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Chapter 4

DETERMINANTS

™ All Mathematical truths are relative and conditional. — C.P. STEINMETZ ™
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
a1 x + b1 y = c 1
a2 x + b2 y = c 2
⎡ a b ⎤ ⎡ x ⎤ ⎡c ⎤
can be represented as ⎢ 1 1 ⎥ ⎢ ⎥ = ⎢ 1 ⎥ . Now, this
⎣ a2 b2 ⎦ ⎣ y ⎦ ⎣ c2 ⎦
system of equations has a unique solution or not, is
determined by the number a1 b2 – a2 b1. (Recall that if
a1 b1 P.S. Laplace
≠ or, a1 b2 – a2 b1 ≠ 0, then the system of linear
a2 b2 (1749-1827)
equations has a unique solution). The number a1 b2 – a2 b1
⎡a b ⎤
which determines uniqueness of solution is associated with the matrix A = ⎢ 1 1 ⎥
⎣ a2 b2 ⎦
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 = [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.

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 Δ.
⎡a b ⎤ a b
If A = ⎢ ⎥ , then determinant of A is written as | A| = = det (A)
⎣c d ⎦ c d
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
⎡ a11 a12 ⎤
Let A= ⎢ ⎥ be a matrix of order 2 × 2,
⎣ a21 a22 ⎦
then the determinant of A is defined as:

det (A) = |A| = Δ = = a11a22 – a21a12

2 4
Example 1 Evaluate .
–1 2

2 4
Solution We have = 2 (2) – 4(–1) = 4 + 4 = 8.
–1 2

x x +1
Example 2 Evaluate
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

4.2.3 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

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.
Consider the determinant of square matrix A = [aij]3 × 3

a 11 a12 a13
i.e., | A | = a21 a22 a23
a31 a32 a33
Expansion along first Row (R1)
Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the
second order determinant obtained by deleting the elements of first row (R1) and first
column (C1) of | A | as a11 lies in R1 and C1,
a22 a23
i.e., (–1)1 + 1 a11
a32 a33
Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second
order determinant obtained by deleting elements of first row (R1) and 2nd column (C2)
of | A | as a12 lies in R1 and C2,
a21 a23
i.e., (–1)1 + 2 a12
a31 a33
Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a ] and the second
13

order determinant obtained by deleting elements of first row (R1) and third column (C3)
of | A | as a13 lies in R1 and C3,
a21 a22
i.e., (–1)1 + 3 a13 a a32
31

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
a22 a23 a21 a23
det A = |A| = (–1)1 + 1 a11 a + (–1)1 + 2 a12
32 a33 a31 a33

1+ 3 a21 a22
+ (–1) a13
a31 a32
or |A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23)
+ a13 (a21 a32 – a31 a22)

106 MATHEMATICS

= a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32
– a13 a31 a22 ... (1)

$Note We shall apply all four steps together.
Expansion along second row (R2)
a 11 a 12 a 13
| A | = a 21 a 22 a 23
a 31 a 32 a 33
Expanding along R2, we get

2+1 a12 a13 a11 a13
| A | = (–1) a21 + (–1)2 + 2 a22
a32 a33 a31 a33

a11 a12
+ (–1) 2 + 3 a23
a31 a32
= – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13)
– a23 (a11 a32 – a31 a12)
| A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
+ a23 a31 a12
= a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32
– a13 a31 a22 ... (2)
Expansion along first Column (C1)
a11 a12 a13
| A | = a21 a22 a23
a31 a32 a33
By expanding along C1, we get

1 + 1 a22 a23 a12 a13
| A | = a11 (–1) + a21 ( −1) 2 + 1
a32 a33 a32 a33

3 + 1 a 12 a13
+ a31 (–1) a22 a23
= a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)

it is easy to verify that A = 2B. Hence. values of | A | in (1). (ii) While expanding. we get –1 3 1 2 1 2 Δ= 4 –0 +0 4 1 4 1 –1 3 = 4 (–1 – 12) – 0 + 0 = – 52 0 sin α – cos α Example 4 Evaluate Δ = – sin α 0 sin β . (3) Clearly. instead of multiplying by (–1)i + j.. So expanding along third column (C3). ⎡2 2⎤ ⎡1 1⎤ (iii) Let A = ⎢ ⎥ and B = ⎢ ⎥ . DETERMINANTS 107 | 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 . 3 1 2 4 Example 3 Evaluate the determinant Δ = –1 3 0 . Observe that. Also ⎣4 0⎦ ⎣2 0⎦ | A | = 0 – 8 = – 8 and | B | = 0 – 2 = – 2. if A = kB where A and B are square matrices of order n. It is left as an exercise to the reader to verify that the values of |A| by expanding along R3. where n = 2 is the order of square matrices A and B. C2 and C3 are equal to the value of | A | obtained in (1). then | A| = kn | B |. 4 1 0 Solution Note that in the third column.. expanding a determinant along any row or column gives same value. (2) and (3) are equal. cos α – sin β 0 . 2. | A | = 4 (– 2) = 22 | B | or | A | = 2n | B |. (2) or (3). where n = 1. we can multiply by +1 or –1 according as (i + j) is even or odd. Then. Remarks (i) For easier calculations. In general. we shall expand the determinant along that row or column which contains maximum number of zeros. two entries are zero.

e.108 MATHEMATICS Solution Expanding along R1. If A= ⎢ ⎥ .e. x 1 4 1 3 x 3 2 Solution We have = x 1 4 1 i. then show that | 2A | = 4 | A | ⎣ 4 2⎦ ⎡1 0 1⎤ ⎢ ⎥ 4.1 Evaluate the determinants in Exercises 1 and 2. –5 –1 cos θ – sin θ x2 – x + 1 x – 1 2. 3 – x2 = 3 – 8 i. Evaluate the determinants 3 –1 –2 3 –4 5 (i) 0 0 –1 (ii) 1 1 –2 3 –5 0 2 3 1 . If A = ⎢ 0 1 2 ⎥ . (i) (ii) sin θ cos θ x +1 x +1 ⎡1 2⎤ 3. we get 0 sin β – sin α sin β – sin α 0 Δ= 0 – sin α – cos α – sin β 0 cos α 0 cos α – sin 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 3 x 3 2 Example 5 Find values of x for which = . x2 = 8 Hence x= ±2 2 EXERCISE 4. 2 4 1. then show that | 3 A | = 27 | A | ⎣⎢ 0 0 4 ⎥⎦ 5.

Find values of x. If = . we have learnt how to expand the determinants. DETERMINANTS 109 0 1 2 2 –1 –2 (iii) –1 0 –3 (iv) 0 2 –1 –2 3 0 3 –5 0 ⎡ 1 1 –2 ⎤ ⎢ ⎥ 6. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 Expanding along first row. Property 1 The value of the determinant remains unchanged if its rows and columns are interchanged. we shall restrict ourselves upto determinants of order 3 only. we get the determinant a1 b1 c1 Δ1 = a2 b2 c2 a3 b3 c3 . then x is equal to 18 x 18 6 (A) 6 (B) ± 6 (C) – 6 (D) 0 4. However. we get b2 b3 b1 b3 b1 b2 Δ = a1 − a2 + a3 c2 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. find | A | ⎢⎣ 5 4 –9 ⎥⎦ 7. These properties are true for determinants of any order.3 Properties of Determinants In the previous section. if 2 4 2x 4 2 3 x 3 (i) = (ii) = 5 1 6 x 4 5 2x 5 x 2 6 2 8. 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. If A = ⎢ 2 1 –3 ⎥ .

