v
All Mathematical truths are relative and conditional. — C.P. STEINMETZ
v
4.1 Introduction
In the previous chapter, we have studied about matrices
and algebra of matrices. We have also learnt that a system
of algebraic equations can be expressed in the form of
matrices. This means, a system of linear equations like
a
1
x + b
1
y = c
1
a
2
x + b
2
y = c
2
can be represented as
1 1 1
2 2 2
a b c
x
a b c
y
=
. Now, this
system of equations has a unique solution or not, is
determined by the number a
1
b
2
– a
2
b
1
. (Recall that if
1 1
2 2
a b
a b
≠
or, a
1
b
2
– a
2
b
1
≠ 0, then the system of linear
equations has a unique solution). The number a
1
b
2
– a
2
b
1
which determines uniqueness of solution is associated with the matrix
1 1
2 2
A
a b
a b
=
and is called the determinant of A or det A. Determinants have wide applications in
Engineering, Science, Economics, Social Science, etc.
In this chapter, we shall study determinants up to order three only with real entries.
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.
4.2 Determinant
To every square matrix A = [a
ij
] of order n, we can associate a number (real or
complex) called determinant of the square matrix A, where a
ij
= (i, j)
th
element of A.
Chapter
4
DETERMINANTS
P.S. Laplace
(1749-1827)
2019-20
104 MATHEMATICS
This may be thought of as a function which associates each square matrix with a
unique number (real or complex). 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. It is also denoted by |A| or det A or ∆.
If A =
a b
c d
, then determinant of A is written as |A| =
a b
c d
= det (A)
Remarks
(i) For matrix A, |A| is read as determinant of A and not modulus of A.
(ii) Only square matrices have determinants.
4.2.1 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
4.2.2 Determinant of a matrix of order two
Let A =
11 12
21 22
a a
a a
be a matrix of order 2 × 2,
then the determinant of A is defined as:
det (A) = |A| = ∆ = = a
11
a
22
– a
21
a
12
Example 1 Evaluate
2 4
–1 2
.
Solution We have
2 4
–1 2
= 2(2) – 4(–1) = 4 + 4 = 8.
Example 2 Evaluate
1
– 1
x x
x x
+
Solution We have
1
– 1
x x
x x
+
= x (x) – (x + 1) (x – 1) = x
2
– (x
2
– 1) = x
2
– x
2
+ 1 = 1
4.2.3 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. This is known as expansion of a determinant along
a row (or a column). There are six ways of expanding a determinant of order
2019-20
DETERMINANTS 105
3 corresponding to each of three rows (R
1
, R
2
and R
3
) and three columns (C
1
, C
2
and
C
3
) giving the same value as shown below.
Consider the determinant of square matrix A = [a
ij
]
3 × 3
i.e., | A | =
21 22 23
31 32 33
a a a
a a a
11 12 13
a a a
Expansion along first Row (R
1
)
Step 1 Multiply first element a
11
of R
1
by (–1)
(1 + 1)
[(–1)
sum of suffixes in a
11
] and with the
second order determinant obtained by deleting the elements of first row (R
1
) and first
column (C
1
) of | A | as a
11
lies in R
1
and C
1
,
i.e., (–1)
1 + 1
a
11
22 23
32 33
a a
a a
Step 2 Multiply 2nd element a
12
of R
1
by (–1)
1 + 2
[(–1)
sum of suffixes in a
12
] and the second
order determinant obtained by deleting elements of first row (R
1
) and 2nd column (C
2
)
of | A | as a
12
lies in R
1
and C
2
,
i.e., (–1)
1 + 2
a
12
21 23
31 33
a a
a a
Step 3 Multiply third element a
13
of R
1
by (–1)
1 + 3
[(–1)
sum of suffixes in a
13
] and the second
order determinant obtained by deleting elements of first row (R
1
) and third column (C
3
)
of | A | as a
13
lies in R
1
and C
3
,
i.e., (–1)
1 + 3
a
13
21 22
31 32
a a
a a
Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three
terms obtained in steps 1, 2 and 3 above is given by
det A = |A| = (–1)
1 + 1
a
11
22 23 21 23
1 2
12
32 33 31 33
(–1)
a a a a
a
a a a a
+
+
+
21 22
1 3
13
31 32
(–1)
a a
a
a a
+
or |A| = a
11
(a
22
a
33
– a
32
a
23
) – a
12
(a
21
a
33
– a
31
a
23
)
+ a
13
(a
21
a
32
– a
31
a
22
)
2019-20
106 MATHEMATICS
= a
11
a
22
a
33
– a
11
a
32
a
23
– a
12
a
21
a
33
+ a
12
a
31
a
23
+ a
13
a
21
a
32
– a
13
a
31
a
22
... (1)
A
Note We shall apply all four steps together.
Expansion along second row (R
2
)
| A | =
11 12 13
31 32 33
a a a
a a a
21 22 23
a a a
Expanding along R
2
,
we get
| A | =
12 13 11 13
2 1 2 2
21 22
32 33 31 33
(–1) (–1)
a a a a
a a
a a a a
+ +
+
11 12
2 3
23
31 32
(–1)
a a
a
a a
+
+
= – a
21
(a
12
a
33
– a
32
a
13
) + a
22
(a
11
a
33
– a
31
a
13
)
– a
23
(a
11
a
32
– a
31
a
12
)
| A | = – a
21
a
12
a
33
+ a
21
a
32
a
13
+ a
22
a
11
a
33
– a
22
a
31
a
13
– a
23
a
11
a
32
+ a
23
a
31
a
12
= a
11
a
22
a
33
– a
11
a
23
a
32
– a
12
a
21
a
33
+ a
12
a
23
a
31
+ a
13
a
21
a
32
– a
13
a
31
a
22
... (2)
Expansion along first Column (C
1
)
| A | =
12 13
22 23
32 33
11
21
31
a
a
a
a a
a a
a a
By expanding along C
1
, we get
| A | =
22 23 12 13
1 1 2 1
11 21
32 33 32 33
(–1) ( 1)
a a a a
a a
a a a a
+ +
+ −
+
12 13
3 1
31
22 23
(–1)
a a
a
a a
+
= a
11
(a
22
a
33
– a
23
a
32
) – a
21
(a
12
a
33
– a
13
a
32
) + a
31
(a
12
a
23
– a
13
a
22
)
2019-20
DETERMINANTS 107
|
A |
= a
11
a
22
a
33
– a
11
a
23
a
32
– a
21
a
12
a
33
+ a
21
a
13
a
32
+ a
31
a
12
a
23
– a
31
a
13
a
22
= a
11
a
22
a
33
– a
11
a
23
a
32
– a
12
a
21
a
33
+ a
12
a
23
a
31
+ a
13
a
21
a
32
– a
13
a
31
a
22
... (3)
Clearly, values of
|A| in (1), (2) and (3) are equal. It is left as an exercise to the
reader to verify that the values of |A| by expanding along R
3
, C
2
and C
3
are equal to the
value of |A| obtained in (1), (2) or (3).
Hence, expanding a determinant along any row or column gives same value.
Remarks
(i) For easier calculations, we shall expand the determinant along that row or column
which contains maximum number of zeros.
(ii) While expanding, instead of multiplying by (–1)
i + j
, we can multiply by +1 or –1
according as (i + j) is even or odd.
(iii) Let A =
2 2
4 0
and B =
1 1
2 0
. Then, it is easy to verify that A = 2B. Also
|A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2.
Observe that, |A| = 4(– 2) = 2
2
|B| or |A| = 2
n
|B|, where n = 2 is the order of
square matrices A and B.
In general, if A = kB where A and B are square matrices of order n, then | A| = k
n
| B |, where n = 1, 2, 3
Example 3 Evaluate the determinant ∆ =
1 2 4
–1 3 0
4 1 0
.
