or column) is multiplied by a
non-zero number k, then the value of the new determinant is k times the value of the original determinant.
)det(...
...
...
...
...
...
)det(
21
21
21
22221
11211
21
ln2211
21
22221
11211
A
aaa
aaa
aaa
aaa
aaa
aaa
aaa
aaa
aaa
aaa
A
nnnn
knkk
knkk
n
n
nnnn
klkl
knkk
n
n
Proof: expand elong the row multiplied by
n
jijij
n
jijij CaCa
11
detdet AB
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Properties (Con.)
4. If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two (or more) determinants.
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3
a +x b c a b c x b c
a +y b c = a b c + y b c
a +z b c a b c z b c
nnananancccnaaanaaa
nnanananbbbnaaanaaa
nnanana
ncnbcbcbnaaanaaa
..21
..21
2....2221
1...1211
..21
..21
2....2221
1...1211
..21
..2211
2....2221
1...1211
Proof: expand along the row containing the sum:
n
j
n
jijijijij
n
jijijij DbDcDcb
1 11
detA
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Properties of Determinants2. If any two rows (or columns) of a determinant are interchanged,
then the value of the determinant is changed by minus sign. Proof: first we proove it for neighbourh rows:
nnnnn
ni
in
i
i
i
i
i
i
n
n
aaaa
a
a
a
a
a
a
a
a
aaaa
aaaa
D
321
)1(3)1(
3
2)1(
2
1)1(
1
2232221
1131211
1
.
.
nnnnn
in
ni
i
i
i
i
i
i
n
n
aaaa
a
a
a
a
a
a
a
a
aaaa
aaaa
D
321
)1(
3
3)1(
2
2)1(
1
1)1(
2232221
1131211
2
.
.
n
jjiij DaD
111
n
jijij DaD
12 1
21 -DD
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Cont. Interchanging 2 rows (columns)
Suppose, that we change the i. and j. rows, and j-i = k.i i ji+1 j ii+2 i+1 i+1…. i+2 i+2I+k=j …. …
Then we need k neighbourh change to position the j. row right below the i. (see 2nd column) One more change and j stands at the earlier position of i, and i is the row right after j. Again, i must be interchanged by its neighbours k times, so the sum of neighbourgh row changes is: k+k+1=2k+1 – an odd number, so because by each change the deterimant changes it sing, after odd number of changes, its changes its sign.
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Properties6. If any two rows (or columns) of a determinant are identical,
then its value is zero.
Proof: if we interchanged the identical rows, the determinant
remains the same, but formally it does changes its sign,
so :D = ( - D ), which is possible only if D=0
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5. The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column).
1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 1 1 2 3
3 3 3 3 3 3 3 3
a b c a +mb - nc b c
a b c = a +mb - nc b c C C + mC - nC
a b c a +mb - nc b c
Applying
Properties (Con.)
nnnn
jnjj
jninjiji
n
nnnn
jnjj
inii
n
aaa
aaa
aaaaaa
aaa
D
aaa
aaa
aaa
aaa
D
21
21
2211
11211
21
21
21
11211
.
Proof:0
0
21
21
21
11211
21
21
21
11211
D
aaa
aaa
aaa
aaa
aaa
aaa
aaa
aaa
D
nnnn
jnjj
jnjj
n
nnnn
jnjj
inii
n
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Properties (Con.)
nn
nnnnnnnn
aaaa
aaa
aa
a
aa
aaa
aa
a
a .........
.
.
0
....
00
.
.
0
....
00
nna..
n2a
n1a
0....
0022
a21
a
00011
a
332211
21
4443
33
2211
21
3332
22
11
If A is an upper (lower) triangular matrix, then the determinant is equal to the product of the elements of the main diagonal:
Proof:
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1. The value of a determinant remains unchanged, if its rows and columns are interchanged. (A matrix and its transpose have the same detereminant)
1 1 1 1 2 3
2 2 2 1 2 3
3 3 3 1 2 3
a b c a a a
a b c = b b b
a b c c c c
i e A A. . '
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Row(Column) Operations
Following are the notations to evaluate a determinant:
Similar notations can be used to denote column operations by replacing R with C.
(i) Ri to denote ith row
(ii) Ri Rj to denote the interchange of ith and jth rows.
(iii) Ri Ri + lRj to denote the addition of l times the elements of jth row to the corresponding elements of ith row.
(iv) lRi to denote the multiplication of all elements of ith row by l.
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Evaluation of DeterminantsIf a determinant becomes zero on putting is the factor of the determinant. x = , then x -
2
3
x 5 2
For example, if Δ = x 9 4 , then at x =2
x 16 8
, because C1 and C2 are identical at x = 2
Hence, (x – 2) is a factor of determinant .