$ Note If R = ith row and C = ith column. then det (A) = det (A′). we get Δ1 = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Hence Δ = Δ 1 Remark It follows from above property that if A is a square matrix. we will symbolically write C ↔ R i i i Let us verify the above property by example. we have 0 4 6 4 6 0 Δ= 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. Property 1 is verified. 2 –3 5 Example 6 Verify Property 1 for Δ = 6 0 4 1 5 –7 Solution Expanding the determinant along first row. where A′ = transpose of A. we get 2 6 1 Δ1 = –3 0 5 (Expanding along first column) 5 4 –7 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 Δ = Δ1 Hence. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3 . then for interchange of row and i columns.110 MATHEMATICS Expanding Δ1 along first column. then sign of determinant changes. Property 2 If any two rows (or columns) of a determinant are interchanged.

$Note We can denote the interchange of rows by R ↔ R and interchange of i j columns by Ci ↔ Cj. R2 ↔ R3. we have 2 –3 5 Δ1 = 1 5 –7 6 0 4 Expanding the determinant Δ1 along first row. the new determinant obtained is given by c1 c2 c3 Δ1 = b1 b2 b3 a1 a2 a3 Expanding along third row. 1 5 –7 2 –3 5 Solution Δ = 6 0 4 = – 28 (See Example 6) 1 5 –7 Interchanging rows R2 and R3 i. 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. 2 –3 5 Example 7 Verify Property 2 for Δ = 6 0 4 .e. 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 have 5 –7 1 –7 1 5 Δ1 = 2 – (–3) +5 0 4 6 4 6 0 = 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30) = 40 + 138 – 150 = 28 . we can verify the result by interchanging any two columns.

then Δ does not change. Then k a1 k b1 k c1 Δ1 = a2 b2 c2 a3 b3 c3 Expanding along first row. by Property 2. 3 2 3 Example 8 Evaluate Δ = 2 2 3 3 2 3 Solution 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 .112 MATHEMATICS Clearly Δ1 = – Δ Hence. Property 4 If each element of a row (or a column) of a determinant is multiplied by a constant k. a1 b1 c1 Verification Let Δ = a2 b2 c2 a3 b3 c3 and Δ1 be the determinant obtained by multiplying the elements of the first row by k. we get Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6) = 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0 Here R1 and R3 are identical. then its value gets multiplied by k. However. then value of determinant is zero. Property 2 is verified. Property 3 If any two rows (or columns) of a determinant are identical (all corresponding elements are same). it follows that Δ has changed its sign Therefore Δ=– Δ or Δ=0 Let us verify the above property by an example. Proof If we interchange the identical rows (or columns) of the determinant Δ.

S. (ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio). a1 + λ1 a2 + λ 2 a3 + λ 3 a1 a2 a3 λ1 λ 2 λ3 For example. DETERMINANTS 113 k a1 k b1 k c1 a1 b1 c1 Hence a2 b2 c2 = k a2 b2 c2 a3 b3 c3 a3 b3 c3 Remarks (i) By this property. = b1 b2 b3 c1 c2 c3 . then its value is zero. then the determinant can be expressed as sum of two (or more) determinants. For example a1 a2 a3 Δ= b1 b2 b3 = 0 (rows R1 and R2 are proportional) k a1 k a2 k a3 102 18 36 Example 9 Evaluate 1 3 4 17 3 6 102 18 36 6(17) 6(3) 6(6) 17 3 6 Solution Note that 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. b1 b2 b3 = b1 b2 b3 + b1 b2 b3 c1 c2 c3 c1 c2 c3 c1 c2 c3 a1 + λ1 a2 + λ 2 a3 + λ 3 Verification L.H. we can take out any common factor from any one row or any one column of a given determinant.

we get Δ = (a1 + λ1) (b2 c3 – c2 b3) – (a2 + λ2) (b1 c3 – b3 c1) + (a3 + λ3) (b1 c2 – b2 c1) = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) + λ1 (b2 c3 – c2 b3) – λ2 (b1 c3 – b3 c1) + λ3 (b1 c2 – b2 c1) (by rearranging terms) a1 a2 a3 λ1 λ 2 λ3 = b1 b2 b3 + b1 b2 b3 = R.114 MATHEMATICS Expanding the determinants along the first row. Symbolically..S. the equimultiples of corresponding elements of other row (or column) are added. 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). . to each element of any row or column of a determinant. Here. Verification a1 a2 a3 a1 + k c1 a2 + k c2 a3 + k c3 Let Δ = b1 b2 b3 and Δ1 = b1 b2 b3 . c1 c2 c3 c1 c2 c3 where Δ1 is obtained by the operation R1 → R1 + kR3 . i. then value of determinant remains the same. c1 c2 c3 c1 c2 c3 Similarly.H. we write this operation as R1 → R1 + k R3.e. a b c Example 10 Show that a + 2 x b + 2 y c + 2 z = 0 x y z a b c a b c a b c Solution We have a + 2 x b + 2 y c + 2 z = a b c + 2x 2 y 2z x y z x y z x y z (by Property 5) =0+0=0 (Using Property 3 and Property 4) Property 6 If. the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + k Cj . we may verify Property 5 for other rows or columns.

we get a a+b a+b+c Δ= 0 a 2a + b 0 0 a Expanding along C1. a a+b a+b+c Example 11 Prove that 2a 3a + 2b 4a + 3b + 2c = a 3 . A similar remark applies to column operations. then Δ1 = kΔ. (ii) If more than one operation like Ri → Ri + kRj is done in one step. again a1 a2 a3 k c1 k c2 k c3 Δ1 = b1 b2 b3 + b1 b2 b3 (Using Property 5) c1 c2 c3 c1 c2 c3 =Δ+0 (since R1 and R3 are proportional) Hence Δ = Δ1 Remarks (i) If Δ1 is the determinant obtained by applying Ri → kRi or Ci → kCi to the determinant Δ. we have a a+b a+b+c Δ= 0 a 2a + b 0 3a 7a + 3b Now applying R3 → R3 – 3R2 . DETERMINANTS 115 Now. care should be taken to see that a row that is affected in one operation should not be used in another operation. we obtain a 2a + b Δ= a +0+0 0 a = a (a2 – 0) = a (a2) = a3 . 3a 6a + 3b 10a + 6b + 3c Solution Applying operations R2 → R2 – 2R1 and R3 → R3 – 3R1 to the given determinant Δ.

prove that x+y y+z z+x Δ= z x y =0 1 1 1 Solution Applying R1 → R1 + R2 to Δ. we get x+y+z x+ y+z x+ y+z Δ= z x y 1 1 1 Since the elements of R1 and R3 are proportional.116 MATHEMATICS Example 12 Without expanding. we get 1 a bc Δ = 0 b − a c ( a − b) 0 c − a b (a − c) Taking factors (b – a) and (c – a) common from R2 and R3. Example 13 Evaluate 1 a bc Δ = 1 b ca 1 c ab Solution Applying R2 → R2 – R1 and R3 → R3 – R1. respectively. we get 1 a bc Δ = (b − a ) (c − a) 0 1 –c 0 1 –b = (b – a) (c – a) [(– b + c)] (Expanding along first column) = (a – b) (b – c) (c – a) b+c a a Example 14 Prove that b c+a b = 4 abc c c a+b b+c a a Solution Let Δ = b c+a b c c a+b . Δ = 0.

