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Chapter
4
he
DETERMINANTS
4
...
We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices
...
— C
...
STEINMETZ
⎡a b ⎤ ⎡ x⎤ ⎡c ⎤
can be represented as ⎢ 1 1 ⎥ ⎢ ⎥ = ⎢ 1 ⎥
...
(Recall that if
a1 b1
or, a1 b2 – a2 b1 ≠ 0, then the system of linear
≠
a2 b2
equations has a unique solution)
...
S
...
Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc
...
Also, we will study various properties of determinants, minors, cofactors and applications
of determinants in finding the area of a triangle, adjoint and inverse of a square matrix,
consistency and inconsistency of system of linear equations and solution of linear
equations in two or three variables using inverse of a matrix
...
2 Determinant
To every square matrix A = [aij] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where aij = (i, j)th element of A
...
If M is the set of square matrices, K is the set of
numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and
k ∈ K, then f (A) is called the determinant of A
...
is
he
a b
⎡a b ⎤
If A = ⎢
⎥ , then determinant of A is written as |A | = c d = det (A)
⎣c d ⎦
Remarks
(i) For matrix A, | A | is read as determinant of A and not modulus of A
...
4
...
2 Determinant of a matrix of order two
bl
4
...
1 Determinant of a matrix of order one
Let A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a
⎡ a11 a12 ⎤
A= ⎢
⎥ be a matrix of order 2 × 2,
⎣ a21 a22 ⎦
then the determinant of A is defined as:
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Let
det (A) = |A| = Δ =
Example 1 Evaluate
Solution We have
2
2 4
–1 2
4
–1 2
Example 2 Evaluate
= a11a22 – a21a12
...
x
x +1
x –1
x
Solution We have
x
x +1
= x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1
x –1
x
4
...
3 Determinant of a matrix of order 3 × 3
Determinant of a matrix of order three can be determined by expressing it in terms of
second order determinants
...
There are six ways of expanding a determinant of order
DETERMINANTS
105
3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and
C3) giving the same value as shown below
...
e
...
e
...
e
...
e
...
(1)
Note We shall apply all four steps together
...
(2)
Expansion along first Column (C1)
a11
| A | = a21
a31
a12
a22
a32
a13
a23
a33
By expanding along C1, we get
1 + 1 a22
| A | = a11 (–1)
a32
3 + 1 a 12
+ a31 (–1)
a22
a23
a
+ a21 (−1) 2 + 1 12
a33
a32
a13
a33
a13
a23
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)
DETERMINANTS
107
he
| A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23
– a31 a13 a22
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22
...
It is left as an exercise to the
reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the
value of | A | obtained in (1), (2) or (3)
...
bl
is
Remarks
(i) For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros
...
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⎡1 1⎤
⎡2 2⎤
(iii) Let A = ⎢
⎥ and B = ⎢ 2 0 ⎥
...
Also
⎣
⎦
⎣4 0⎦
| A | = 0 – 8 = – 8 and | B | = 0 – 2 = – 2
...
In general, if A = kB where A and B are square matrices of order n, then | A| = kn
| B |, where n = 1, 2, 3
1 2 4
Example 3 Evaluate the determinant Δ = –1 3 0
...
So expanding along third
column (C3), we get
Δ= 4
–1 3
1 2
1 2
–0
+0
4 1
4 1
–1 3
= 4 (–1 – 12) – 0 + 0 = – 52
0
sin α – cos α
0
sin β
...
x 1
4 1
4 1
bl
3 – x2 = 3 – 8
x2 = 8
i
...
i
...
x= ±2 2
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Hence
EXERCISE 4
...
1
...
(i)
sin θ
cos θ
(ii)
x2 – x + 1 x – 1
x +1
x +1
⎡1 2⎤
3
...
If A = ⎢ 0 1 2 ⎥ , then show that | 3 A | = 27 | A |
⎢0 0 4⎥
⎣
⎦
5
...
If
2 4
he
⎡ 1 1 –2 ⎤
⎢
⎥
6
...
Find values of x, if
(i)
109
4
...
In this section,
we will study some properties of determinants which simplifies its evaluation by obtaining
maximum number of zeros in a row or a column
...
However, we shall restrict ourselves upto determinants of
order 3 only
...
a1
a2
a3
Verification Let Δ = b1
c1
b2
c2
b3
c3
Expanding along first row, we get
Δ = a1
b2
c2
b3
b b
b b
− a2 1 3 + a3 1 2
c3
c1 c3
c1 c2
= a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)
By interchanging the rows and columns of Δ, we get the determinant
a1
Δ1 = a2
a3
b1
b2
b3
c1
c2
c3
110
MATHEMATICS
Expanding Δ1 along first column, we get
Δ1 = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)
Hence Δ = Δ 1
he
Remark It follows from above property that if A is a square matrix, then
det (A) = det (A′), where A′ = transpose of A
...