Solution Note that in the third column, two entries are zero. So expanding along third
column (C
3
), we get
∆ =
–1 3 1 2 1 2
4 – 0 0
4 1 4 1 –1 3
+
= 4 (–1 – 12) – 0 + 0 = – 52
Example 4 Evaluate ∆ =
0 sin – cos
– sin 0 sin
cos – sin 0
α α
α β
α β
.
2019-20
108 MATHEMATICS
Solution Expanding along R
1
, we get
∆ =
0 sin – sin sin –sin 0
0 – sin – cos
– sin 0 cos 0 cos – sin
β α β α
α α
β α α β
= 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0)
= sin α sin β cos α – cos α sin α sin β = 0
Example 5 Find values of x for which
3 3 2
1 4 1
x
x
=
.
Solution We have
3 3 2
1 4 1
x
x
=
i.e. 3 – x
2
= 3 – 8
i.e. x
2
= 8
Hence x =
2 2
±
EXERCISE 4.1
Evaluate the determinants in Exercises 1 and 2.
1.
2 4
–5 –1
2. (i)
cos – sin
sin cos
θ θ
θ θ
(ii)
2
– 1 – 1
1 1
x x x
x x
+
+ +
3. If A =
1 2
4 2
, then show that | 2A | = 4 | A |
4. If A =
1 0 1
0 1 2
0 0 4
, then show that | 3 A | = 27 | A |
5. Evaluate the determinants
(i)
3 –1 –2
0 0 –1
3 –5 0
(ii)
3 – 4 5
1 1 –2
2 3 1
2019-20
DETERMINANTS 109
(iii)
0 1 2
–1 0 –3
–2 3 0
(iv)
2 –1 –2
0 2 –1
3 –5 0
6. If A =
1 1 –2
2 1 –3
5 4 –9
, find | A |
7. Find values of x, if
(i)
2 4 2 4
5 1 6
x
x
=
(ii)
2 3 3
4 5 2 5
x
x
=
8. If
2 6 2
18 18 6
x
x
=
, then x is equal to
(A) 6 (B) ± 6 (C) – 6 (D) 0
4.3 Properties of Determinants
In the previous section, we have learnt how to expand the determinants. 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. These properties are true for
determinants of any order. However, we shall restrict ourselves upto determinants of
order 3 only.
Property 1 The value of the determinant remains unchanged if its rows and columns
are interchanged.
Verification Let ∆ =
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
Expanding along first row, we get
∆ =
2 3 1 3
1 2
1 2 3
2 3 1 3
1 2
b b b b
b b
a a a
c c c c
c c
− +
= a
1
(b
2
c
3
– b
3
c
2
) – a
2
(b
1
c
3
– b
3
c
1
) + a
3
(b
1
c
2
– b
2
c
1
)
By interchanging the rows and columns of ∆, we get the determinant
∆
1
=
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
2019-20
110 MATHEMATICS
Expanding ∆
1
along first column, we get
∆
1
= a
1
(b
2
c
3
– c
2
b
3
) – a
2
(b
1
c
3
– b
3
c
1
) + a
3
(b
1
c
2
– b
2
c
1
)
Hence ∆ = ∆
1
Remark It follows from above property that if A is a square matrix, then
det (A) = det (A′), where A′ = transpose of A.
A
Note If R
i
= ith row and
C
i
= ith column, then for interchange of row and
columns, we will symbolically write C
i
↔ R
i
Let us verify the above property by example.
Example 6 Verify Property 1 for ∆ =
2 –3 5
6 0 4
1 5 –7
Solution Expanding the determinant along first row, we have
∆ =
0 4 6 4 6 0
2 – (–3) 5
5 –7 1 –7 1 5
+
= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
By interchanging rows and columns, we get
∆
1
=
2 6 1
–3 0 5
5 4 –7
(Expanding along first column)
=
0 5 6 1 6 1
2 – (–3) 5
4 –7 4 –7 0 5
+
= 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0)
= – 40 – 138 + 150 = – 28
Clearly ∆ = ∆
1
Hence, Property 1 is verified.
Property 2 If any two rows (or columns) of a determinant are interchanged, then sign
of determinant changes.
Verification Let ∆ =
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
2019-20
DETERMINANTS 111
Expanding along first row, we get
∆ = a
1
(b
2
c
3
– b
3
c
2
) – a
2
(b
1
c
3
– b
3
c
1
) +
a
3
(b
1
c
2
– b
2
c
1
)
Interchanging first and third rows, the new determinant obtained is given by
∆
1
=
1 2 3
1 2 3
1 2 3
c c c
b b b
a a a
Expanding along third row, we get
∆
1
= a
1
(c
2
b
3
– b
2
c
3
) – a
2
(c
1
b
3
– c
3
b
1
) + a
3
(b
2
c
1
– b
1
c
2
)
= – [a
1
(b
2
c
3
– b
3
c
2
) – a
2
(b
1
c
3
– b
3
c
1
) + a
3
(b
1
c
2
– b
2
c
1
)]
Clearly ∆
1
= – ∆
Similarly, we can verify the result by interchanging any two columns.
A
Note We can denote the interchange of rows by R
i
↔ R
j
and interchange of
columns by C
i
↔ C
j
.
Example 7 Verify Property 2 for ∆ =
2 –3 5
6 0 4
1 5 –7
.
Solution ∆ =
2 –3 5
6 0 4
1 5 –7
= – 28 (See Example 6)
Interchanging rows R
2
and R
3
i.e., R
2
↔ R
3
, we have
∆
1
=
2 –3 5
1 5 –7
6 0 4
Expanding the determinant ∆
1
along first row, we have
∆
1
=
5 –7 1 –7 1 5
2 – (–3) 5
0 4 6 4 6 0
+
= 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30)
= 40 + 138 – 150 = 28
2019-20
112 MATHEMATICS
Clearly ∆
1
= – ∆
Hence, Property 2 is verified.
Property 3 If any two rows (or columns) of a determinant are identical (all corresponding
elements are same), then value of determinant is zero.
Proof If we interchange the identical rows (or columns) of the determinant ∆, then ∆
does not change. However, by Property 2, it follows that ∆ has changed its sign
Therefore ∆ = – ∆
or ∆ = 0
Let us verify the above property by an example.
Example 8 Evaluate ∆ =
3 2 3
2 2 3
3 2 3
Solution Expanding along first row, we get
∆ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0
Here R
1
and R
3
are identical.
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.
Verification Let ∆ =
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
and ∆
1
be the determinant obtained by multiplying the elements of the first row by k.
Then
∆
1
=
1 1 1
2 2 2
3 3 3
k a k b k c
a b c
a b c
Expanding along first row, we get
∆
1
= k a
1
(b
2
c
3
– b
3
c
2
) – k b
1
(a
2
c
3
– c
2
a
3
) + k c
1
(a
2
b
3
– b
2
a
3
)
= k [a
1
(b
2
c
3
– b
3
c
2
) – b
1
(a
2
c
3
– c
2
a
3
) + c
1
(a
2
b
3
– b
2
a
3
)]
= k ∆
2019-20
DETERMINANTS 113
Hence
1 1 1
2 2 2
3 3 3
k a k b k c
a b c
a b c
= k
1 1 1
2 2 2
3 3 3
a b c
a b c
a b c
Remarks
(i) By this property, we can take out any common factor from any one row or any
one column of a given determinant.
(ii) If corresponding elements of any two rows (or columns) of a determinant are
proportional (in the same ratio), then its value is zero. For example
∆ =
1 2 3
1 2 3
1 2 3
a a a
b b b
k a k a k a
= 0 (rows R
1
and R
2
are proportional)
Example 9 Evaluate
102 18 36
1 3 4
17 3 6
Solution Note that
6(17) 6(3) 6(6) 17 3 6102 18 36
1 3 4 1 3 4 6 1 3 4 0
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.