we get 0 –2c –2b Δ= b c+a b c c a+b Expanding along R1. then z z2 1 + z3 show that 1 + xyz = 0 Solution We have x x2 1 + x3 Δ= y y 2 1 + y3 z z2 1 + z3 x x2 1 x x2 x3 = y y2 1 + y y2 y 3 (Using Property 5) z z2 1 z z2 z3 1 x x2 1 x x2 = (−1) 1 y y 2 + xyz 1 y 2 y2 (Using C3 ↔ C2 and then C1 ↔ C2) 2 2 1 z z 1 z z 1 x x2 = 1 y y 2 (1+ xyz ) 1 z z2 . z are different and Δ = y y 2 1 + y 3 = 0 . we obtain c+a b b b b c+a Δ= 0 – (–2 c ) + (–2b) c a+b c a+b c c = 2 c (a b + b2 – bc) – 2 b (b c – c2 – ac) = 2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc = 4 abc x x2 1 + x3 Example 15 If x. y. DETERMINANTS 117 Applying R1 → R1 – R2 – R3 to Δ.

we get 1 x x2 Δ = (1+xyz ) (y –x ) (z –x) 0 1 y+x 0 1 z+x = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1) Since Δ = 0 and x.b.c common from R1. R2 and R3. we get 1 + xyz = 0 Example 16 Show that 1+ a 1 1 ⎛ 1 1 1⎞ 1 1+ b 1 = abc ⎜ 1 + + + ⎟ = abc + bc + ca + ab ⎝ a b c⎠ 1 1 1+ c Solution Taking out factors a. we get 1 1 1 +1 a a a 1 1 1 L. y – z ≠ 0.e. x – y ≠ 0. we have 1 1 1 1 1 1 1 1 1 1+ + + 1+ + + 1+ + + a b c a b c a b c 1 1 1 Δ = abc +1 b b b 1 1 1 +1 c c c . = abc +1 b b b 1 1 1 +1 c c c Applying R1→ R1 + R2 + R3.H.. i. z are all different. y. z – x ≠ 0.118 MATHEMATICS 1 x x2 = (1 + xyz ) 0 y−x y2 − x2 (Using R2 → R2–R1 and R3 → R3– R1) 0 z−x z −x 2 2 Taking out common factor (y – x) from R2 and (z – x) from R3.S.

then apply C →C –aC. DETERMINANTS 119 1 1 1 ⎛ 1 1 1⎞ 1 1 1 = abc ⎜ 1+ + + ⎟ +1 ⎝ a b c⎠ b b b 1 1 1 +1 c c c Now applying C2 → C2 – C1. C3 → C3 – C1.H. 3 8 75 = 0 4.S. 1 1 2 3 3 2 1 1 3 EXERCISE 4. b−c c−a a −b = 0 z c z+c c−a a−b b−c 2 7 65 1 bc a ( b + c ) 3. prove that: x a x+a a −b b −c c − a 1.2 Using the property of determinants and without expanding in Exercises 1 to 7. 1 ca b ( c + a ) = 0 5 9 86 1 ab c ( a + b ) b+c q+r y+z a p x 5. ⎝ a b c⎠ $ Note Alternately try by applying C → C – C and C → C – C . y b y +b = 0 2. c+a r+ p z+x = 2 b q y a+b p+q x+ y c r z . we get 1 0 0 ⎛ 1 1 1⎞ 1 Δ = abc ⎜ 1+ + + ⎟ 1 0 ⎝ a b c⎠ b 1 0 1 c ⎛ 1 1 1⎞ = abc ⎜1 + + + ⎟ ⎡⎣1(1 – 0 )⎤⎦ ⎝ a b c⎠ ⎛ 1 1 1⎞ = abc ⎜1+ + + ⎟ = abc + bc + ca + ab = R.

(i) 2b b−c−a 2c 2c c−a −b x + y + 2z x y = 2( x + y + z ) 3 (ii) z y + z + 2x y z x z + x + 2y . (i) 2 x x + 4 2x = ( 5 x + 4 )( 4 − x ) 2 2x 2x x + 4 y+k y y (ii) y y+k y = k 2 (3y + k ) y y y+k a −b −c 2a 2a 2b = ( a + b + c ) 3 11. ba −b 2 bc = 4 a 2 b 2 c 2 b c 0 ca cb −c 2 By using properties of determinants. − a 0 −c = 0 7. in Exercises 8 to 14. show that: 1 a a2 8. y y2 zx = (x – y) (y – z) (z – x) (xy + yz + zx) z z2 xy x + 4 2x 2x 10. (i) 1 b b = ( a − b )( b − c )( c − a ) 2 1 c c2 1 1 1 (ii) a b c = ( a − b )( b − c )( c − a )( a + b + c ) a3 b3 c3 x x2 yz 9.120 MATHEMATICS 0 a −b −a2 ab ac 6.

. is given by the expression [x (y –y ) + x2 (y3–y1) + 2 1 2 3 x3 (y1–y2)]. Now this expression can be written in the form of a determinant as x1 y1 1 1 Δ= x2 y2 1 . y1). y3). Let A be a square matrix of order 3 × 3. 2ab 1− a + b 2 2 2a = 1 + a2 + b2 2b −2a 1 − a2 − b2 a2 + 1 ab ac 14. x2 1 x = 1 − x3 x x2 1 1 + a 2 − b2 2ab −2b ( ) 3 13. then | kA | is equal to (A) k| A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A | 16. (B) Determinant is a number associated to a matrix.. (D) None of these 4. ab b +1 2 bc =1 + a 2 + b2 + c 2 ca cb c2 + 1 Choose the correct answer in Exercises 15 and 16. (1) 2 x3 y3 1 Remarks (i) Since area is a positive quantity. (x2. we always take the absolute value of the determinant in (1).. we have studied that the area of a triangle whose vertices are 1 (x1.4 Area of a Triangle In earlier classes. 15. (C) Determinant is a number associated to a square matrix. Which of the following is correct (A) Determinant is a square matrix. DETERMINANTS 121 1 x x2 ( ) 2 12. y2) and (x3.

Solution The area of triangle is given by 3 8 1 1 Δ= –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. use both positive and negative values of the determinant for calculation. area of triangle ABP is zero (Why?). 8). 2). Example 17 Find the area of the triangle whose vertices are (3. since the area of the triangle ABD is 3 sq. 2 EXERCISE 4. 1).. 7). –8) . 0). 3) (ii) (2. = ± 3 . 0) is a point such that area of triangle ABD is 3sq units. units. k = ∓ 2. 2 which is the equation of required line AB. (3. we have 1 3 1 1 0 0 1 =±3 2 k 0 1 − 3k This gives. (10.3 1. (iii) The area of the triangle formed by three collinear points is zero. i. Also. (4. So 0 0 1 1 1 3 1 =0 2 x y 1 1 This gives ( y – 3 x ) = 0 or y = 3x. y) be any point on AB. (–1. –3). Find area of the triangle with vertices at the point given in each of the following : (i) (1.122 MATHEMATICS (ii) If area is given. (6. 2) and (5. (1. (– 4. Solution Let P (x.e. 8) (iii) (–2. Then. 1). 0) using determinants and find k if D(k. 0).