0 4
6 4
6 0
– (–3)
+5
5 –7
1 –7
1 5
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Δ= 2
= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
By interchanging rows and columns, we get
2 6 1
Δ1 = –3 0 5
5 4 –7
= 2
(Expanding along first column)
0 5
6
1
6 1
– (–3)
+5
4 –7
4 –7
0 5
= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
Clearly
Δ = Δ1
Hence, Property 1 is verified
...
a1
Verification Let Δ = b1
c1
a2
b2
c2
a3
b3
c3
DETERMINANTS
111
Expanding along first row, we get
Δ = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)
Interchanging first and third rows, the new determinant obtained is given by
c2
c3
Δ1 = b1
a1
b2
a2
b3
a3
he
c1
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Expanding along third row, we get
Δ1 = a1 (c2 b3 – b2 c3) – a2 (c1 b3 – c3 b1) + a3 (b2 c1 – b1 c2)
= – [a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)]
Clearly Δ1 = – Δ
Similarly, we can verify the result by interchanging any two columns
...
2 –3 5
Example 7 Verify Property 2 for Δ = 6 0 4
...
e
...
Property 3 If any two rows (or columns) of a determinant are identical (all corresponding
elements are same), then value of determinant is zero
...
However, by Property 2, it follows that Δ has changed its sign
bl
3 2 3
Example 8 Evaluate Δ = 2 2 3
3 2 3
is
Therefore
Δ=–Δ
or
Δ=0
Let us verify the above property by an example
...
Property 4 If each element of a row (or a column) of a determinant is multiplied by a
constant k, then its value gets multiplied by k
...
Then
k a1
Δ1 = a2
a3
k b1
b2
b3
k c1
c2
c3
Expanding along first row, we get
Δ1 = k a1 (b2 c3 – b3 c2) – k b1 (a2 c3 – c2 a3) + k c1 (a2 b3 – b2 a3)
= k [a1 (b2 c3 – b3 c2) – b1 (a2 c3 – c2 a3) + c1 (a2 b3 – b2 a3)]
=k Δ
DETERMINANTS
k a1
Hence
k b1
b2
b3
c2
c3
b1
c1
= k a2
a3
b2
b3
c2
c3
k c1
a2
a3
a1
113
he
Remarks
(i) By this property, we can take out any common factor from any one row or any
one column of a given determinant
...
For example
102 18 36
3 4
Example 9 Evaluate 1
17 3 6
102 18 36 6(17) 6(3) 6(6)
17 3 6
3 4 = 1
3
4 =6 1 3 4 =0
Solution Note that 1
17 3 6
17
3
6
17 3 6
(Using Properties 3 and 4)
Property 5 If some or all elements of a row or column of a determinant are expressed
as sum of two (or more) terms, then the determinant can be expressed as sum of two
(or more) determinants
...
H
...
=
c1
a1
a3 + λ 3
b3
= b1
c1
c3
a2 + λ 2
b2
c2
a3 + λ 3
b3
c3
a2
b2
c2
a3
λ1 λ 2
b3 + b1 b2
c3
c1 c2
λ3
b3
c3
114
MATHEMATICS
a2
b2
c2
a3
λ1 λ 2
b3 + b1 b2
c3
c1 c2
λ3
b3 = R
...
S
...
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a
b
c
Example 10 Show that a + 2 x b + 2 y c + 2 z = 0
x
y
z
a
b
c
Solution We have a + 2 x b + 2 y c + 2 z
x
y
z
a b c
a b
c
= a b c + 2x 2 y 2z
x y z
x
y
z
=0+0=0
(by Property 5)
(Using Property 3 and Property 4)
Property 6 If, to each element of any row or column of a determinant, the equimultiples
of corresponding elements of other row (or column) are added, then value of determinant
remains the same, i
...
, the value of determinant remain same if we apply the operation
Ri → Ri + kRj or Ci → Ci + k Cj
...
Here, we have multiplied the elements of the third row (R3) by a constant k and
added them to the corresponding elements of the first row (R1)
...
DETERMINANTS
115
Now, again
a1
Δ1 = b1
c1
a2
b2
a3
k c1
b3 + b1
c2
c3
c1
k c2
b2
c2
=Δ+0
(since R1 and R3 are proportional)
he
Hence
k c3
b3 (Using Property 5)
c3
Δ = Δ1
is
Remarks
(i) If Δ1 is the determinant obtained by applying Ri → kRi or Ci → kCi to the
determinant Δ, then Δ1 = kΔ
...
A similar remark applies to column operations
...