For example,
1 1 2 2 3 3
1 2 3
1 2 3
a a a
b b b
c c c
+ λ + λ + λ
=
1 2 3 1 2 3
1 2 3 1 2 3
1 2 3 1 2 3
a a a
b b b b b b
c c c c c c
λ λ λ
+
Verification L.H.S. =
1 1 2 2 3 3
1 2 3
1 2 3
a a a
b b b
c c c
+ λ + λ + λ
2019-20
114 MATHEMATICS
Expanding the determinants along the first row, we get
∆ = (a
1
+ λ
1
) (b
2
c
3
– c
2
b
3
) – (
a
2
+ λ
2
) (b
1
c
3
– b
3
c
1
)
+ (a
3
+ λ
3
) (b
1
c
2
– b
2
c
1
)
= a
1
(b
2
c
3
– c
2
b
3
) – a
2
(b
1
c
3
– b
3
c
1
) + a
3
(b
1
c
2
– b
2
c
1
)
+ λ
1
(b
2
c
3
– c
2
b
3
) – λ
2
(b
1
c
3
– b
3
c
1
) + λ
3
(b
1
c
2
– b
2
c
1
)
(by rearranging terms)
=
1 2 3 1 2 3
1 2 3 1 2 3
1 2 3 1 2 3
a a a
b b b b b b
c c c c c c
λ λ λ
+
= R.H.S.
Similarly, we may verify Property 5 for other rows or columns.
Example 10 Show that
2 2 2 0
a b c
a x b y c z
x y z
+ + + =
Solution We have
2 2 2
a b c
a x b y c z
x y z
+ + +
=
2 2 2
a b c a b c
a b c x y z
x y z x y z
+
(by Property 5)
= 0 + 0 = 0 (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.e., the value of determinant remain same if we apply the operation
R
i
→ R
i
+ kR
j
or C
i
→ C
i
+ kC
j
.
Verification
Let ∆ =
1 2 3
1 2 3
1 2 3
a a a
b b b
c c c
and ∆
1
=
1 1 2 2 3 3
1 2 3
1 2 3
a k c a k c a k c
b b b
c c c
+ + +
,
where ∆
1
is obtained by the operation R
1
→ R
1
+ kR
3
.
Here, we have multiplied the elements of the third row (R
3
) by a constant k and
added them to the corresponding elements of the first row (R
1
).
Symbolically, we write this operation as R
1
→ R
1
+ k R
3
.
2019-20
DETERMINANTS 115
Now, again
∆
1
=
1 2 3 1 2 3
1 2 3 1 2 3
1 2 3 1 2 3
a a a k c k c k c
b b b b b b
c c c c c c
+
(Using Property 5)
= ∆ + 0 (since R
1
and R
3
are proportional)
Hence ∆ = ∆
1
Remarks
(i) If ∆
1
is the determinant obtained by applying R
i
→ kR
i
or C
i
→ kC
i
to the
determinant ∆, then ∆
1
= k∆.
(ii) If more than one operation like R
i
→ R
i
+ kR
j
is done in one step, care should be
taken to see that a row that is affected in one operation should not be used in
another operation. A similar remark applies to column operations.
Example 11 Prove that
3
2 3 2 4 3 2
3 6 3 10 6 3
a a b a b c
a a b a b c a
a a b a b c
+ + +
+ + + =
+ + +
.
Solution Applying operations R
2
→ R
2
– 2R
1
and R
3
→ R
3
– 3R
1
to the given
determinant ∆, we have
∆ =
0 2
0 3 7 3
a a b a b c
a a b
a a b
+ + +
+
+
Now applying R
3
→ R
3
– 3R
2
, we get
∆ =
0 2
0 0
a a b a b c
a a b
a
+ + +
+
Expanding along C
1
, we obtain
∆ =
2
0
a a b
a
a
+
+ 0 + 0
= a (a
2
– 0) = a (a
2
) = a
3
2019-20
116 MATHEMATICS
Example 12 Without expanding, prove that
∆ =
0
1 1 1
x y y z z x
z x y
+ + +
=
Solution Applying R
1
→ R
1
+ R
2
to ∆, we get
∆ =
1 1 1
x y z x y z x y z
z x y
+ + + + + +
Since the elements of R
1
and R
3
are proportional, ∆ = 0.
Example 13 Evaluate
∆ =
1
1
1
a b c
b c a
c a b
Solution Applying R
2
→ R
2
– R
1
and R
3
→ R
3
– R
1
, we get
∆ =
1
0 ( )
0 ( )
a b c
b a c a b
c a b a c
− −
− −
Taking factors (b – a) and (c – a) common from R
2
and R
3
, respectively, we get
∆ =
1
( ) ( ) 0 1 –
0 1 –
a b c
b a c a c
b
− −
= (b – a) (c – a) [(– b + c)] (Expanding along first column)
= (a – b) (b – c) (c – a)
Example 14 Prove that
4
b c a a
b c a b abc
c c a b
+
+ =
+
Solution Let ∆ =
b c a a
b c a b
c c a b
+
+
+
2019-20
DETERMINANTS 117
Applying R
1
→ R
1
– R
2
– R
3
to ∆, we get
∆ =
0 –2 –2
c b
b c a b
c c a b
+
+
Expanding along R
1
, we obtain
∆ =
0 – (–2 )
c a b b b
c
c a b c a b
+
+ +
(–2 )
b c a
b
c c
+
+
= 2 c (a b + b
2
– bc) – 2 b (b c – c
2
– ac)
= 2 a b c + 2 cb
2
– 2 bc
2
– 2 b
2
c + 2 bc
2
+ 2 abc
= 4 abc
Example 15 If x, y, z are different and
2 3
2 3
2 3
1
1 0
1
x x x
y y y
z z z
+
∆= + =
+
, then
show that 1 + xyz = 0
Solution We have
∆ =
2 3
2 3
2 3
1
1
1
x x x
y y y
z z z
+
+
+
=
2 2 3
2 2 3
2 2 3
1
1
1
x x x x x
y y y y y
z z z z z
+
(Using Property 5)
=
2 2
2 2 2
2 2
1 1
( 1) 1 1
1 1
x x x x
y y xyz y y
z z z z
− +
(Using C
3
↔C
2
and then C
1
↔ C
2
)
=
2
2
2
1
1 (1 )
1
x x
y y xyz
z z
+
2019-20
118 MATHEMATICS
=
( )
2
2 2
2 2
1
1 0
0
x x
xyz y x y x
z x z x
+ − −
− −
(Using R
2
→R
2
–R
1
and R
3
→ R
3
–R
1
)
Taking out common factor (y – x) from R
2
and (
z – x) from R
3
, we get
∆ =
2
1
(1+ ) ( – ) ( – ) 0 1
0 1
x x
xyz y x z x y x
z x
+
+
= (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C
1
)
Since ∆ = 0 and x, y, z are all different, i.e., x – y ≠ 0, y – z ≠ 0, z – x ≠ 0, we get
1 + xyz = 0
Example 16 Show that
1 1 1
1 1 1
1 1 1 1
1 1 1
a
b abc abc bc ca ab
a b c
c
+
+ = + + + = + + +
+
Solution Taking out factors a,b,c common from R
1
, R
2
and R
3
, we get
L.H.S. =
1 1 1
1
1 1 1
1
1 1 1
1
a a a
abc
b b b
c c c
+
+
+
Applying R
1
→ R
1
+ R
2
+ R
3
, we have
∆ =
1 1 1 1 1 1 1 1 1
1 1 1
1 1 1
1
1 1 1
1
a b c a b c a b c
abc
b b b
c c c
+ + + + + + + + +
+
+
2019-20
DETERMINANTS 119
=
1 1 1
1 1 1 1 1 1
1+ 1
1 1 1
1
abc
a b c b b b
c c c
+ + +
+
Now applying C
2
→ C
2
– C
1
, C
3
→ C
3
– C
1
, we get
∆ =
1 0 0
1 1 1 1
1+ 1 0
1
0 1
abc
a b c b
c
+ +
=
( )
1 1 1
1 1 1 – 0
abc
a b c
+ + +
=
1 1 1
1+abc
a b c
+ +
= abc + bc + ca + ab = R.H.S.