0). 5. (i) Find equation of line joining (1. (ii) Find equation of line joining (3. Show that points A (a. we will learn to write the expansion of a determinant in compact form using minors and cofactors. units and vertices are (i) (k. B (b. 6) using determinants. If area of triangle is 35 sq units with vertices (2. 2) (ii) (–2. Find values of k if area of triangle is 4 sq. 2) and (3. 7 8 Definition 2 Cofactor of an element aij . C (c. –2 (D) 12. DETERMINANTS 123 2. 0). k) 4. (0. 4). Then k is (A) 12 (B) –2 (C) –12. (0. 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. 3) using determinants. So M11 = Minor of a11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2 . b + c). denoted by Aij is defined by Aij = (–1)i + j Mij .5 Minors and Cofactors In this section. 4) and (k. (4. Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of order n – 1. 1 –2 Example 20 Find minors and cofactors of all the elements of the determinant 4 3 Solution Minor of the element aij is Mij Here a11 = 1. a + b) are collinear. c + a). Definition 1 Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. (5. its minor M23 is given by 1 2 M23 = = 8 – 14 = – 6 (obtained by deleting R2 and C3 in Δ). –2 4. – 6). (0. Minor of an element aij is denoted by Mij. 0). 4). where Mij is minor of aij . 1) and (9. 3.

124 MATHEMATICS M22 = Minor of the element a22 = 1 Now. R3. where Aij is cofactor of aij = sum of product of elements of R1 with their corresponding cofactors Similarly. a21 in the determinant a11 a12 a13 Δ = a21 a22 a23 a31 a32 a33 Solution By definition of minors and cofactors. Δ can be calculated by other five ways of expansion that is along R2. So A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3 A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4 A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2 A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 Example 21 Find minors and cofactors of the elements a11. Hence Δ = sum of the product of elements of any row (or column) with their corresponding cofactors. C1. . in Example 21. C2 and C3. we have a22 a23 Minor of a11 = M11 = = a22 a33– a23 a32 a32 a33 Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32 a12 a13 Minor of a21 = M21 = = a12 a33 – a13 a32 a32 a33 Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 Remark Expanding the determinant Δ. we have a22 a23 a21 a23 a21 a22 Δ = (–1) a11 a 1+1 1+2 a + (–1) a12 a 1+3 a + (–1) a13 a31 a32 32 33 31 33 = a11 A11 + a12 A12 + a13 A13. For example. $Note If elements of a row (or column) are multiplied with cofactors of any other row (or column). along R1. then their sum is zero. cofactor of aij is Aij.

Example 22 Find minors and cofactors of the elements of the determinant 2 –3 5 6 0 4 and verify that a11 A31 + a12 A32 + a13 A33= 0 1 5 –7 0 4 Solution We have M11 = = 0 –20 = –20. A22 = (–1)2+2 (–19) = –19 1 –7 2 –3 M23 = = 10 + 3 = 13. A23 = (–1)2+3 (13) = –13 1 5 –3 5 M31 = = –12 – 0 = –12. A21 = (–1)2+1 (– 4) = 4 5 –7 2 5 M22 = = –14 – 5 = –19. A11 = (–1)1+1 (–20) = –20 5 –7 6 4 M12 = = – 42 – 4 = – 46. DETERMINANTS 125 Δ = a11 A21 + a12 A22 + a13 A23 a12 a13 a11 a13 a11 a12 = a11 (–1)1+1 + a12 (–1)1+2 + a13 (–1)1+3 a32 a33 a31 a33 a31 a32 a11 a12 a13 = a11 a12 a13 = 0 (since R and R are identical) 1 2 a31 a32 a33 Similarly. A12 = (–1)1+2 (– 46) = 46 1 –7 6 0 M13 = = 30 – 0 = 30. A13 = (–1)1+3 (30) = 30 1 5 –3 5 M21 = = 21 – 25 = – 4. A31 = (–1)3+1 (–12) = –12 0 4 . we can try for other rows and columns.

1 2 3 1 x yz 4. evaluate Δ = 1 y zx . (i) 0 1 0 (ii) 3 5 –1 0 0 1 0 1 2 5 3 8 3.6 Adjoint and Inverse of a Matrix In the previous chapter. Using Cofactors of elements of second row. If Δ = a21 a22 a23 and Aij is Cofactors of aij .126 MATHEMATICS 2 5 M32 = = 8 – 30 = –22. A33 = (–1)3+3 (18) = 18 6 0 Now a11 = 2. In this section.e.. i.4 Write Minors and Cofactors of the elements of following determinants: 2 –4 a c 1. . 1 z xy a11 a12 a13 5. A–1 we shall first define adjoint of a matrix. we have studied inverse of a matrix. a13 = 5. (i) (ii) 0 3 b d 1 0 0 1 0 4 2. evaluate Δ = 2 0 1 . a12 = –3. then value of Δ is given by a31 a32 a33 (A) a11 A31+ a12 A32 + a13 A33 (B) a11 A11+ a12 A21 + a13 A31 (C) a21 A11+ a22 A12 + a23 A13 (D) a11 A11+ a21 A21 + a31 A31 4. A32 = (–1)3+2 (–22) = 22 6 4 2 –3 and M33 = = 0 + 18 = 18. A32 = 22. A33 = 18 So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 EXERCISE 4. A31 = –12. Using Cofactors of elements of third column. we shall discuss the condition for existence of inverse of a matrix. To find inverse of a matrix A.

i. DETERMINANTS 127 4.. then A(adj A) = (adj A) A = A I . Theorem 1 If A be any given square matrix of order n. We state the following theorem without proof. A21 = –3. A12 = –1. where I is the identity matrix of order n .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. A22 = 2 ⎡ A11 A 21 ⎤ ⎡ 4 –3⎤ Hence adj A = ⎢ ⎥ =⎢ ⎥ ⎣ A12 A 22 ⎦ ⎣ –1 2 ⎦ Remark For a square matrix of order 2. where Aij is the cofactor of the element aij .6. ⎡ a11 a12 a13 ⎤ Let A = ⎢⎢ a21 a22 a23 ⎥⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎡ A11 A12 A13 ⎤ ⎡ A11 A 21 A 31 ⎤ Then adj A = Transpose of ⎢⎢A 21 A 22 A 23 ⎥⎥ = ⎢ A12 A 22 A 32 ⎥⎥ ⎢ ⎢⎣ A31 A32 A 33 ⎥⎦ ⎢⎣ A13 A 23 A33 ⎦⎥ ⎡ 2 3⎤ Example 23 Find adj A for A = ⎢ ⎥ ⎣1 4⎦ Solution We have A11 = 4.e. given by ⎡ 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. Adjoint of the matrix A is denoted by adj A.

the determinant of matrix A = ⎢ 4 8 ⎥ is zero ⎣ ⎦ Hence A is a singular matrix. that is. where A and B are square matrices of the same order ⎡A 0 0⎤ ⎢ ⎥ Remark We know that (adj A) A = A I = ⎢ 0 A 0⎥ ⎢⎣ 0 0 A ⎥⎦ . then AB and BA are also nonsingular matrices of the same order. then adj A = ⎢ A12 A 22 A32 ⎥⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎢⎣ A13 A 23 A 33 ⎥⎦ Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to | A | and otherwise zero. 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. ⎣3 4 ⎦ 3 4 Hence A is a nonsingular matrix We state the following theorems without proof. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants. AB = A B . Definition 5 A square matrix A is said to be non-singular if A ≠ 0 ⎡1 2 ⎤ 1 2 Let A= ⎢ ⎥ . Then A = = 4 – 6 = – 2 ≠ 0.128 MATHEMATICS Verification ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A 31 ⎤ ⎢a ⎥ ⎢ Let A = ⎢ 21 a22 a23 ⎥ . Theorem 2 If A and B are nonsingular matrices of the same order. we have ⎡A 0 0⎤ ⎡ 1 0 0⎤ ⎢ ⎥ A (adj A) = ⎢ 0 A 0 ⎥ = A ⎢⎢ 0 1 0⎥⎥ = A I ⎢⎣ 0 0 A ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ Similarly. ⎡1 2⎤ For example.