3a 6a + 3b 10a + 6b + 3c
Solution Applying operations R2 → R2 – 2R1 and R3 → R3 – 3R1 to the given
determinant Δ, we have
a a+b a+b+c
a
2a + b
Δ= 0
0
3a
7 a + 3b
Now applying R3 → R3 – 3R2 , we get
a a+b a+b+c
a
2a + b
Δ= 0
0
0
a
Expanding along C1, we obtain
Δ= a
a 2a + b
0
a
+0+0
= a (a2 – 0) = a (a2) = a3
116
MATHEMATICS
Example 12 Without expanding, prove that
x+ y
z
Δ=
1
y+z
x
1
z+x
y =0
1
x+ y+z
x
1
x+ y+z
y
1
Since the elements of R1 and R3 are proportional, Δ = 0
...
e
...
H
...
= abc
b
b
b
1
1
1
+1
c
c
c
Applying R1→ R1 + R2 + R3, we have
1+
Δ = abc
1 1 1
1 1 1
1 1 1
1+ + +
1+ + +
+ +
a b c
a b c
a b c
1
1
1
+1
b
b
b
1
1
1
+1
c
c
c
DETERMINANTS
1
1
1
1
b
1
c
0
is
1 0
0 1
bl
1 1 1⎞
⎛
Δ = abc ⎜ 1+ + + ⎟
a b c⎠
⎝
0
he
1
⎛ 1 1 1⎞ 1 1
+1
= abc ⎜ 1+ + + ⎟
b
⎝ a b c⎠ b b
1
1
1
+1
c
c
c
Now applying C2 → C2 – C1, C3 → C3 – C1, we get
1
119
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⎛ 1 1 1⎞
= abc ⎜1 + + + ⎟ ⎡1(1 – 0 ) ⎤
⎣
⎦
⎝ a b c⎠
⎛ 1 1 1⎞
= abc ⎜1+ + + ⎟ = abc + bc + ca + ab = R
...
S
...
EXERCISE 4
...
x a
y b
z c
3
...
b+c q+r
c+a r+ p
a+b p+q
c−a
p
q
r
x
y
z
b−c c−a a−b = 0
c−a a−b b−c
4
...
1 bc a ( b + c )
1 ca b ( c + a ) = 0
1 ab c ( a + b )
120
MATHEMATICS
6
...
ab
ac
ba
−b 2
bc = 4 a 2 b 2 c 2
ca
a −b
0
cb
−c 2
he
By using properties of determinants, in Exercises 8 to 14, show that:
1 a a2
2
8
...
x2
2
z
xy
x + 4 2x
2x
2
10
...
(i)
2c
2c
c−a−b
x + y + 2z
z
(ii)
z
x
y + z + 2x
x
y
3
y
= 2( x + y + z)
z + x + 2y
DETERMINANTS
x2
1
x = 1 − x3
x2
1
(
)
1 + a 2 − b2
2ab
2ab
1− a + b
2b
−2a
2
a2 + 1
b2 + 1
bc
cb
(
= 1 + a2 + b2
)
3
1 − a2 − b2
c2 + 1
=1 + a 2 + b 2 + c 2
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ab
2a
ac
ca
14
...
2
he
x2
x
12
...
15
...
Which of the following is correct
(A) Determinant is a square matrix
...
(C) Determinant is a number associated to a square matrix
...
4 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
1
[x (y –y ) + x2 (y3–y1) +
2 1 2 3
x3 (y1–y2)]
...
(1)
Remarks
(i) Since area is a positive quantity, we always take the absolute value of the
determinant in (1)
...
(iii) The area of the triangle formed by three collinear points is zero
...
Solution The area of triangle is given by
=
1
61
( 3 + 72 – 14 ) =
2
2
is
1
⎡3 ( 2 – 1) – 8 ( – 4 – 5 ) + 1( – 4 – 10 ) ⎤
⎦
2⎣
bl
=
he
3 8 1
1
–4 2 1
Δ=
2
5 1 1
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Example 18 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants
and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units
...
Then, area of triangle ABP is zero (Why?)
...
Also, since the area of the triangle ABD is 3 sq
...
e
...
2
EXERCISE 4
...
Find area of the triangle with vertices at the point given in each of the following :
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)
DETERMINANTS
123
is
4
...
Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear
...
Find values of k if area of triangle is 4 sq
...
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
...
5
...
Then k is
(A) 12
(B) –2
(C) –12, –2
(D) 12, –2
bl
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors
...
Minor of an element aij is
denoted by Mij
...
1 2 3
Example 19 Find the minor of element 6 in the determinant Δ = 4 5 6
7 8 9
Solution Since 6 lies in the second row and third column, its minor M23 is given by
M23 =
1 2
= 8 – 14 = – 6 (obtained by deleting R2 and C3 in Δ)
...
Example 20 Find minors and cofactors of all the elements of the determinant
Solution Minor of the element aij is Mij
Here a11 = 1
...