A
Note Alternately try by applying C
1
→ C
1
– C
2
and C
3
→ C
3
– C
2
, then apply
C
1
→ C
1
– a C
3
.
EXERCISE 4.2
Using the property of determinants and without expanding in Exercises 1 to 7, prove
that:
1.
0
x a x a
y b y b
z c z c
+
+ =
+
2.
0
a b b c c a
b c c a a b
c a a b b c
− − −
− − − =
− − −
3.
2 7 65
3 8 75 0
5 9 86
=
4.
(
)
( )
( )
1
1 0
1
bc a b c
ca b c a
ab c a b
+
+ =
+
5.
2
b c q r y z a p x
c a r p z x b q y
a b p q x y c r z
+ + +
+ + + =
+ + +
2019-20
120 MATHEMATICS
6.
0
0 0
0
a b
a c
b c
−
− − =
7.
2
2 2 2 2
2
4
a ab ac
ba b bc a b c
ca cb c
−
− =
−
By using properties of determinants, in Exercises 8 to 14, show that:
8. (i)
( )( )( )
2
2
2
1
1
1
a a
b b a b b c c a
c c
= − − −
(ii)
( )( )( )( )
3 3 3
1 1 1
a b c a b b c c a a b c
a b c
= − − − + +
9.
2
2
2
x x yz
y y zx
z z xy
= (x – y) (y – z) (z – x) (xy + yz + zx)
10. (i)
( )( )
2
4 2 2
2 4 5 4 4
2 2 4
x + x x
x x + 2x x x
x x x +
= + −
(ii)
( )
2
3
y + k y y
y y + k y k y k
y y y + k
= +
11. (i)
( )
3
2 2
2 2
2 2
a b c a a
b b c a b a b c
c c c a b
− −
− − = + +
− −
(ii)
( )
3
2
2 2
2
x y z x y
z y z x y x y z
z x z x y
+ +
+ + = + +
+ +
2019-20
DETERMINANTS 121
12.
( )
2
2
2 3
2
1
1 1
1
x x
x x x
x x
= −
13.
( )
2 2
3
2 2 2 2
2 2
1 2 2
2 1 2 1
2 2 1
a b ab b
ab a b a a b
b a a b
+ − −
− + = + +
− − −
14.
2
2 2 2 2
2
1
1 1
1
a ab ac
ab b bc a b c
ca cb c
+
+ = + + +
+
Choose the correct answer in Exercises 15 and 16.
15. Let A be a square matrix of order 3 × 3, then |kA| is equal to
(A) k|A| (B) k
2
|A| (C) k
3
|A| (D) 3k|A|
16. Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these
4.4 Area of a Triangle
In earlier classes, we have studied that the area of a triangle whose vertices are
(x
1
, y
1
), (x
2
, y
2
) and (x
3
, y
3
), is given by the expression
1
2
[x
1
(y
2
–y
3
) + x
2
(y
3
–y
1
) +
x
3
(y
1
–y
2
)]. Now this expression can be written in the form of a determinant as
∆ =
1 1
2 2
3 3
1
1
1
2
1
x y
x y
x y
... (1)
Remarks
(i) Since area is a positive quantity, we always take the absolute value of the
determinant in (1).
2019-20
122 MATHEMATICS
(ii) If area is given, use both positive and negative values of the determinant for
calculation.
(iii) The area of the triangle formed by three collinear points is zero.
Example 17
Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1).
Solution The area of triangle is given by
∆ =
3 8 1
1
4 2 1
2
5 1 1
–
=
( ) ( ) ( )
1
3 2 –1 – 8 – 4 – 5 1 – 4 –10
2
+
=
( )
1 61
3 72 14
2 2
–+ =
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.
Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So
0 0 1
1
1 3 1
2
1
x y
= 0
This gives
( )
1
3
2
y – x
= 0 or y = 3x,
which is the equation of required line AB.
Also, since the area of the triangle ABD is 3 sq. units, we have
1 3 1
1
0 0 1
2
0 1
k
= ± 3
This gives,
3
3
2
k
−
= ±
, i.e., k =
∓
2.
EXERCISE 4.3
1. 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)
2019-20
DETERMINANTS 123
2. Show that points
A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
3. Find values of k if area of triangle is 4 sq. units and vertices are
(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)
4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.
5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2
4.5 Minors and Cofactors
In this section, we will learn to write the expansion of a determinant in compact form
using minors and cofactors.
Definition 1 Minor of an element a
ij
of a determinant is the determinant obtained by
deleting its
ith row and jth column in which element a
ij
lies. Minor of an element a
ij
is
denoted by M
ij
.
Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of
order n – 1.
Example 19
Find the minor of element 6 in the determinant
1 2 3
4 5 6
7 8 9
∆ =
Solution Since 6 lies in the second row and third column, its minor M
23
is given by
M
23
=
1 2
7 8
= 8 – 14 = – 6 (obtained by deleting R
2
and C
3
in ∆).
Definition 2 Cofactor of an element a
ij
, denoted by A
ij
is defined by
A
ij
= (–1)
i + j
M
ij
, where M
ij
is minor of a
ij
.
Example 20 Find minors and cofactors of all the elements of the determinant
1 –2
4 3
Solution Minor of the element a
ij
is M
ij
Here a
11
= 1. So M
11
= Minor of a
11
= 3
M
12
= Minor of the element a
12
= 4
M
21
= Minor of the element a
21
= –2
2019-20
124 MATHEMATICS
M
22
= Minor of the element a
22
= 1
Now, cofactor of a
ij
is A
ij
. So
A
11
= (–1)
1 + 1
M
11
= (–1)
2
(3) = 3
A
12
= (–1)
1 + 2
M
12
= (–1)
3
(4) = – 4
A
21
= (–1)
2 + 1
M
21
= (–1)
3
(–2) = 2
A
22
= (–1)
2 + 2
M
22
= (–1)
4
(1) = 1
Example 21 Find minors and cofactors of the elements a
11
, a
21
in the determinant
∆ =
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
Solution By definition of minors and cofactors, we have
Minor of a
11
= M
11
=
22 23
32 33
a a
a a
= a
22
a
33
– a
23
a
32
Cofactor of a
11
= A
11
= (–1)
1+1
M
11
= a
22
a
33
– a
23
a
32
Minor of a
21
= M
21
=
12 13
32 33
a a
a a
= a
12
a
33
– a
13
a
32
Cofactor of a
21
= A
21
= (–1)
2+1
M
21
= (–1) (a
12
a
33
– a
13
a
32
) = – a
12
a
33
+ a
13
a
32
Remark Expanding the determinant ∆, in Example 21, along R
1
, we have
∆ = (–1)
1+1
a
11
22 23
32 33
a a
a a
+ (–1)
1+2
a
12
21 23
31 33
a a
a a
+ (–1)
1+3
a
13
21 22
31 32
a a
a a
= a
11
A
11
+ a
12
A
12
+ a
13
A
13
, where A
ij
is cofactor of a
ij
= sum of product of elements of R
1
with their corresponding cofactors
Similarly, ∆ can be calculated by other five ways of expansion that is along R
2
, R
3
,
C
1
, C
2
and C
3
.