Hence A is nonsingular. DETERMINANTS 129 Writing determinants of matrices on both sides. there exists a square matrix B of order n such that AB = BA = I Now AB = I. |(adj A)| = | A | 2 In general. if A is a square matrix of order n. Proof Let A be invertible matrix of order n and I be the identity matrix of order n. Conversely. |(adj A)| |A| = | A |3 (1) i. AB = A B ) This gives A ≠ 0. we have A 0 0 (adj A) A = 0 A 0 0 0 A 1 0 0 3 i.e. let A be nonsingular.e. |(adj A)| |A| = A 0 1 0 (Why?) 0 0 1 i. Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. where B = adj A |A| 1 Thus A is invertible and A–1 = adj A |A| ⎡1 3 3⎤ ⎢ ⎥ Example 24 If A = ⎢1 4 3⎥ . then | adj (A) | = | A |n – 1. Also find A–1. Then. then verify that A adj A = | A | I. So AB = I or A B =1 (since I =1. ⎢⎣1 3 4⎥⎦ Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0 .e. Then A ≠ 0 Now A (adj A) = (adj A) A = A I (Theorem 1) ⎛ 1 ⎞ ⎛ 1 ⎞ or A⎜ adj A ⎟ = ⎜ adj A ⎟ A = I ⎝|A| ⎠ ⎝ |A| ⎠ 1 or AB = BA = I.

A–1 and B–1 both exist and are given by 1 ⎡ − 4 −3⎤ −1 ⎡3 2⎤ A–1 = − .I ⎣⎢ 0 0 1 ⎥⎦ ⎣⎢ 0 0 1 ⎥⎦ ⎡ 7 −3 −3⎤ ⎡ 7 −3 −3⎤ −1 1 1⎢ ⎥ ⎢ ⎥ Also A = adj A = ⎢ −1 1 0 ⎥ = ⎢ −1 1 0 ⎥ A 1 ⎢⎣ −1 0 1 ⎥⎦ ⎢⎣ −1 0 1 ⎥⎦ ⎡2 3 ⎤ ⎡ 1 −2 ⎤ Example 25 If A = ⎢ ⎥ and B = ⎢ ⎥ .B = ⎢ 11 ⎢⎣ −1 2 ⎥⎦ ⎥ ⎣1 1⎦ .A23 = 0. A21 = –3. A13 = –1. A = –11 ≠ 0 and B = 1 ≠ 0. A32 = 0. –1 –1 –1 ⎣ 1 − 4 ⎦ ⎣ − 1 3 ⎦ ⎡2 3⎤ ⎡ 1 −2 ⎤ ⎡ −1 5 ⎤ Solution We have AB = ⎢ ⎥ ⎢ −1 3 ⎥ = ⎢ 5 −14 ⎥ ⎣1 − 4⎦ ⎣ ⎦ ⎣ ⎦ Since.130 MATHEMATICS Now A11 = 7. Therefore. A31 = –3. A12 = –1. A33 = 1 ⎡ 7 −3 −3⎤ ⎢ ⎥ Therefore adj A = ⎢ −1 1 0 ⎥ ⎢⎣ −1 0 1 ⎥⎦ ⎡1 3 3⎤ ⎡ 7 −3 −3⎤ ⎢ ⎥⎢ ⎥ Now A (adj A) = ⎢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 ⎤ ⎡1 0 0 ⎤ ⎢ ⎥ ⎢0 1 0 ⎥ = ⎢ 0 1 0 ⎥ = (1) ⎢ ⎥ = A . AB = –11 ≠ 0. A22 = 1. (AB)–1 exists and is given by 1 1 ⎡ −14 −5⎤ = 1 ⎡14 5⎤ adj (AB) = − ⎢ ⎢ ⎥ 11 ⎣ −5 −1⎥⎦ (AB)–1 = 11 ⎣ 5 1⎦ AB Further. then verify that (AB) = B A .

A = ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎣1 2 ⎦ ⎣1 2 ⎦ ⎣ 4 7 ⎦ ⎡ 7 12⎤ ⎡ 8 12⎤ ⎡ 1 0 ⎤ ⎡ 0 0⎤ Hence A 2 − 4A + I = ⎢ ⎥− ⎢ ⎥+⎢ ⎥ =⎢ ⎥=O ⎣ 4 7 ⎦ ⎣ 4 8 ⎦ ⎣ 0 1 ⎦ ⎣ 0 0⎦ Now A2 – 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 ⎡ 4 0 ⎤ ⎡2 3 ⎤ ⎡ 2 −3 ⎤ or A–1 = 4I – A = ⎢ ⎥ −⎢ ⎥ = ⎢ ⎥ ⎣0 4 ⎦ ⎣1 2 ⎦ ⎣ −1 2 ⎦ ⎡ 2 −3 ⎤ Hence A −1 = ⎢ ⎥ ⎣ −1 2 ⎦ EXERCISE 4. DETERMINANTS 131 1 ⎡3 2⎤ ⎡ −4 −3⎤ 1 ⎡ −14 −5⎤ 1 ⎡14 5⎤ Therefore B−1 A −1 = − ⎢ ⎥ ⎢ ⎥ =− ⎢ ⎥ = ⎢ 11 ⎣1 1 ⎦ ⎣ −1 2 ⎦ 11 ⎣ −5 −1⎦ 11 ⎣ 5 1⎥⎦ Hence (AB)–1 = B–1 A–1 ⎡ 2 3⎤ Example 26 Show that the matrix A = ⎢ ⎥ satisfies the equation A2 – 4A + I = O. find A–1. ⎡ 2 3 ⎤ ⎡ 2 3 ⎤ ⎡ 7 12⎤ Solution We have A 2 = A. ⎢ 3 0 −2⎥ ⎣ ⎦ ⎢⎣1 0 3 ⎥⎦ . ⎢ −4 −6 ⎥ 4. ⎡ 1 −1 2⎤ ⎡ 1 2⎤ ⎢ ⎥ 1. Using this equation. ⎢ 2 3 5⎥ ⎣ ⎦ ⎢⎣ −2 0 1 ⎥⎦ Verify A (adj A) = (adj A) A = | A | I in Exercises 3 and 4 ⎡1 −1 2 ⎤ ⎡2 3⎤ ⎢ ⎥ 3. ⎢ 3 4⎥ 2.5 Find adjoint of each of the matrices in Exercises 1 and 2. ⎣ 1 2⎦ where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix.