So
A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3
A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4
he
A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2
A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1
a12
a22
a32
a13
a23
a33
bl
a11
Δ = a21
a31
is
Example 21 Find minors and cofactors of the elements a11, a21 in the determinant
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Solution By definition of minors and cofactors, we have
Minor of a11 = M11 =
a22
a32
a23
= a22 a33– a23 a32
a33
Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32
Minor of a21 = M21 =
a12
a32
a13
= a12 a33 – a13 a32
a33
Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32
Remark Expanding the determinant Δ, in Example 21, along R1, we have
a21 a22
a22 a23
a21 a23
1+1
1+2
1+3
Δ = (–1) a11 a
a + (–1) a12 a
a + (–1) a13 a31 a32
32
33
31
33
= a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij
= sum of product of elements of R1 with their corresponding cofactors
Similarly, Δ can be calculated by other five ways of expansion that is along R2, R3,
C1, C2 and C3
...
Note If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero
...
he
a11
Example 22 Find minors and cofactors of the elements of the determinant
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2 –3 5
6 0 4 and verify that a11 A31 + a12 A32 + a13 A33= 0
1 5 –7
Solution We have M11 =
0 4
= 0 –20 = –20; A11 = (–1)1+1 (–20) = –20
5 –7
M12 =
6 4
= – 42 – 4 = – 46;
1 –7
A12 = (–1)1+2 (– 46) = 46
M13 =
6
1
A13 = (–1)1+3 (30) = 30
M21 =
–3 5
= 21 – 25 = – 4;
5 –7
A21 = (–1)2+1 (– 4) = 4
M22 =
2 5
= –14 – 5 = –19;
1 –7
A22 = (–1)2+2 (–19) = –19
M23 =
2
1
–3
= 10 + 3 = 13;
5
A23 = (–1)2+3 (13) = –13
M31 =
–3 5
= –12 – 0 = –12;
0 4
A31 = (–1)3+1 (–12) = –12
0
= 30 – 0 = 30;
5
MATHEMATICS
M32 =
2
6
5
= 8 – 30 = –22;
4
and
M33 =
2
6
Now
a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18
So
a11 A31 + a12 A32 + a13 A33
–3
= 0 + 18 = 18;
0
A32 = (–1)3+2 (–22) = 22
A33 = (–1)3+3 (18) = 18
EXERCISE 4
...
(i)
bl
Write Minors and Cofactors of the elements of following determinants:
1 0 0
2
...
Using Cofactors of elements of second row, evaluate Δ = 2 0 1
...
Using Cofactors of elements of third column, evaluate Δ = 1 y
1 z
a11
5
...
xy
a13
a23 and Aij is Cofactors of aij , then value of Δ is given by
a33
(A) a11 A31+ a12 A32 + a13 A33
(C) a21 A11+ a22 A12 + a23 A13
(B) a11 A11+ a12 A21 + a13 A31
(D) a11 A11+ a21 A21 + a31 A31
4
...
In this section, we shall
discuss the condition for existence of inverse of a matrix
...
e
...
DETERMINANTS
127
Then
a22
a32
a13 ⎤
a23 ⎥
⎥
a33 ⎥
⎦
⎡ A11
adj A = Transpose of ⎢ A 21
⎢
⎢ A 31
⎣
A12
A 22
A 32
A13 ⎤ ⎡ A11
A 23 ⎥ = ⎢ A12
⎥ ⎢
A 33 ⎥ ⎢ A13
⎦ ⎣
A 21
A 22
A 23
A 31 ⎤
A32 ⎥
⎥
A 33 ⎥
⎦
is
Let
a12
bl
⎡ a11
A = ⎢ a21
⎢
⎢ a31
⎣
he
4
...
1 Adjoint of a matrix
Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of
the matrix [Aij]n × n, where Aij is the cofactor of the element aij
...
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⎡ 2 3⎤
Example 23 Find adj A for A = ⎢
⎥
⎣1 4⎦
Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2
⎡ A11 A 21 ⎤ ⎡ 4 –3⎤
adj A = ⎢
⎥ =⎢
⎥
⎣ A12 A 22 ⎦ ⎣ –1 2 ⎦
Remark For a square matrix of order 2, given by
Hence
⎡ a11 a12 ⎤
A= ⎢
⎥
⎣ a21 a22 ⎦
The adj A can also be obtained by interchanging a11 and a22 and by changing signs
of a12 and a21, i
...
,
We state the following theorem without proof
...
For example, the determinant of matrix A =
1 2
is zero
4 8
Hence A is a singular matrix
...
Then A = 3 4 = 4 – 6 = – 2 ≠ 0
...
Let
Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA
are also nonsingular matrices of the same order
...
e
...
e
...
e
...
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Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix
...
Then, there exists a square matrix B of order n such that AB = BA = I
Now
This gives
AB = I
...
Hence A is nonsingular
...