Hence ∆ = sum of the product of elements of any row (or column) with their
corresponding cofactors.
A
Note If elements of a row (or column) are multiplied with cofactors of any
other row (or column), then their sum is zero. For example,
2019-20
DETERMINANTS 125
∆ = a
11
A
21
+ a
12
A
22
+ a
13
A
23
= a
11
(–1)
1+1
12 13
32 33
a a
a a
+ a
12
(–1)
1+2
11 13
31 33
a a
a a
+ a
13
(–1)
1+3
11 12
31 32
a a
a a
=
11 12 13
11 12 13
31 32 33
a a a
a a a
a a a
= 0 (since R
1
and R
2
are identical)
Similarly, we can try for other rows and columns.
Example 22 Find minors and cofactors of the elements of the determinant
2 3 5
6 0 4
1 5 7
–
–
and verify that a
11
A
31
+ a
12
A
32
+ a
13
A
33
= 0
Solution We have M
11
=
0 4
5 7
–
= 0 –20 = –20; A
11
= (–1)
1+1
(–20) = –20
M
12
=
6 4
1 7
–
= – 42 – 4 = – 46; A
12
= (–1)
1+2
(– 46) = 46
M
13
=
6 0
1 5
= 30 – 0 = 30; A
13
= (–1)
1+3
(30) = 30
M
21
=
3 5
5 7
–
–
= 21 – 25 = – 4; A
21
= (–1)
2+1
(– 4) = 4
M
22
=
2 5
1 7
–
= –14 – 5 = –19; A
22
= (–1)
2+2
(–19) = –19
M
23
=
2 3
1 5
–
= 10 + 3 = 13; A
23
= (–1)
2+3
(13) = –13
M
31
=
3 5
0 4
–
= –12 – 0 = –12; A
31
= (–1)
3+1
(–12) = –12
2019-20
126 MATHEMATICS
M
32
=
2 5
6 4
= 8 – 30 = –22;
A
32
= (–1)
3+2
(–22) = 22
and M
33
=
2 3
6 0
–
= 0 + 18 = 18; A
33
= (–1)
3+3
(18) = 18
Now a
11
= 2, a
12
= –3, a
13
= 5; A
31
= –12, A
32
= 22, A
33
= 18
So a
11
A
31
+ a
12
A
32
+ a
13
A
33
= 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0
EXERCISE 4.4
Write Minors and Cofactors of the elements of following determinants:
1. (i)
2 4
0 3
–
(ii)
a c
b d
2. (i)
1 0 0
0 1 0
0 0 1
(ii)
1 0 4
3 5 1
0 1 2
–
3. Using Cofactors of elements of second row, evaluate ∆ =
5 3 8
2 0 1
1 2 3
.
4. Using Cofactors of elements of third column, evaluate ∆ =
1
1
1
x yz
y zx
z xy
.
5. If ∆ =
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
and A
ij
is Cofactors of a
ij
, then value of ∆ is given by
(A) a
11
A
31
+ a
12
A
32
+ a
13
A
33
(B) a
11
A
11
+ a
12
A
21
+ a
13
A
31
(C) a
21
A
11
+ a
22
A
12
+ a
23
A
13
(D) a
11
A
11
+ a
21
A
21
+ a
31
A
31
4.6 Adjoint and Inverse of a Matrix
In the previous chapter, we have studied inverse of a matrix. In this section, we shall
discuss the condition for existence of inverse of a matrix.
To find inverse of a matrix A, i.e., A
–1
we shall first define adjoint of a matrix.
2019-20
DETERMINANTS 127
4.6.1 Adjoint of a matrix
Definition 3
The adjoint of a square matrix A = [a
ij
]
n × n
is defined as the transpose of
the matrix [A
ij
]
n × n
, where A
ij
is the cofactor of the element a
ij
. Adjoint of the matrix A
is denoted by adj A.
Let
11 12 13
21 22 23
31 32 33
A =
a a a
a a a
a a a
Then
11 12 13
21 22 23
31 32 33
A A A
A =Transposeof A A A
A A A
adj
11 21 31
12 22 32
13 23 33
A A A
= A A A
A A A
Example 23
2 3
Find A for A =
1 4
adj
Solution We have A
11
= 4, A
12
= –1, A
21
= –3, A
22
= 2
Hence adj A =
11 21
12 22
A A
4 –3
=
A A
–1 2
Remark For a square matrix of order 2, given by
A =
11 12
21 22
a a
a a
The adj A can also be obtained by interchanging a
11
and a
22
and by changing signs
of a
12
and a
21
, i.e.,
We state the following theorem without proof.
Theorem 1 If A be any given square matrix of order n, then
A(adj A) = (adj A) A =
A I
,
where I is the identity matrix of order n
2019-20
128 MATHEMATICS
Verification
Let A =
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
, then adj A =
11 21 31
12 22 32
13 23 33
A A A
A A A
A A A
Since sum of product of elements of a row (or a column) with corresponding
cofactors is equal to |A| and otherwise zero, we have
A (adj A) =
A 0 0
0 A 0
0 0 A
=
A
1 0 0
0 1 0
0 0 1
=
A
I
Similarly, we can show (adj A) A =
A
I
Hence A (adj A) = (adj A) A =
A
I
Definition 4 A square matrix A is said to be singular if
A
= 0.
For example, the determinant of matrix A =
1 2
4 8
is zero
Hence A is a singular matrix.
Definition 5 A square matrix A is said to be non-singular if
A
≠ 0
Let A =
1 2
3 4
. Then
A
=
1 2
3 4
= 4 – 6 = – 2 ≠ 0.
Hence A is a nonsingular matrix
We state the following theorems without proof.
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.
Theorem 3 The determinant of the product of matrices is equal to product of their
respective determinants, that is,
AB
=
A
B
, where A and B are square matrices of
the same order
Remark We know that (adj A) A =
A
I =
A
A
A
A
0 0
0 0
0 0
0
≠,
2019-20
DETERMINANTS 129
Writing determinants of matrices on both sides, we have
( A) A
adj
=
A 0 0
0 A 0
0 0 A
i.e. |(adj A)| |A| =
3
1 0 0
A 0 1 0
0 0 1
(Why?)
i.e. |(adj A)| |A| = |A|
3
(1)
i.e. |(adj A)| = |A|
2
In general, if A is a square matrix of order n, then |adj(A)| = |A|
n – 1
.
Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix.
Proof Let A be invertible matrix of order n and I be the identity matrix of order n.
Then, there exists a square matrix B of order n such that AB = BA = I
Now AB = I. So
AB
=
I
or
A
B
= 1
(since I 1, AB A B )
= =
This gives
A
≠ 0. Hence A is nonsingular.
Conversely, let A be nonsingular. Then
A
≠ 0
Now A (adj A) = (adj A) A =
A
I (Theorem 1)
or A
1 1
A A A I
| A | | A |
adj adj
= =
or AB = BA = I, where B =
1
A
| A |
adj
Thus A is invertible and A
–1
=
1
A
| A |
adj
Example 24 If A =
1 3 3
1 4 3
1 3 4
, then verify that A adj A = |A| I. Also find A
–1
.
Solution We have
A
= 1
(16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0
2019-20
130 MATHEMATICS
Now A
1
1
= 7, A
12
= –1, A
13
= –1, A
21
= –3, A
22
= 1,A
23
= 0, A
31
= –3, A
32
= 0,
A
33
= 1
Therefore adj A =
7 3 3
1 1 0
1 0 1
− −
−
−
Now A (adj A) =
1 3 3 7 3 3
1 4 3 1 1 0
1 3 4 1 0 1
− −
−
−
=
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
0 0 1
= (1)
1 0 0
0 1 0
0 0 1
=
A
. I
Also A
–1
1
A
A
a d j
=
=
7 3 3
1
1 1 0
1
1 0 1
− −
−
−
=
7 3 3
1 1 0
1 0 1
− −
−
−
Example 25 If A =
2 3 1 2
and B
1 4 1 3
−
=
− −
, then verify that (AB)
–1
= B
–1
A
–1
.