find A–1. ⎢ ⎥ 7. show that A – 5A + 7I = O. For the matrix A = ⎢ 1 2 −3⎥ ⎢⎣ 2 −1 3 ⎥⎦ Show that A3– 6A2 + 5A + 11 I = O. ⎢ 0 2 −3⎥ ⎢⎣ 5 2 −1⎥⎦ ⎢⎣ −7 2 1⎥⎦ ⎣⎢ 3 −2 4 ⎥⎦ ⎡1 0 0 ⎤ ⎢ 0 cos α sin α ⎥ 11. If A = ⎢ ⎥ . ⎡ 2 −1 1 ⎤ ⎢ ⎥ 16. Verify that (AB) = B A . –1 –1 –1 ⎣2 5⎦ ⎣ 7 9⎦ ⎡ 3 1⎤ 13. Hence. Let A = ⎢ ⎥ and B = ⎢ ⎥ . 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. ⎢ 0 2 4⎥ ⎣ ⎦ ⎣ −3 2 ⎦ ⎢⎣ 0 0 5⎥⎦ ⎡1 0 0 ⎤ ⎡ 2 1 3⎤ ⎡ 1 −1 2 ⎤ ⎢3 3 0 ⎥ ⎢ ⎥ ⎢ ⎥ 8. 2 ⎣ 1 1 ⎦ ⎡1 1 1 ⎤ ⎢ ⎥ 15. find the numbers a and b such that A + aA + bI = O. then det (A–1) is equal to 1 (A) det (A) (B) det (A) (C) 1 (D) 0 . If A is an invertible matrix of order 2. ⎢ ⎥ 9. If A = ⎢ −1 2 −1⎥ ⎣⎢ 1 −1 2 ⎥⎦ Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 17. ⎢ 4 −1 0⎥ 10.132 MATHEMATICS Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. ⎢4 3 ⎥ 6. 2 –1 ⎣ −1 2 ⎦ ⎡3 2 ⎤ 14. ⎡ 1 2 3⎤ ⎡ 2 −2 ⎤ ⎡ −1 5 ⎤ ⎢ ⎥ 5. Hence find A . For the matrix A = ⎢ ⎥ . ⎢ ⎥ ⎢⎣ 0 sin α − cos α ⎥⎦ ⎡3 7⎤ ⎡ 6 8⎤ 12.

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. ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ ⎢a b2 c2 ⎥⎥ ⎢ y⎥ ⎢ ⎥ ⎢ 2 ⎢ ⎥ = ⎢ d2 ⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ Case I If A is a nonsingular matrix. we restrict ourselves to the system of linear equations having unique solutions only..7 Applications of Determinants and Matrices In this section. X = ⎢ y ⎥ and B = ⎢ d ⎥ Let A = ⎢ 2 2 2⎥ ⎢ ⎥ ⎢ 2⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d 3 ⎥⎦ Then. Consistent system A system of equations is said to be consistent if its solution (one or more) exists. 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. This method of solving system of equations is known as Matrix Method. Consider the system of equations a1 x + b1 y + c 1 z = d1 a2 x + b2 y + c 2 z = d 2 a3 x + b3 y + c 3 z = d 3 ⎡ a1 b1 c1 ⎤ ⎡x⎤ ⎡ d1 ⎤ ⎢ a b c ⎥ .7. i. . then its inverse exists. DETERMINANTS 133 4. $ Note In this chapter. 4. AX = B.e. the system of equations can be written as. Now AX = B or A (AX) = A–1 B –1 (premultiplying by A–1) or (A–1A) X = A–1 B (by associative property) –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.

A = –11 ≠ 0. where ⎡ 3 −2 3 ⎤ ⎡ x⎤ ⎡ 8⎤ A = ⎢ 2 1 −1⎥ . y = – 1 Example 28 Solve the following system of equations by matrix method. X = ⎢⎢ y ⎥⎥ and B = ⎢ ⎥ ⎢ 1⎥ ⎢ ⎥ ⎢⎣ 4 −3 2 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 4⎥⎦ We see that A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0 . (O being zero matrix). 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. then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. X = ⎢ ⎥ and B = ⎢ ⎥ ⎣3 2⎦ ⎣ y⎦ ⎣7 ⎦ Now. 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.134 MATHEMATICS Case II If A is a singular matrix. If (adj A) B ≠ O. Hence. where ⎡2 5⎤ ⎡x⎤ ⎡1 ⎤ A= ⎢ ⎥ . ⎣ ⎦ Hence x = 3. then solution does not exist and the system of equations is called inconsistent. If (adj A) B = O. A is nonsingular matrix and so has a unique solution. we calculate (adj A) B.e. In this case. then | A | = 0. 1 ⎡ 2 −5 ⎤ A–1 = − 11 ⎢⎣ −3 2 ⎥⎦ Note that 1 ⎡ 2 −5 ⎤ ⎡ 1 ⎤ Therefore X = A–1B = – 11 ⎢⎣ −3 2 ⎥⎦ ⎢⎣ 7⎥⎦ ⎡ x⎤ 1 ⎡ −33⎤ ⎡ 3 ⎤ ⎢ y⎥ = − ⎢ = 11 ⎣ 11 ⎥⎦ ⎢⎣ −1⎥⎦ i.

Now A11 = –1. y = 2 and z = 3. A22 = – 6. Solution Let first. we get double of the second number. 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. A33 = 7 ⎡ −1 − 5 −1⎤ 1 ⎢ ⎥ Therefore A = − ⎢ −8 − 6 9 ⎥ –1 17 ⎢⎣ −10 1 7 ⎥⎦ ⎡ −1 − 5 −1⎤ ⎡ 8 ⎤ 1 ⎢ ⎥ ⎢ ⎥ X = A B = − ⎢ −8 − 6 9 ⎥ ⎢ 1 ⎥ –1 So 17 ⎣⎢ −10 1 7 ⎥⎦ ⎢⎣ 4⎥⎦ ⎡ x⎤ ⎡ −17 ⎤ ⎡ 1 ⎤ ⎢ y⎥ 1 ⎢ ⎥ ⎢ ⎥ i. A32 = 9. Example 29 The sum of three numbers is 6. A12 = – (0 – 3) = 3. A23 = 1 A31 = –1. Represent it algebraically and find the numbers using matrix method. respectively. Now we find adj A A11 = 1 (1 + 6) = 7. where ⎡1 1 1⎤ ⎡ x⎤ ⎡6⎤ ⎢ ⎥ ⎢ y⎥ ⎢11⎥ A = ⎢0 1 3⎥ . second and third numbers be denoted by x. If we multiply third number by 3 and add second number to it. A13 = –10 A21 = –5. Then. y and z. A13 = – 1 A21 = – (1 + 2) = – 3. A33 = (1 – 0) = 1 . DETERMINANTS 135 Hence. By adding first and third numbers. according to given conditions.e. A23 = – (– 2 – 1) = 3 A31 = (3 – 1) = 2. we get 11. A32 = – (3 – 0) = – 3. A22 = 0. A is nonsingular and so its inverse exists. A12 = – 8. X = ⎢ ⎥ and B = ⎢ ⎥ ⎢⎣1 –2 1⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 0 ⎥⎦ Here A = 1 (1 + 6) – (0 – 3) + ( 0 – 1) = 9 ≠ 0 . ⎢ ⎥ = − 17 ⎢ −34 ⎥ = ⎢ 2 ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ −51⎥⎦ ⎢⎣ 3 ⎥⎦ Hence x = 1.