0
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Sign System for Expansion of Determinant
Sign System for order 2 and order 3 are given by
+ – ++ –
, – + –– +
+ – +
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42 1 6 6×7 1 6
i 28 7 4 = 4×7 7 4
14 3 2 2×7 3 2
1
6 1 6
=7 4 7 4 Taking out 7 common from C
2 3 2
Example-1
6 -3 2
2 -1 2
-10 5 2
42 1 6
28 7 4
14 3 2
Find the value of the following determinants
(i) (ii)
Solution :
1 3= 7×0 C and C are identical
= 0
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Example –1 (ii)6 -3 2
2 -1 2
-10 5 2
(ii)
3 2 3 2
1 2 1 2
5 2 5 2
1
1 2
3 3 2
( 2) 1 1 2 Taking out 2 common from C
5 5 2
( 2) 0 C and C are identical
0
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Evaluate the determinant1 a b+c1 b c+a1 c a+b
Solution :
3 2 3
1 a b+c 1 a a+b+c
1 b c+a = 1 b a+b+c Applying c c +c
1 c a+b 1 c a+b+c
3
1 a 1
= a+b+c 1 b 1 Taking a+b+c common from C
1 c 1
Example - 2
1 3= a+ b + c ×0 C and C are identical
= 0
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2 2 2
a b c
We have a b c
bc ca ab
21 1 2 2 2 3
(a-b) b- c c
= (a-b)(a+b) (b- c)(b+c) c Applying C C - C and C C - C
-c(a-b) -a(b- c) ab
2
1 2
1 1 cTaking a- b and b- c common
=(a- b)(b- c) a+b b+c cfrom C and C respectively
-c -a ab
Example - 3
bc
2 2 2
a b c
a b c
ca ab
Evaluate the determinant:
Solution:
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21 1 2
0 1 c
=(a- b)(b- c) -(c- a) b+c c Applying c c - c
-(c- a) -a ab
2
0 1 c
=-(a- b)(b- c)(c- a) 1 b+c c
1 -a ab
22 2 3
0 1 c
= - (a- b)(b- c)(c- a) 0 a+b+c c - ab Applying R R - R
1 -a ab
Now expanding along C1 , we get(a-b) (b-c) (c-a) [- (c2 – ab – ac – bc – c2)]= (a-b) (b-c) (c-a) (ab + bc + ac)
Solution Cont.
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Without expanding the determinant,
prove that 3
3x+y 2x x
4x+3y 3x 3x =x
5x+6y 4x 6x
3x+y 2x x 3x 2x x y 2x x
L.H.S= 4x+3y 3x 3x = 4x 3x 3x + 3y 3x 3x
5x+6y 4x 6x 5x 4x 6x 6y 4x 6x
3 2
3 2 1 1 2 1
=x 4 3 3 +x y 3 3 3
5 4 6 6 4 6
Example-4
Solution :
3 21 2
3 2 1
=x 4 3 3 +x y×0 C and C are identical in I I determinant
5 4 6
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Solution Cont.
31 1 2
1 2 1
=x 1 3 3 Applying C C - C
1 4 6
32 2 1 3 3 2
1 2 1
=x 0 1 2 ApplyingR R - R and R R - R
0 1 3
31
3
= x ×(3- 2) Expanding along C
=x =R.H.S.
3
3 2 1
=x 4 3 3
5 4 6
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Prove that : = 0 , where w is cube root of unity.
3 5
3 4
5 5
1 ω ω
ω 1 ω
ω ω 1
3 5 3 3 2
3 4 3 3
5 5 3 2 3 2
1 ω ω 1 ω ω .ω
L.H.S = ω 1 ω = ω 1 ω .ω
ω ω 1 ω .ω ω .ω 1
2
3
2 2
1 2
1 1 ω
= 1 1 ω ω =1
ω ω 1
=0=R.H.S. C and C are identical
Example -5
Solution :
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Example-6
2
x+a b c
a x+b c =x (x+a+b+c)
a b x+C
Prove that :
1 1 2 3
x+a b c x+a+b+c b c
L.H.S= a x+b c = x+a+b+c x+b c
a b x+C x+a+b+c b x+c
Applying C C +C +C
Solution :
1
1 b c
= x+a+b+c 1 x+b c
1 b x+c
Taking x+a+b+c commonfrom C
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Solution cont.
2 2 1 3 3 1
1 b c
=(x+a+b+c) 0 x 0
0 0 x
Applying R R -R and R R -R
Expanding along C1 , we get
(x + a + b + c) [1(x2)] = x2 (x + a + b + c)
= R.H.S