Solution We have AB =
2 3 1 2 1 5
1 4 1 3 5 14
− −
=
− − −
Since,
AB
= –11 ≠ 0, (AB)
–1
exists and is given by
(AB)
–1
=
14 5
1 1
(AB)
5 1
AB 11
adj
− −
=−
− −
14 5
1
5 1
11
=
Further,
A
= –11 ≠ 0 and
B
= 1 ≠ 0. Therefore, A
–1
and B
–1
both exist and are given by
A
–1
=
−
− −
−
=
−
1
11
4 3
1 2
3 2
1 1
1
,B
2019-20
DETERMINANTS 131
Therefore
B A
− −
= −
− −
−
1 1
1
11
3 2
1 1
4 3
1 2
= −
− −
− −
1
11
14 5
5 1
14 5
1
5 1
11
=
Hence (AB)
–1
= B
–1
A
–1
Example 26 Show that the matrix A =
2 3
1 2
satisfies the equation A
2
– 4A + I
= O,
where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A
–1
.
Solution We have
2
2 3 2 3 7 12
A A.A
1 2 1 2 4 7
= = =
Hence
2
7 12 8 12 1 0
A 4A I
4 7 4 8 0 1
− + = − +
0 0
O
0 0
= =
Now A
2
– 4A + I = O
Therefore A A – 4A = – I
or A A (A
–1
) – 4 A A
–1
= – I A
–1
(Post multiplying by A
–1
because |A| ≠ 0)
or A (A
A
–1
) – 4I = – A
–1
or AI – 4I = – A
–1
or A
–1
= 4I – A =
4 0 2 3 2 3
0 4 1 2 1 2
−
− =
−
Hence
1
2 3
A
1 2
−
−
=
−
EXERCISE 4.5
Find adjoint of each of the matrices in Exercises 1 and 2.
1.
1 2
3 4
2.
1 1 2
2 3 5
2 0 1
−
−
Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4
3.
2 3
4 6− −
4.
1 1 2
3 0 2
1 0 3
−
−
2019-20
132 MATHEMATICS
Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11.
5.
2 2
4 3
−
6.
−
−
1 5
3 2
7.
1 2 3
0 2 4
0 0 5
8.
1 0 0
3 3 0
5 2 1−
9.
2 1 3
4 1 0
7 2 1
−
−
10.
1 1 2
0 2 3
3 2 4
−
−
−
11.
1 0 0
0 cos sin
0 sin cos
α α
α − α
12. Let A =
3 7
2 5
and B =
6 8
7 9
. Verify that (AB)
–1
= B
–1
A
–1
.
13. If A =
3 1
1 2−
, show that A
2
– 5A + 7I = O. Hence find A
–1
.
14. For the matrix A =
3 2
1 1
, find the numbers a and b such that A
2
+ aA + bI = O.
15. For the matrix A =
1 1 1
1 2 3
2 1 3
−
−
Show that A
3
– 6A
2
+ 5A + 11 I = O. Hence, find A
–1
.
16. If A =
2 1 1
1 2 1
1 1 2
−
− −
−
Verify that A
3
– 6A
2
+ 9A – 4I = O and hence find A
–1
17. Let A be a nonsingular square matrix of order 3 × 3. Then |adj A| is equal to
(A) |A| (B) |A|
2
(C) |A|
3
(D) 3|A|
18. If A is an invertible matrix of order 2, then det (A
–1
) is equal to
(A) det (A) (B)
1
det (A)
(C) 1 (D) 0
2019-20
DETERMINANTS 133
4.7 Applications of Determinants and Matrices
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.
Consistent system A system of equations is said to be consistent if its solution (one
or more) exists.
Inconsistent system A system of equations is said to be inconsistent if its solution
does not exist.
A
Note In this chapter, we restrict ourselves to the system of linear equations
having unique solutions only.
4.7.1 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.
Consider the system of equations
a
1
x + b
1
y + c
1
z = d
1
a
2
x + b
2
y + c
2
z = d
2
a
3
x + b
3
y + c
3
z = d
3
Let A =
1 1 1 1
2 2 2 2
3 3 3 3
, X and B
a b c x d
a b c y d
a b c z d
= =
Then, the system of equations can be written as, AX = B, i.e.,
1 1 1
2 2 2
3 3 3
a b c x
a b c y
a b c z
=
1
2
3
d
d
d
Case I If A is a nonsingular matrix, then its inverse exists. Now
AX = B
or A
–1
(AX) = A
–1
B (premultiplying by A
–1
)
or (A
–1
A) X = A
–1
B (by associative property)
or I X = A
–1
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. This method of solving system of equations is known as
Matrix Method.
2019-20
134 MATHEMATICS
Case II If A is a singular matrix, then |A| = 0.
In this case, we calculate (adj A) B.
If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the
system of equations is called inconsistent.
If (adj A) B = O, then system may be either consistent or inconsistent according
as the system have either infinitely many solutions or no solution.
Example 27 Solve the system of equations
2x + 5y = 1
3x + 2y = 7
Solution
The system of equations can be written in the form AX = B, where
A =
2 5 1
,X and B
3 2 7
x
y
= =
Now,
A
= –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution.
Note that A
–1
=
−
−
−
1
11
2 5
3 2
Therefore X = A
–1
B = –
1
11
2 5
3 2
1
7
−
−
i.e.
x
y
=
−
−
=
−
1
11
33
11
3
1
Hence x = 3, y = – 1
Example 28 Solve the following system of equations by matrix method.
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 8
A 2 1 1 , X and B 1
4 3 2 4
x
y
z
−
= − = =
−
We see that
A
= 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0
2019-20
DETERMINANTS 135
Hence, A is nonsingular and so its inverse exists. Now
A
11
= –1, A
12
= – 8, A
13
= –10
A
21
= –5, A
22
= – 6, A
23
= 1
A
31
= –1, A
32
= 9, A
33
= 7
Therefore A
–1
=
1 5 1
1
8 6 9
17
10 1 7
− − −
− − −
−
So X =
–1
1 5 1 8
1
A B = 8 6 9 1
17
10 1 7 4
− − −
− − −
−
i.e.
x
y
z
=
17 1
1
34 2
17
51 3
−
− − =
−
Hence x = 1, y = 2 and z = 3.
Example 29 The sum of three numbers is 6. If we multiply third number by 3 and add
second number to it, we get 11. By adding first and third numbers, we get double of the
second number. Represent it algebraically and find the numbers using matrix method.
Solution Let first, second and third numbers be denoted by x, y and z, respectively.
Then, according to given conditions, we have
x + y + z = 6
y + 3z = 11
x + z = 2y or x – 2y + z = 0
This system can be written as A X = B, where
A =
1 1 1
0 1 3
1 2 1
, X =
x
y
z
and B =
6
11
0
Here
(
)
(
)
A 1 1 6 – (0 – 3) 0 –1 9 0
= + + = ≠
. Now we find adj A
A
11
= 1 (1 + 6) = 7, A
12
= – (0 – 3) = 3, A
13
= – 1
A
21
= – (1 + 2) = – 3, A
22
= 0, A
23
= – (– 2 – 1) = 3
A
31
= (3 – 1) = 2, A
32
= – (3 – 0) = – 3, A
33
= (1 – 0) = 1
2019-20
136 MATHEMATICS
Hence adj
A =
7 –3 2
3 0 –3
–1 3 1
Thus A
–1
=
1
A
adj (A) =
7 3 2
1
3 0 3
9
1 3 1
–
–
–
Since X = A
–1
B
X =
7 3 2 6
1
3 0 3 11
9
1 3 1 0
–
–
–
or
x
y
z
=
1
9
42 33 0
18 0 0
6 33 0
− +
+ +
− + +
=
1
9
9
18
27
=
1
2
3
Thus x = 1, y = 2, z = 3
EXERCISE 4.6
Examine the consistency of the system of equations in Exercises 1 to 6.