Then A ≠ 0
Now
A (adj A) = (adj A) A = A I
or
⎛ 1
⎞ ⎛ 1
⎞
adj A ⎟ = ⎜
adj A ⎟ A = I
A⎜
|A|
|A|
⎝
⎠ ⎝
⎠
or
AB = BA = I, where B =
Thus
A is invertible and A–1 =
(Theorem 1)
1
adj A
|A|
1
adj A
|A|
1 3 3
Example 24 If A = 1 4 3 , then verify that A adj A = | A | I
...
1 3 4
Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
130
MATHEMATICS
Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0,
A33 = 1
Therefore
⎡1 3 3 ⎤ ⎡ 7 −3 −3⎤
⎢
⎥⎢
⎥
A (adj A) = ⎢1 4 3 ⎥ ⎢ −1 1 0 ⎥
⎢1 3 4 ⎥ ⎢ −1 0 1 ⎥
⎣
⎦⎣
⎦
is
Now
he
⎡ 7 −3 −3⎤
⎢
⎥
adj A = ⎢ −1 1 0 ⎥
⎢ −1 0 1 ⎥
⎣
⎦
bl
⎡ 7 − 3 − 3 −3 + 3 + 0 −3 + 0 + 3 ⎤
⎢
⎥
= ⎢ 7 − 4 − 3 −3 + 4 + 0 − 3 + 0 + 3 ⎥
⎢ 7 − 3 − 4 −3 + 3 + 0 −3 + 0 + 4 ⎥
⎣
⎦
⎡1 0 0 ⎤
⎢0 1 0 ⎥
⎢
⎥ = A
...
⎣1 − 4 ⎦
⎣
⎦
3⎤
⎡2
Solution We have AB = ⎢
⎥
⎣1 − 4 ⎦
Since,
⎡ 1 −2 ⎤ ⎡ − 1 5 ⎤
⎢ −1 3 ⎥ = ⎢ 5 −14 ⎥
⎣
⎦ ⎣
⎦
AB = –11 ≠ 0, (AB)–1 exists and is given by
(AB)–1 =
1
1 ⎡ −14 −5⎤ = 1 ⎡14 5⎤
adj (AB) = − ⎢
⎢
⎥
AB
11 ⎣ −5 −1⎥ 11 ⎣ 5 1 ⎦
⎦
Further, A = –11 ≠ 0 and B = 1 ≠ 0
...
Using this equation, find A–1
...
A = ⎢
⎥⎢
⎥ =⎢
⎥
⎣1 2⎦ ⎣1 2 ⎦ ⎣ 4 7 ⎦
he
Example 26 Show that the matrix A =
⎡ 7 12 ⎤ ⎡8 12 ⎤ ⎡1 0 ⎤ ⎡0 0 ⎤
A 2 − 4A + I = ⎢
⎥− ⎢
⎥+⎢
⎥ =⎢
⎥=O
⎣ 4 7 ⎦ ⎣ 4 8 ⎦ ⎣0 1 ⎦ ⎣0 0 ⎦
Now
Therefore
or
or
or
A2 – 4A + I = O
A A – 4A = – I
A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0)
A (A A–1) – 4I = – A–1
AI – 4I = – A–1
or
⎡ 4 0 ⎤ ⎡2 3⎤
⎡ 2 −3 ⎤
A–1 = 4I – A = ⎢
⎥ − ⎢ 1 2 ⎥ = ⎢ −1 2 ⎥
⎣0 4⎦ ⎣
⎦
⎣
⎦
Hence
⎡ 2 −3 ⎤
A −1 = ⎢
⎥
⎣ −1 2 ⎦
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Hence
EXERCISE 4
...
1
...
1
2
2
1 2
3 5
0 1
Verify A (adj A) = (adj A) A = | A | I in Exercises 3 and 4
3
...
1
3
1
1
0
0
2
2
3
132
MATHEMATICS
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11
...
1 0
3 3
5 2
6
...
0
0
1
1 5
3 2
2
4
7
1 3
1 0
2 1
7
...
1 2 3
0 2 4
0 0 5
1
0
3
1
2
2
2
3
4
he
5
...
⎢
⎥
⎢ 0 sin α − cos α ⎥
⎣
⎦
3 7
6 8
12
...
Verify that (AB)–1 = B–1 A–1
...
If A =
3 1
, show that A2 – 5A + 7I = O
...
1 2
14
...
1 1
1
15
...
Hence, find A–1
...
If A =
2
1
1
1
2
1
1
1
2
Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
17
...
Then | adj A | is equal to
(C) | A | 3
(D) 3 | A |
(A) | A |
(B) | A | 2
–1
18
...
7 Applications of Determinants and Matrices
he
In this section, we shall discuss application of determinants and matrices for solving the
system of linear equations in two or three variables and for checking the consistency of
the system of linear equations
...
Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist
...
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4
...
1 Solution of system of linear equations using inverse of a matrix
Let us express the system of linear equations as matrix equations and solve them using
inverse of the coefficient matrix
...
e
...