z = 3 EXERCISE 4. 2x – y = 5 3. 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. 7. in Exercises 7 to 14. x + y + z = 1 5. x – y + z = 4 3 3x + 2y = 5 x – 2y – z = 2x + y – 3z = 0 2 3y – 5z = 9 x+y+z=2 13. 2x + 3y +3 z = 5 14. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10. y = 2. x + 3y = 5 2x + 3y = 3 x+y=4 2x + 6y = 8 4. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12 . 2x – y = –2 9. 1. using matrix method. 2x + y + z = 1 12. 3x–y – 2z = 2 6. 5x + 2y = 3 11. 5x + 2y = 4 8.6 Examine the consistency of the system of equations in Exercises 1 to 6.136 MATHEMATICS ⎡ 7 –3 2 ⎤ ⎢ ⎥ Hence adj A = ⎢ 3 0 –3⎥ ⎢⎣ –1 3 1 ⎥⎦ ⎡ 7 –3 2 ⎤ 1 1 ⎢ ⎥ Thus A –1 = adj (A) = ⎢ 3 0 –3⎥ A 9 ⎢⎣ –1 3 1 ⎥⎦ Since X = A–1 B ⎡ 7 –3 2 ⎤ ⎡ 6 ⎤ 1⎢ ⎥⎢ ⎥ X = ⎢ 3 0 –3⎥ ⎢11⎥ 9 ⎢⎣ –1 3 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ x⎤ ⎡ 42 − 33 + 0 ⎤ ⎡9⎤ ⎡ 1⎤ ⎢ y⎥ ⎢ 1 18 + 0 + 0 ⎥ ⎢ 1 18 ⎥ ⎢ ⎥ or ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ = ⎢ 2⎥ ⎢⎣ z ⎥⎦ 9 ⎢ −6 + 33 + 0⎥ 9 ⎢ 27 ⎥ ⎢ 3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Thus x = 1. x + 2y = 2 2.

Using A solve the system of equations ⎢⎣ 1 1 –2 ⎥⎦ 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. DETERMINANTS 137 ⎡ 2 –3 5⎤ ⎢3 2 – 4⎥ –1 –1 15. 4 kg wheat and 6 kg rice is Rs 90. The cost of 2 kg onion. If A = ⎢ ⎥ . The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70. show that value of the determinant a b c Δ = b c a is negative. Miscellaneous Examples Example 30 If a. The cost of 4 kg onion. c a b Solution Applying C1 → C1 + C2 + C3 to the given determinant.and R3 →R3 –R1) 0 a–b b–c = (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1) = (a + b + c)(– a2 – b2 – c2 + ab + bc + ca) –1 = (a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) 2 –1 = (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] 2 which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2 + (c – a)2 > 0) . 3 kg wheat and 2 kg rice is Rs 60. find A . Find cost of each item per kg by matrix method. we get a+b+c b c 1 b c Δ = a + b + c c a = (a + b + c) 1 c a a+b+c a b 1 a b 1 b c = (a + b + c) 0 c – b a – c (Applying R2→ R2–R1. c are positive and unequal. b.

y. we get x ( y+ z) 2 x2 y x2 z 1 y ( x+ z ) 2 Δ= xy 2 y2 z xyz z ( x+ y ) 2 xz 2 yz 2 Taking common factors x. c. respectively. we have ( y + z )2 x2 – ( y + z ) 2 x2 − ( y + z ) 2 Δ= y2 ( x + z )2 − y 2 0 z 2 0 ( x + y )2 – z 2 .P. b. R2 → yR2 . C3 → C3– C1. we obtain 0 0 0 3y + 5 6 y + 8 9 y + b = 0 (Since 2b = a + c) 4 y + 6 7 y + 9 10 y + c Example 32 Show that ( y+ z ) 2 xy zx ( x+ z ) 2 Δ= xy yz = 2xyz (x + y + z)3 ( x+ y ) 2 xz yz Solution Applying R1 → xR1. z from C1 C2 and C3. are in A. R3 → z R3 to Δ and dividing by xyz. we get ( y+ z) 2 x2 x2 xyz (x+ z) 2 Δ= y2 y2 xyz ( x+ y) 2 z2 z2 Applying C2 → C2– C1.138 MATHEMATICS Example 31 If a. find value of 2y + 4 5y + 7 8y + a 3y + 5 6 y + 8 9 y + b 4 y + 6 7 y + 9 10 y + c Solution Applying R1 → R1 + R3 – 2R2 to the given determinant.

we have (y + z) x – ( y + z) x – ( y + z) 2 Δ = (x + y + z)2 y 2 (x+ z) – y 0 z2 0 ( x + y) – z Applying R1 → R1 – (R2 + R3). DETERMINANTS 139 Taking common factor (x + y + z) from C2 and C3. we get y ⎝ z ⎠ 2 yz 0 0 y2 Δ = (x + y + z)2 y2 x+ z z z2 z2 x+ y y Finally expanding along R1. we have 2 yz –2z –2y Δ = (x + y + z)2 y2 x− y+z 0 z2 0 x+ y –z 1 ⎛ 1 ⎞ Applying C2 → (C2 + C1) and C3 → ⎜ C3 + C1 ⎟ . 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 ⎤ ⎡ –2 0 1 ⎤ ⎢ ⎥ ⎢ ⎥ Example 33 Use product ⎢0 2 –3⎥ ⎢ 9 2 –3⎥ to solve the system of equations ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 ⎡1 –1 2⎤ ⎡ –2 0 1 ⎤ ⎢ –3⎥ ⎢ 9 –3⎥ Solution Consider the product ⎢ 0 2 ⎥ ⎢ 2 ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 – 2 ⎥⎦ .

given system of equations can be written. in matrix form. we get a (1 − x 2 ) c (1 − x 2 ) p (1 − x 2 ) Δ= ax + b cx + d px + q u v w a c p = (1 − x ) ax + b cx + d px + q 2 u v w .140 MATHEMATICS ⎡ − 2 − 9 + 12 0 − 2 + 2 1 + 3 − 4⎤ ⎡ 1 0 0⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ 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 –1 2 ⎤ ⎡ –2 0 1 ⎤ –1 Hence ⎢ 0 2 –3 = ⎢ 9 2 –3⎥ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ Now. y = 5 and z = 3 Example 34 Prove that a + bx c + dx p + qx a c p Δ = ax + b cx + d px + q = (1 − x ) b d2 q u v w u v w Solution Applying R1 → R1 – x R2 to Δ. as follows ⎡ 1 –1 2 ⎤ ⎡ x ⎤ ⎡ 1 ⎤ ⎢ 0 2 –3⎥ ⎢ y ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢1 ⎥ ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 2⎥⎦ −1 ⎡ x ⎤ ⎡1 −1 2 ⎤ ⎡ 1 ⎤ ⎡ –2 0 1 ⎤ ⎡1⎤ ⎢ y⎥ ⎢ or ⎢ ⎥ = ⎢0 2 −3⎥⎥ ⎢⎢ 1 ⎥⎥ = ⎢⎢ 9 2 –3⎥⎥ ⎢⎢1⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 3 −2 4 ⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 6 1 –2⎥⎦ ⎢⎣ 2⎥⎦ ⎡ −2 + 0 + 2⎤ ⎡ 0 ⎤ ⎢ ⎥ ⎢ ⎥ = ⎢ 9 + 2 − 6 ⎥ = ⎢5 ⎥ ⎢⎣ 6 + 1 − 4 ⎥⎦ ⎢⎣ 3⎥⎦ Hence x = 0.