1. x + 2y = 2 2. 2x – y = 5 3. x + 3y = 5
2x + 3y = 3 x + y = 4 2x + 6y = 8
4. x + y + z = 1 5. 3x–y – 2z = 2 6. 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.
7. 5x + 2y = 4 8. 2x – y = –2 9. 4x – 3y = 3
7x + 3y = 5 3x + 4y = 3 3x – 5y = 7
10. 5x + 2y = 3 11. 2x + y + z = 1 12. x – y + z = 4
3x + 2y = 5 x – 2y – z =
3
2
2x + y – 3z = 0
3y – 5z = 9 x + y + z = 2
13. 2x + 3y +3 z = 5 14. x – y + 2z = 7
x – 2y + z = – 4 3x + 4y – 5z = – 5
3x – y – 2z = 3 2x – y + 3z = 12
2019-20
DETERMINANTS 137
15. If A =
2 –3 5
3 2 – 4
1 1 –2
, find A
–1
. Using A
–1
solve the system of equations
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ` 60. The cost of 2 kg onion,
4 kg wheat and 6 kg rice is ` 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice
is ` 70. Find cost of each item per kg by matrix method.
Miscellaneous Examples
Example 30
If a, b, c
are positive and unequal, show that value of the determinant
∆ =
a b c
b c a
c a b
is negative.
Solution Applying C
1
→ C
1
+ C
2
+ C
3
to the given determinant, we get
∆ =
a b c b c
a b c c a
a b c a b
+ +
+ +
+ +
= (a + b + c)
1
1
1
b c
c a
a b
= (a + b + c)
1
0 – –
0 – –
b c
c b a c
a b b c
(ApplyingR
2
→R
2
–R
1
,andR
3
→R
3
–R
1
)
= (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C
1
)
= (a + b + c)(– a
2
– b
2
– c
2
+ ab + bc + ca)
=
1
2
–
(a + b + c) (2a
2
+ 2b
2
+ 2c
2
– 2ab – 2bc – 2ca)
=
1
2
–
(a + b + c) [(a – b)
2
+ (b – c)
2
+ (c – a)
2
]
which is negative (since a + b + c > 0 and (a – b)
2
+ (b – c)
2
+ (c – a)
2
> 0)
2019-20
138 MATHEMATICS
Example 31 If a, b, c, are in A.P, find value of
2 4 5 7 8
3 5 6 8 9
4 6 7 9 10
y y y a
y y y b
y y y c
+ + +
+ + +
+ + +
Solution Applying R
1
→ R
1
+ R
3
– 2R
2
to the given determinant, we obtain
0 0 0
3 5 6 8 9
4 6 7 9 10
y y y b
y y y c
+ + +
+ + +
= 0 (Since 2b = a + c)
Example 32 Show that
∆ =
(
)
( )
( )
2
2
2
y z xy zx
xy x z yz
xz yz x y
+
+
+
= 2xyz (x + y + z)
3
Solution Applying R
1
→ xR
1
, R
2
→ yR
2
,
R
3
→ zR
3
to ∆ and dividing by xyz, we get
∆ =
(
)
( )
( )
2
2 2
2
2 2
2
2 2
1
+
+
+
x y z x y x z
xy y x z y z
xyz
xz yz z x y
Taking common factors x, y, z from C
1
C
2
and C
3
, respectively, we get
∆ =
(
)
( )
( )
2
2 2
2
2 2
2
2 2
y z x x
xyz
y x z y
xyz
z z x y
+
+
+
Applying C
2
→ C
2
– C
1
, C
3
→ C
3
– C
1
,
we have
∆ =
( ) ( ) ( )
( )
( )
2 2 2
2 2
2
2 2
2
2 2
–
0
0 –
y z x y z x y z
y x z y
z x y z
+ + − +
+ −
+
2019-20
DETERMINANTS 139
Taking common factor (x + y + z) from C
2
and C
3
, we have
∆ = (x + y + z)
2
(
)
(
)
(
)
( )
( )
2
2
2
y
0
0
z x – y z x – y z
y x z – y
z x y – z
+ + +
+
+
Applying R
1
→ R
1
– (R
2
+ R
3
), we have
∆ = (x + y + z)
2
2
2
2 2 2
+ 0
0 – z
yz – z – y
y x y z
z x y
−
+
Applying C
2
→ (C
2
+
1
y
C
1
) and
3 3 1
1
C C C
→ +
z
, we get
∆ = (x + y + z)
2
2
2
2
2
2 0 0
+
+
yz
y
y x z
z
z
z x y
y
Finally expanding along R
1
, we have
∆ = (x + y + z)
2
(2yz) [(x + z) (x + y) – yz] = (x + y + z)
2
(2yz) (x
2
+ xy + xz)
= (x + y + z)
3
(2xyz)
Example 33 Use product
1 1 2
0 2 3
3 2 4
2 0 1
9 2 3
6 1 2
to solve the system of equations
x – y + 2z = 1
2y – 3z = 1
3x – 2y + 4z = 2
Solution Consider the product
1 1 2 2 0 1
0 2 3 9 2 3
3 2 4 6 1 2
– –
– –
– –
2019-20
140 MATHEMATICS
=
2 9 12 0 2 2 1 3 4
0 18 18 0 4 3 0 6 6
6 18 24 0 4 4 3 6 8
− − + − + + −
+ − + − − +
− − + − + + −
=
1 0 0
0 1 0
0 0 1
Hence
1 1 2
0 2 3
3 2 4
2 0 1
9 2 3
6 1 2
1
–
–
–
–
–
–
–
=
Now, given system of equations can be written, in matrix form, as follows
1 –1 2
0 2 –3
3 –2 4
x
y
z
=
1
1
2
or
x
y
z
=
1
1 1 2 1
0 2 3 1
3 2 4 2
−
−
−
−
=
2 0 1
9 2 3
6 1 2
1
1
2
=
2 0 2 0
9 2 6 5
6 1 4 3
− + +
+ − =
+ −
Hence x = 0, y = 5 and z = 3
Example 34 Prove that
∆ =
2
(1 )
a bx c dx p qx a c p
ax b cx d px q x b d q
u v w u v w
+ + +
+ + + = −
Solution Applying R
1
→ R
1
– x R
2
to ∆, we get
∆ =
2 2 2
(1 ) (1 ) (1 )
a x c x p x
ax b cx d px q
u v w
− − −
+ + +
=
2
(1 )
a c p
x ax b cx d px q
u v w
− + + +
2019-20
DETERMINANTS 141
Applying R
2
→ R
2
– x R
1
, we get
∆ =
2
(1 )
a c p
x b d q
u v w
−
Miscellaneous Exercises on Chapter 4
1. Prove that the determinant
sin cos
–sin – 1
cos 1
x
x
x
θ θ
θ
θ
is independent of θ.
2. Without expanding the determinant, prove that
2 2 3
2 2 3
2 2 3
1
1
1
=
a a bc a a
b b ca b b
c c ab c c
.
3. Evaluate
cos cos cos sin – sin
– sin cos 0
sin cos sin sin cos
α β α β α
β β
α β α β α
.