Now
AX = B
(premultiplying by A–1)
or
A (AX) = A–1 B
(by associative property)
or
(A–1A) X = A–1 B
–1
or
IX=A B
or
X = A–1 B
This matrix equation provides unique solution for the given system of equations as
inverse of a matrix is unique
...
–1
134
MATHEMATICS
he
Case II If A is a singular matrix, then | A | = 0
...
If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent
...
Example 27 Solve the system of equations
is
2x + 5y = 1
3x + 2y = 7
Solution The system of equations can be written in the form AX = B, where
bl
⎡2 5⎤
⎡ x⎤
⎡1 ⎤
A= ⎢
⎥ , X = ⎢ y ⎥ and B = ⎢ 7 ⎥
⎣ 3 2⎦
⎣ ⎦
⎣ ⎦
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Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution
...
e
...
3x – 2y + 3z = 8
2x + y – z = 1
4x – 3y + 2z = 4
Solution The system of equations can be written in the form AX = B, where
⎡ 3 −2 3 ⎤
⎡ x⎤
⎢ 2 1 −1⎥ , X = ⎢ y ⎥ and B =
A= ⎢
⎥
⎢ ⎥
⎢ 4 −3 2 ⎥
⎢z⎥
⎣
⎦
⎣ ⎦
⎡8 ⎤
⎢1 ⎥
⎢ ⎥
⎢ 4⎥
⎣ ⎦
We see that
A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
DETERMINANTS
Hence, A is nonsingular and so its inverse exists
...
e
...
Example 29 The sum of three numbers is 6
...
By adding first and third numbers, we get double of the
second number
...
Solution Let first, second and third numbers be denoted by x, y and z, respectively
...
Now we find adj A
A12 = – (0 – 3) = 3,
A22 = 0,
A32 = – (3 – 0) = – 3,
A13 = – 1
A23 = – (– 2 – 1) = 3
A33 = (1 – 0) = 1
Thus
Since
⎡ 7 –3 2 ⎤
⎢
⎥
adj A = ⎢ 3 0 –3⎥
⎢ –1 3
1⎥
⎣
⎦
⎡ 7 –3 2 ⎤
1
1⎢
⎥
A –1 =
adj (A) = ⎢ 3 0 –3⎥
A
9
⎢ –1 3 1 ⎥
⎣
⎦
–1
X=A B
bl
⎡ 7 –3 2 ⎤ ⎡ 6 ⎤
1⎢
⎥⎢ ⎥
X = ⎢ 3 0 –3⎥ ⎢11⎥
9
⎢ –1 3 1 ⎥ ⎢ 0 ⎥
⎣
⎦⎣ ⎦
he
Hence
MATHEMATICS
is
136
9
1
18 = 2
27
3
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x
⎡ 42 − 33 + 0 ⎤
1 ⎢ 18 + 0 + 0 ⎥ 1
y =
⎢
⎥ =
9 ⎢ −6 + 33 + 0 ⎥ 9
z
⎣
⎦
x = 1, y = 2, z = 3
or
Thus
EXERCISE 4
...
1
...
2x – y = 5
3
...
x + y + z = 1
5
...
5x – y + 4z = 5
2x + 3y + 2z = 2
2y – z = –1
2x + 3y + 5z = 2
ax + ay + 2az = 4
3x – 5y = 3
5x – 2y + 6z = –1
Solve system of linear equations, using matrix method, in Exercises 7 to 14
...
5x + 2y = 4
8
...
4x – 3y = 3
7x + 3y = 5
3x + 4y = 3
3x – 5y = 7
10
...
2x + y + z = 1
12
...
2x + 3y +3 z = 5
14
...
If A = ⎢
⎥ , find A
...
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60
...
The cost of 6 kg onion 2 kg wheat and 3 kg
rice is Rs 70
...
Example 30 If a, b, c are positive and unequal, show that value of the determinant
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a b c
Δ = b c a is negative
...