Evaluate – sin β cos β 0 . Solve the equation x x+a x = 0. we get a c p Δ = (1 − x ) b d q 2 u v w Miscellaneous Exercises on Chapter 4 x sin θ cos θ 1. Prove that a + ab 2 b2 ac = 4a2b2c2 ab b 2 + bc c2 ⎡ 3 –1 1 ⎤ ⎡ 1 2 –2 ⎤ ⎢ –15 6 –5⎥ and B = ⎢ –1 3 0 ⎥ . prove that b b 2 ca = 1 b 2 b3 . If a. x+a x x 5. Without expanding the determinant. sin α cos β sin α sin β cos α 4. If A = ⎢ –1 ⎥ ⎢ ⎥ ( ) ⎢⎣ 5 –2 2 ⎥⎦ ⎢⎣ 0 –2 1 ⎥⎦ . a ≠ 0 x x x+a a 2 bc ac + c 2 6. find AB –1 7. DETERMINANTS 141 Applying R2 → R2 – x R1. c c2 ab 1 c2 c3 cos α cos β cos α sin β – sin α 3. Prove that the determinant – sin θ – x 1 is independent of θ. and b+ c c+ a a+ b Δ = c + a a + b b + c = 0. b and c are real numbers. cos θ 1 x a a2 bc 1 a2 a3 2. a+ b b+ c c+ a Show that either a + b + c = 0 or a = b = c.

sin β cos β cos ( β + δ ) = 0 3 6 + 3 p 10 + 6 p + 3 q sin γ cos γ cos ( γ + δ ) 16.142 MATHEMATICS ⎡ 1 –2 1⎤ ⎢ 1⎥ . where p is any scalar. z z2 1 + pz 3 3a – a+ b – a+ c 13. Evaluate y x+ y x x+ y x y 1 x y 10. y y 2 1 + py 3 = (1 + pxyz) (x – y) (y – z) (z – x). prove that: α α2 β+γ 11. Evaluate 1 x + y y 1 x x+ y Using properties of determinants in Exercises 11 to 15. 2 3+ 2 p 4 + 3 p + 2q = 1 15. Solve the system of equations 2 3 10 + + =4 x y z . Let A = ⎢ –2 3 ⎥ ⎢⎣ 1 1 5⎥⎦ (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A x y x+ y 9. –b+ a 3b – b + c = 3(a + b + c) (ab + bc + ca) –c+ a – c+ b 3c 1 1+ p 1+ p+ q sin α cos α cos ( α + δ ) 14. β β 2 γ + α = (β – γ) (γ – α) (α – β) (α + β + γ) γ γ 2 α +β x x 2 1 + px 3 12. Verify that 8.

If a. where 0 ≤ θ ≤ 2π. then the inverse of matrix A = ⎢ 0 y 0⎥ is ⎢ ⎥ ⎢⎣ 0 0 z ⎥⎦ ⎡ x −1 0 0 ⎤ ⎡ x −1 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ (A) ⎢ 0 y −1 0 ⎥ (B) xyz ⎢ 0 y −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 0 z −1 ⎦ ⎣ 0 0 z −1 ⎦ ⎡ x 0 0⎤ ⎡1 0 0 ⎤ 1 ⎢ 0 y 0 ⎥⎥ 1 ⎢ (C) (D) 0 1 0 ⎥⎥ xyz ⎢ xyz ⎢ ⎢⎣ 0 0 z ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎡ 1 sin θ 1 ⎤ ⎢ − sin θ 1 sin θ⎥⎥ . b. c. 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. y. z are nonzero real numbers. 4) (D) Det (A) ∈ [2. Let A = ⎢ ⎢⎣ −1 − sin θ 1 ⎥⎦ (A) Det (A) = 0 (B) Det (A) ∈ (2. ∞) (C) Det (A) ∈ (2. Then 19. then the determinant x + 2 x + 3 x + 2a x + 3 x + 4 x + 2b is x + 4 x + 5 x + 2c (A) 0 (B) 1 (C) x (D) 2x ⎡ x 0 0⎤ 18. 4] . If x. 17. are in A.P.

 If we multiply each element of a row or a column of a determinant by constant k.144 MATHEMATICS Summary  Determinant of a matrix A = [a11]1× 1 is given by | a11| = a11 ⎡a a12 ⎤  Determinant of a matrix A = ⎢ 11 a22 ⎥⎦ is given by ⎣ a21 a11 a12 A = a21 a22 = a11 a22 – a12 a21 ⎡ a1 b1 c1 ⎤ ⎢ c2 ⎥⎥ is given by (expanding along R1)  Determinant of a matrix A = ⎢ a2 b2 ⎢⎣ a3 b3 c3 ⎥⎦ a1 b1 c1 b c2 a2 c2 a2 b2 A = a2 b2 c2 = a1 2 − b1 + c1 b3 c3 a3 c3 a3 b3 a3 b3 c3 For any square matrix A.  |A′| = |A|.  Multiplying a determinant by k means multiply elements of only one row (or one column) by k. the |A| satisfy following properties.A = k 3 A  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. then k . where A′ = transpose of A.  If any two rows or any two columns are identical or proportional.  If A = [aij ]3×3 . . then value of determinant is multiplied by k. then value of determinant is zero. then sign of determinant changes. then value of determinant remains same.  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.  If we interchange any two rows (or columns).

y3) is given by x1 y1 1 1 Δ= x2 y2 1 2 x3 y3 1  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.  A square matrix A has inverse if and only if A is non-singular. then these equations can be written as A X = B. then their sum is zero. where ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ A = ⎢⎢ a2 b2 c2 ⎥ . where B is square matrix. A = a11 A11 + a12 A12 + a13 A13.  Cofactor of aij of given by Aij = (– 1)i + j Mij  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. DETERMINANTS 145  Area of a triangle with vertices (x1.  A square matrix A is said to be singular or non-singular according as | A | = 0 or | A | ≠ 0. where A is square matrix of order n. then B is called inverse of A. For example. X = ⎢ y ⎥ and B= ⎢⎢ d 2 ⎥⎥ ⎥ ⎢ ⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ .  If AB = BA = I. (x2. then adj A = ⎢ A12 A 22 A32 ⎥ . y1).  If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column). 1  A –1 = ( adj A) A  If a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3. where A ij is ⎢ ⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎢⎣ A13 A 23 A33 ⎥⎦ cofactor of aij  A (adj A) = (adj A) A = | A | I. Also A–1 = B or B–1 = A and hence (A–1)–1 = A. a11 A21 + a12 A22 + a13 A23 = 0 ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A31 ⎤ ⎢ ⎥  If A = ⎢ a21 a22 a23 ⎥ . y2) and (x3.

pp 30. “The Fakudoi and Determinants in Japanese Mathematics. ‘T. 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. 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. then there exists no solution (iii) | A | = 0 and (adj A) B = 0.  A system of equation is consistent or inconsistent according as its solution exists or not. then system may or may not be consistent. Vendermonde was the first to recognise determinants as independent functions. gave general method of expanding a determinant in terms of its complementary minors.  For a square matrix A in matrix equation AX = B (i) | A | ≠ 0. Gauss used determinants in his theory of numbers. there exists unique solution (ii) | A | = 0 and (adj A) B ≠ 0. of the Tokyo Math. Seki Kowa. In 1801. after this the word determinant received its final acceptance..Marie Binet. He used the word ‘determinant’ in its present sense. The next great contributor was Jacques . Hayashi.” in the proc.Philippe .146 MATHEMATICS  Unique solution of equation AX = B is given by X = A–1 B. (1812) who stated the theorem relating to the product of two matrices of m-columns and n- rows. The arrangement of rods was precisely that of the numbers in a determinant. The greatest contributor to the theory was Carl Gustav Jacob Jacobi. Soc. China. 93. He gave the proof of multiplication theorem more satisfactory than Binet’s. Laplace (1772). therefore. where A ≠ 0 . The Chinese. Also on the same day. . early developed the idea of subtracting columns and rows as in simplification of a determinant ‘Mikami. which for the special case of m = n reduces to the multiplication theorem. 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. V. 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. Cauchy (1812) presented one on the same subject.