4. If a, b and c are real numbers, and
∆ =
b c c a a b
c a a b b c
a b b c c a
+ + +
+ + +
+ + +
= 0,
Show that either a + b + c = 0 or a = b = c.
5. Solve the equation
0
x a x x
x x a x
x x x a
+
+ =
+
, a ≠ 0
6. Prove that
2 2
2 2
2 2
+
+
+
a bc ac c
a ab b ac
ab b bc c
= 4a
2
b
2
c
2
7. If A
–1
=
( )
1
3 1 1 1 2 2
15 6 5 and B 1 3 0 , find AB
5 2 2 0 2 1
–
– –
– – –
– –
=
2019-20
142 MATHEMATICS
8. Let A =
1 2 1
2 3 1
1 1 5
. Verify that
(i) [adj A]
–1
= adj (A
–1
) (ii) (A
–1
)
–1
= A
9. Evaluate
x y x y
y x y x
x y x y
+
+
+
10. Evaluate
1
1
1
x y
x y y
x x+ y
+
Using properties of determinants in Exercises 11 to 15, prove that:
11.
2
2
2
α α β + γ
β β γ + α
γ γ α + β
= (β – γ) (γ – α) (α – β) (α + β + γ)
12.
2 3
2 3
2 3
1
1
1
+
+
+
x x px
y y py
z z pz
= (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar.
13.
3
3
a 3c
a – a+ b – a+ c
–b a b – b c
– c – c+b
+ +
+
= 3(a + b + c) (ab + bc + ca)
14.
1 1 1
2 3 2 4 3 2
3 6 3 10 6 3
p p q
p p q
p p q
+ + +
+ + +
+ + +
= 1 15.
(
)
( )
( )
sin cos cos
sin cos cos 0
sin cos cos
α α α + δ
β β β + δ =
γ γ γ + δ
16. Solve the system of equations
2 3 10
4
+ + =
x y z
2019-20
DETERMINANTS 143
4 6 5
1
+ =
–
x y z
6 9 20
2
+ =
–
x y z
Choose the correct answer in Exercise 17 to 19.
17. If a, b, c, are in A.P, then the determinant
2 3 2
3 4 2
4 5 2
x x x a
x x x b
x x x c
+ + +
+ + +
+ + +
is
(A) 0 (B) 1 (C) x (D) 2x
18. If x, y, z are nonzero real numbers, then the inverse of matrix
0 0
A 0 0
0 0
x
y
z
=
is
(A)
1
1
1
0 0
0 0
0 0
x
y
z
−
−
−
(B)
1
1
1
0 0
0 0
0 0
x
xyz y
z
−
−
−
(C)
0 0
1
0 0
0 0
x
y
xyz
z
(D)
1 0 0
1
0 1 0
0 0 1
xyz
19. Let A =
1 sin 1
sin 1 sin
1 sin 1
θ
− θ θ
− − θ
, where 0 ≤ θ ≤ 2π. Then
(A) Det(A) = 0 (B) Det(A) ∈ (2, ∞)
(C) Det(A) ∈ (2, 4) (D) Det(A) ∈ [2, 4]
2019-20
144 MATHEMATICS
Summary
®
Determinant of a matrix A = [a
1
1
]
1×1
is given by |a
1
1
| = a
1
1
®
Determinant of a matrix
A =
a a
a a
11 12
21 22
is given by
11 12
21 22
A
a a
a a
=
= a
11
a
22
– a
12
a
21
®
Determinant of a matrix
A =
a b c
a b c
a b c
1 1 1
2 2 2
3 3 3
is given by (expanding along R
1
)
1 1 1
2 2 2 2 2 2
2 2 2 1 1 1
3 3 3 3 3 3
3 3 3
A
a b c
b c a c a b
a b c a b c
b c a c a b
a b c
= = − +
For any square matrix A, the |A| satisfy following properties.
®
|A′| = |A|, where A′ = transpose of A.
®
If we interchange any two rows (or columns), then sign of determinant
changes.
®
If any two rows or any two columns are identical or proportional, then value
of determinant is zero.
®
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.
®
Multiplying a determinant by k means multiply elements of only one row
(or one column) by k.
®
If
3
3 3
A [ ] , then .A A
ij
a k k
×
= =
®
If elements of a row or a column in a determinant can be expressed as sum
of two or more elements, then the given determinant can be expressed as
sum of two or more determinants.
®
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.
2019-20
DETERMINANTS 145
®
Area of a triangle with vertices (x
1
, y
1
), (x
2
, y
2
) and (
x
3
, y
3
) is given by
1 1
2 2
3 3
1
1
1
2
1
x y
x y
x y
∆=
®
Minor of an element a
ij
of the determinant of matrix A is the determinant
obtained by deleting i
th
row and j
th
column and denoted by M
ij
.
®
Cofactor of a
ij
of given by A
ij
= (– 1)
i+j
M
ij
®
Value of determinant of a matrix A is obtained by sum of product of elements
of a row (or a column) with corresponding cofactors. For example,
A
= a
11
A
11
+ a
12
A
12
+ a
13
A
13
.
®
If elements of one row (or column) are multiplied with cofactors of elements
of any other row (or column), then their sum is zero. For example, a
11
A
21
+ a
12
A
22
+ a
13
A
23
= 0
®
If
11 12 13
21 22 23
31 32 33
A ,
a a a
a a a
a a a
=
then
11 21 31
12 22 32
13 23 33
A A A
A A A A
A A A
adj
=
,
where A
ij
is
cofactor of a
ij
®
A (adj A) = (adj A) A = |A| I, where A is square matrix of order n.
®
A square matrix A is said to be singular or non-singular according as
|A| = 0 or |A| ≠ 0.
®
If AB = BA = I, where B is square matrix, then B is called inverse of A.
Also A
–1
= B or B
–1
= A and hence (A
–1
)
–1
= A.
®
A square matrix A has inverse if and only if A is non-singular.
®
–1
1
A ( A)
A
adj
=
®
If a
1
x
+ b
1
y + c
1
z = d
1
a
2
x
+ b
2
y + c
2
z = d
2
a
3
x
+ b
3
y + c
3
z = d
3
,
then these equations can be written as A X = B, where
1 1 1 1
2 2 2 2
3 3 3 3
A , X = and B=
a b c x d
a b c y d
a b c z d
=
2019-20
146 MATHEMATICS
®
Unique solution of equation AX = B is given by X = A
–1
B, where
A 0
≠
.
®
A system of equation is consistent or inconsistent according as its solution
exists or not.
®
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.
Historical Note
The Chinese method of representing the coefficients of the unknowns of
several linear equations by using rods on a calculating board naturally led to the
discovery of simple method of elimination. The arrangement of rods was precisely
that of the numbers in a determinant. The Chinese, therefore, early developed the
idea of subtracting columns and rows as in simplification of a determinant
Mikami, China, pp 30, 93.
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. But he used this device only in eliminating a
quantity from two equations and not directly in the solution of a set of simultaneous
linear equations. T. Hayashi, “The Fakudoi and Determinants in Japanese
Mathematics,” in the proc. of the Tokyo Math. Soc., V.
Vendermonde was the first to recognise determinants as independent functions.
He may be called the formal founder. Laplace (1772), gave general method of
expanding a determinant in terms of its complementary minors. In 1773 Lagrange
treated determinants of the second and third orders and used them for purpose
other than the solution of equations. In 1801, Gauss used determinants in his
theory of numbers.
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 n-
rows, which for the special case of m = n reduces to the multiplication theorem.
Also on the same day, Cauchy (1812) presented one on the same subject. He
used the word ‘determinant’ in its present sense. He gave the proof of multiplication
theorem more satisfactory than Binet’s.
The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after
this the word determinant received its final acceptance.
2019-20