P, find value of
2y + 4 5y + 7 8y + a
3y + 5 6 y + 8 9 y + b
he
4 y + 6 7 y + 9 10 y + c
Solution Applying R1 → R1 + R3 – 2R2 to the given determinant, we obtain
(Since 2b = a + c)
is
0
0
0
3y + 5 6 y + 8 9 y + b = 0
4 y + 6 7 y + 9 10 y + c
( y+ z )
xy
zx
( x+ z )
xy
2
yz
= 2xyz (x + y + z)3
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Δ=
2
bl
Example 32 Show that
xz
( x+ y )
yz
2
Solution Applying R1 → xR1, R2 → yR2 , R3 → z R3 to Δ and dividing by xyz, we get
x y z
Δ=
1
xyz
xy 2
2
x2 y
y x z
xz 2
yz 2
x2 z
2
y2 z
z x y
2
Taking common factors x, y, z from C1 C2 and C3, respectively, we get
( y + z)
Δ=
xyz
xyz
2
x2
y2
( x + z)
z2
x2
z2
2
y2
( x + y)
2
Applying C2 → C2– C1, C3 → C3– C1, we have
( y + z )2
Δ=
x2 – ( y + z )
y2
( x + z )2 − y 2
0
z2
0
( x + y )2 – z 2
2
x2 − ( y + z )
2
DETERMINANTS
139
Taking common factor (x + y + z) from C2 and C3, we have
(y + z)
Δ = (x + y + z)
2
y
x – ( y + z)
( x + z) – y
2
2
z2
0
x – ( y + z)
0
( x + y) – z
1
C ) and C3
y 1
–2y
0
x+ y –z
is
–2z
x− y+z
0
1
C1 , we get
z
C3
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Applying C2 → (C2 +
2 yz
y2
z2
bl
Δ = (x + y + z)
2
he
Applying R1 → R1 – (R2 + R3), we have
2 yz
y2
x z
z2
Δ = (x + y + z)2
0
z2
y
0
y2
z
x y
Finally expanding along R1, we have
Δ = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x2 + xy + xz)
= (x + y + z)3 (2xyz)
1 –1 2
Example 33 Use product 0 2 –3
3 –2 4
–2 0 1
9 2 –3 to solve the system of equations
6 1 –2
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
⎡1
⎢
Solution Consider the product ⎢ 0
⎢3
⎣
–1
2
–2
2⎤
–3⎥
⎥
4⎥
⎦
⎡ –2
⎢9
⎢
⎢6
⎣
0
2
1
1 ⎤
–3 ⎥
⎥
–2 ⎥
⎦
140
MATHEMATICS
1 0 0
⎡ − 2 − 9 + 12 0 − 2 + 2 1 + 3 − 4⎤
⎢ 0 + 18 − 18 0 + 4 − 3 0 − 6 + 6⎥
= ⎢
⎥ = 0 1 0
⎢ − 6 − 18 + 24 0 − 4 + 4 3 + 6 − 8⎥
0 0 1
⎣
⎦
–1
he
1 –1 2
–2 0 1
Hence
0 2 –3
9 2 –3
3 –2 4
6 1 –2
Now, given system of equations can be written, in matrix form, as follows
bl
is
⎡1 –1 2 ⎤ ⎡ x ⎤ ⎡1 ⎤
⎢ 0 2 –3⎥ ⎢ y ⎥ ⎢ ⎥
⎢
⎥ ⎢ ⎥ = ⎢1 ⎥
⎢ 3 –2 4 ⎦ ⎢ z ⎥ ⎢ 2⎥
⎥ ⎣ ⎦ ⎣ ⎦
⎣
−1
x
–2 0 1
⎡1 −1 2 ⎤ ⎡ 1 ⎤
⎢0
⎥ ⎢1 ⎥
y =
2 −3⎥ ⎢ ⎥ = 9 2 –3
⎢
⎢ 3 −2 4 ⎥ ⎢ 2 ⎥
z
6 1 –2
⎣
⎦ ⎣ ⎦
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or
⎡ −2 + 0 + 2 ⎤ ⎡0 ⎤
⎢
⎥ ⎢ ⎥
= ⎢ 9 + 2 − 6 ⎥ = ⎢5 ⎥
⎢ 6 + 1 − 4 ⎦ ⎢ 3⎥
⎥ ⎣ ⎦
⎣
x = 0, y = 5 and z = 3
Hence
Example 34 Prove that
a + bx c + dx p + qx
a c p
2
Δ = ax + b cx + d px + q = (1 − x ) b d q
u
v
w
u v w
Solution Applying R1 → R1 – x R2 to Δ, we get
Δ=
a (1 − x 2 ) c (1 − x 2 )
ax + b
cx + d
u
v
a
c
= (1 − x ) ax + b cx + d
u
v
2
p (1 − x 2 )
px + q
w
p
px + q
w
1
1
2
DETERMINANTS
141
Applying R2 → R2 – x R1, we get
a c p
Δ = (1 − x ) b d q
u v w
Miscellaneous Exercises on Chapter 4
he
2
sin θ cos θ
1 is independent of θ
...
Prove that the determinant – sin θ – x
cos θ
1
x
bc
ca
ab
1 a2
1 b2
1 c2
a3
b3
...
Without expanding the determinant, prove that b b 2
c c2
is
x
cos α cos β cos α sin β – sin α
– sin β
cos β
0
...
Evaluate
sin α cos β sin α sin β cos α
4
...
x+a
5
...
Prove that a ab
2
ab
b bc
c2
–1 1 ⎤
⎡ 3
⎡ 1 2 –2 ⎤
⎢ –15 6 –5⎥ and B = ⎢ –1 3 0 ⎥ , find AB –1
( )
7
...
Let A = –2 3 1
...
Evaluate 1 x + y
1
x
he
x+ y
x
y
y
x+ y
x
y
y
x+ y
bl
x
9
...
α α2
β β2
γ γ2
β+γ
γ + α = (β – γ) (γ – α) (α – β) (α + β + γ)
α +β
12
...
3a
– a+ b – a+ c
–b+ a
3b
– b + c = 3(a + b + c) (ab + bc + ca)
– c + a – c+ b
3c
14
...
z 2 1 pz 3
16
...
sin α cos α cos ( α + δ )
sin β cos β cos ( β + δ ) = 0
sin γ cos γ cos ( γ + δ )
DETERMINANTS
6
x
5
z
1
9 20
–
y
z
2
he
4 6
–
x y
143
Choose the correct answer in Exercise 17 to 19
...
If a, b, c, are in A
...
If x, y, z are nonzero real numbers, then the inverse of matrix A = ⎢ 0 y 0 ⎥ is
⎢
⎥
⎢0 0 z ⎥
⎣
⎦
⎡ x −1
⎢
(A) ⎢ 0
⎢
⎣ 0
0
y −1
0
0 ⎤
⎥
0 ⎥
⎥
z −1 ⎦
⎡ x 0 0⎤
1 ⎢
0 y 0⎥
(C)
⎥
xyz ⎢
⎢0 0 z ⎥
⎣
⎦
⎡ x −1
⎢
(B) xyz ⎢ 0
⎢
⎣ 0
0
y −1
0
0 ⎤
⎥
0 ⎥
⎥
z −1 ⎦
⎡1 0 0 ⎤
1 ⎢
0 1 0⎥
(D)
⎥
xyz ⎢
⎢0 0 1 ⎥
⎣
⎦
sin θ 1 ⎤
⎡ 1
⎢ − sin θ
1 sin θ ⎥ , where 0 ≤ θ ≤ 2π
...
Let A = ⎢
⎥
⎢ −1 − sin θ 1 ⎥
⎣
⎦
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4]
144
MATHEMATICS
Summary
Determinant of a matrix A = [a11]1 × 1 is given by | a11| = a11
Determinant of a matrix A
b1
b2
b3
he
is given by
a12
= a11 a22 – a12 a21
a22
a1
a2
a3
c1
b
c2 = a1 2
b3
c3
b1
b2
b3
c2
c1
c2 is given by (expanding along R1)
c3
a2
c2
a2
b2
a3
b3
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a1
A = a2
a3
a11
a21
a12
a22
is
A =
a11
a21
bl
Determinant of a matrix A
c3
− b1
a3
c3
+ c1
For any square matrix A, the |A| satisfy following properties
...
If we interchange any two rows (or columns), then sign of determinant
changes
...
If we multiply each element of a row or a column of a determinant by constant
k, then value of determinant is multiplied by k
...
If A = [aij ]3×3 , then k
...
If to each element of a row or a column of a determinant the equimultiples of
corresponding elements of other rows or columns are added, then value of
determinant remains same
...
is
he
Minor of an element aij of the determinant of matrix A is the determinant
obtained by deleting ith row and jth column and denoted by Mij
...
For example,
bl
If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero
...
A square matrix A is said to be singular or non-singular according as
| A | = 0 or | A | ≠ 0
...
Also A–1 = B or B–1 = A and hence (A–1)–1 = A
...
A –1 =
1
(adj A)
A
If
a 1 x + b1 y + c 1 z = d1
a2 x + b2 y + c 2 z = d2
a3 x + b3 y + c3 z = d3,
then these equations can be written as A X = B, where
⎡ a1
A = ⎢ a2
⎢
⎢ a3
⎣
b1
b2
b3
c1 ⎤
⎡ x⎤
⎡ d1 ⎤
⎥ , X = ⎢ y ⎥ and B= ⎢ d ⎥
c2 ⎥
⎢ ⎥
⎢ 2⎥
⎢ d3 ⎥
⎢ ⎦
c3 ⎥
⎣z⎥
⎦
⎣ ⎦
146
MATHEMATICS
is
Historical Note
he
Unique solution of equation AX = B is given by X = A–1 B, where A ≠ 0
...
For a square matrix A in matrix equation AX = B
(i) | A | ≠ 0, there exists unique solution
(ii) | A | = 0 and (adj A) B ≠ 0, then there exists no solution
(iii) | A | = 0 and (adj A) B = 0, then system may or may not be consistent
...
The arrangement of rods was precisely
that of the numbers in a determinant
...
Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth
century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of
determinants and of their expansion
...
‘T
...
of the Tokyo Math
...
, V
...
He may be called the formal founder
...
In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations
...
The next great contributor was Jacques - Philippe - Marie Binet, (1812) who
stated the theorem relating to the product of two matrices of m-columns and nrows, which for the special case of m = n reduces to the multiplication theorem
...
He
used the word ‘determinant’ in its present sense
...
The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance